While it comes in handy in solving one-dimensional (1D) and two-dimensional (2D) problems, it may not be as such when solving three-dimensional problems.
\(\\[5pt] {\small \lambda \ }\) is a constant value.
\(\\[8pt] \hspace{3em} (\vec{r} \ – \ \vec{a}) \cdot \vec{n} \ \hspace{0.7em} \ = \ 0 \)
\(\\[8pt] \hspace{3em} \vec{r} \ \cdot \vec{n} \ – \ \vec{a} \ \cdot \vec{n} \ = \ 0 \)
\(\\[8pt] \hspace{3em} \vec{r} \ \cdot \vec{n} \hspace{3.3em} \ = \ \vec{a} \ \cdot \vec{n} \)
with \({\small \ (x,y,z) \ } \) are the cartesian coordinates of any points on the plane and \({\small \ (a,b,c)} \ \) are the cartesian components of the normal vector \( \ {\small \vec{n} \ }\). Then the cartesian form is:
with \( \ {\small d \ = \ \vec{a} \ \cdot \vec{n} \ }\) (the dot product of vector \( \ {\small \vec{a} \ }\) and \( \ {\small \vec{n} \ }\) ).
I have put together some of the questions I received in the comment section below. You can try these questions also to further your understanding on this topic.
To check your answer, you can look through the solutions that I have posted either in Youtube videos or Instagram posts.
You can subscribe, like or follow my youtube channel and IG account. I will keep updating my IG daily post, preferably.
Furthermore, you can find some examples and more practices below! =).
Try some of the examples below and if you need any help, just look at the solution I have written. Cheers ! =) .
\(\\[1pt]\)
\({\small 1.\enspace}\)
9709/03/SP/20 – Specimen Paper 2020 Pure Maths 3 No 8 \(\\[1pt]\)
\(\\[1pt]\)
In the diagram,
OABC is a pyramid in which
OA = 2 units,
OB = 4 units and
OC = 2 units. The edge
OC is vertical, the base
OAB is horizontal and angle \({\small \ AOB \ = \ 90^{\large{\circ}}}\). Unit vectors
i,
j and
k are parallel to
OA,
OB and
OC respectively. The midpoints of
AB and
BC are
M and
N respectively.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{a}).\hspace{0.8em}}\) Express the vectors \({\small \ \overrightarrow{ON} \ }\) and \({\small \ \overrightarrow{CM} \ }\) in terms of
i,
j and
k.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{b}).\hspace{0.8em}}\) Calculate the angle between the directions of \({\small \ \overrightarrow{ON} \ }\) and \({\small \ \overrightarrow{CM} \ }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{c}).\hspace{0.8em}}\) Show that the length of the perpendicular from
M to
ON is \( {\small \ {\large \frac{3}{5} } \sqrt 5 }\).
Check out my solution here:
\(\\[1pt]\)
https://www.instagram.com/p/CMTbkCxl8aJ/ \(\\[1pt]\)
Follow my instagram for daily updates on many more solutions on other topics too! \(\\[1pt]\)
\({\small 2.\enspace}\)
9709/11/O/N/16 – Paper 11 Oct Nov 2020 Pure Maths 1 No 9 \(\\[1pt]\)
\(\\[1pt]\)
The diagram shows a cuboid
OABCDEFG with a horizontal base
OABC in which
OA = 4 cm and
AB = 15 cm. The height
OD of the cuboid is 2 cm. The point
X on
AB is such that
AX = 5 cm and the point
P on
DG is such that
DP =
p cm, where
p is a constant. Unit vectors
i,
j and
k are parallel to
OA,
OC and
OD respectively.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}\) Find the possible values of
p such that angle \({\small \ OPX \ = \ 90^{\large{\circ}}}\).
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}\) For the case where
p = 9, find the unit vector in the direction of \({\small \ \overrightarrow{XP} }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{iii}).\hspace{0.5em}}\) A point
Q lies on the face
CBFG and is such that
XQ is parallel to AG. Find \({\small \ \overrightarrow{XQ} }\).
https://www.instagram.com/p/CMGNblWlqfB/ \(\\[1pt]\)
Follow my instagram for daily updates on many more solutions on other topics too! \(\\[1pt]\)
\({\small 3.\enspace}\) The points
A and
B have position vectors
i + 2
j –
k and 3
i +
j +
k respectively. The line
l has equation
r = 2
i +
j +
k + \({\small \mu}\)(
i +
j + 2
k).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\) Show that
l does not intersect the line passing through
A and
B.
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}\) The plane
m is perpendicular to
AB and passes through the mid-point of
AB. The plane
m intersects the line
l at the point
P. Find the equation of
m and the position vector of
P.
\({\small \hspace{1.2em}\textrm{(i)}. \enspace }\) First, we need to construct the vector line equation of the line passing through A and B, let’s name it line p.
\(\\[1pt]\)
\( {\small \hspace{1.2em} \vec{{r}_{p}} \ = \ \vec{a} \ + \ \lambda(\vec{b} \ – \ \vec{a}) }\)
\(\\[1pt]\)
\( {\small \hspace{1.2em} \vec{{r}_{p}} \ = \ \begin{pmatrix}
1 \\[1pt]
2 \\[1pt]
-1
\end{pmatrix} \ + \ \lambda \begin{pmatrix}
3 \ – \ 1 \\[1pt]
1 \ – \ 2 \\[1pt]
1 \ – \ (-1)
\end{pmatrix} }\)
\(\\[1pt]\)
\( {\small \hspace{2.4em} = \ \begin{pmatrix}
1 \\[1pt]
2 \\[1pt]
-1
\end{pmatrix} \ + \ \lambda \begin{pmatrix}
2 \\[1pt]
-1 \\[1pt]
2
\end{pmatrix} }\)
\(\\[1pt]\)
\(\\[5pt]\) or, \( \ {\small \vec{{r}_{p}} \ = \ \textbf{i} \ + \ 2\textbf{j} \ – \ \textbf{k} \ + \ \lambda (2\textbf{i} \ – \ \textbf{j} \ + \ 2\textbf{k}) }\).
\(\\[1pt]\)
The two lines, p and l can intersect if there is a point that satisfy both line equations.
\(\\[1pt]\)
We equates \( {\small \vec{{r}_{p}} }\) and \( {\small \vec{{r}_{l}}}\) to get 3 simultaneous equations from each row.
\(\\[1pt]\)
\( {\small \hspace{5em} \vec{{r}_{p}} \quad = \quad \vec{{r}_{l}} }\)
\(\\[1pt]\)
\( {\small \hspace{1.2em} \begin{pmatrix}
1 \ + \ 2 \lambda \\[1pt]
2 \ – \ \lambda \\[1pt]
-1 \ + \ 2 \lambda
\end{pmatrix} \ = \ \begin{pmatrix}
2 \ + \ \mu \\[1pt]
1 \ + \ \mu \\[1pt]
1 \ + \ 2\mu
\end{pmatrix} }\)
\(\\[1pt]\)
\(\\[10pt]\) Compare the first row of the above matrix,
\(\\[10pt]\hspace{1.2 em} {\small 1 \ + \ 2 \lambda \ = \ 2 \ + \ \mu} \)
\(\\[10pt]\hspace{1.7 em} {\small 2 \lambda \ – \ \mu \ = \ 2 \ – \ 1} \)
\(\\[17pt]\hspace{1.7 em} {\small 2 \lambda \ – \ \mu \ = \ 1} \quad \ldots \quad {\small (1)} \)
\(\\[10pt]\) Compare the second row,
\(\\[10pt]\hspace{2 em} {\small 2 \ – \ \lambda \ = \ 1 \ + \ \mu} \)
\(\\[10pt]\hspace{1.2 em} {\small – \ \lambda \ – \ \mu \ = \ 1 \ – \ 2} \)
\(\\[17pt]\hspace{1.5 em} {\small \lambda \ + \ \mu \ = \ 1} \quad \ldots \quad {\small (2)} \)
\(\\[10pt]\) Compare the third row,
\(\\[10pt]\hspace{1.2 em} {\small -1 \ + \ 2\lambda \ = \ 1 \ + \ 2\mu} \)
\(\\[10pt]\hspace{1.8 em} {\small 2\lambda \ – \ 2\mu \ = \ 1 \ + \ 1} \)
\(\\[10pt]\hspace{1.8 em} {\small 2\lambda \ – \ 2\mu \ = \ 2} \)
\(\\[10pt]\hspace{2.7 em} {\small \lambda \ – \ \mu \ = \ 1} \quad \ldots \quad {\small (3)} \)
\(\\[1pt]\)
Next, we choose any two equations from the above to solve for \({\small \lambda}\) and \({\small \mu}\) and then check whether the acquired \({\small \lambda}\) and \({\small \mu}\) values also satisfy the third equation.
\(\\[1pt]\)
\(\\[10pt]\hspace {1.2em}\) (1) + (2),
\(\\[10pt]\hspace{2 em} { \small 3\lambda \ = \ 2 } \)
\(\\[15pt]\hspace{2.4 em} {\small \lambda \ = \ {\large \frac{2}{3}} }\)
\(\\[10pt]\hspace{1.7 em} {\small \lambda \ + \ \mu \ = \ 1} \quad \ldots \quad {\small (2)} \)
\(\\[15pt]\hspace{1.4 em} {\small {\large \frac{2}{3}} \ + \ \mu \ = \ 1} \)
\(\\[15pt]\hspace{3.8 em} {\small \mu \ = \ 1 \ – \ {\large \frac{2}{3}} } \)
\(\hspace{3.8 em} {\small \mu \ = \ {\large \frac{1}{3}} } \)
\(\\[1pt]\)
Using equation (1) and (2), we find \({\small \lambda \ = \ {\large\frac{2}{3}}}\) and \({\small \mu \ = \ {\large\frac{1}{3}}}\). Then, we substitute these values to equation (3) and check if the equation is still satisfied.
\(\\[1pt]\)
\(\\[10pt]\hspace{2.9 em} {\small \lambda \ – \ \mu \quad = \quad 1} \quad \ldots \quad {\small (3)} \)
\(\\[10pt]\hspace{2.5 em} {\small {\large \frac{2}{3}} \ – \ {\large \frac{1}{3}} \quad ? \quad 1} \)
\(\\[17pt]\hspace{4 em} {\small {\large \frac{1}{3}} \quad \ne \quad 1} \)
Since, the equation (3) is not satisfied then both lines does not intersect.
\(\\[1pt]\)
\({\small \hspace{1.2em}(ii). \enspace }\) There are two questions for this part, the first question is to find a plane equation from its normal vector and a point lies on it.
\(\\[1pt]\)
The second is to find a point of intersection between a line and a plane.
\(\\[1pt]\)
Let’s tackle the first question at the moment.
\(\\[1pt]\)
Since plane m is perpendicular to AB, the normal vector of plane m is the same as the direction vector of line p.
\(\\[1pt]\)
Also, we can find the midpoint of AB that lies on plane m.
\(\\[1pt]\)
Afterwards, we can write the general plane equation.
\(\\[1pt]\)
\(\\[7pt]\) The midpoint of AB,
\(\\[15pt]\hspace{1.2 em} { \small = \ \big( {\large \frac{x_{A} \ + \ x_{B}}{2}}, {\large \frac{y_{A} \ + \ y_{B}}{2}}, {\large \frac{z_{A} \ + \ z_{B}}{2}} \big) }\)
\(\\[15pt]\hspace{1.2 em} { \small = \ \big( {\large \frac{1 \ + \ 3}{2}}, {\large \frac{2 \ + \ 1}{2}}, {\large \frac{-1 \ + \ 1}{2}} \big) }\)
\(\\[15pt]\hspace{1.2 em} { \small = \ \big( 2, \ {\large \frac{3}{2}}, \ 0 \big) }\)
\(\\[15pt]\hspace{1.2 em} { \small = \ 2\textbf{i} \ + \ {\large\frac{3}{2}}\textbf{j} }\).
The normal vector of plane m is \({\small \ 2\textbf{i} \ – \ \textbf{j} \ + \ 2\textbf{k}}\) (the direction vector of \( \ {\small \vec{{r}_{p}} }\)).
\(\\[1pt]\)
\(\\[7pt]\) The plane equation in vector form,
\(\\[10pt] \hspace{3em} (\vec{r} \ – \ \vec{a}) \cdot \vec{n} \ = \ 0 \)
\(\\[15pt] {\small \big( \vec{r} \ – \ ( 2\textbf{i} \ + \ \frac{3}{2}\textbf{j} ) \big) \ \cdot \ (2\textbf{i} \ – \ \textbf{j} \ + \ 2\textbf{k}) \ = \ 0 }\)
\(\\[20pt] \hspace{1.2em} {\small \big( \vec{r} \ – \ 2\textbf{i} \ – \ \frac{3}{2}\textbf{j} \big) \ \cdot \ (2\textbf{i} \ – \ \textbf{j} \ + \ 2\textbf{k}) \ = \ 0 }\)
\(\\[7pt]\) Or, in cartesian form,
\(\\[10pt] \hspace{3em} {\small ax \ + \ by \ + \ cz \ = \ d }\)
with \(\\[27pt] \ {\small \begin{pmatrix}
a \\[1pt]
b \\[1pt]
c
\end{pmatrix} = \begin{pmatrix}
2 \\[1pt]
-1 \\[1pt]
2 \\[1pt]
\end{pmatrix} }\) \(\\[7pt]\) and
\(\\[10pt] \hspace{1.2em} {\small d \ = \ \vec{a} \ \cdot \vec{n} \ }\)
\(\\[30pt]\hspace{1.8em} \ {\small = \begin{pmatrix}
2 \\[1pt]
{\large \frac{3}{2}} \\[1pt]
0
\end{pmatrix} \cdot \begin{pmatrix}
2 \\[1pt]
-1 \\[1pt]
2
\end{pmatrix} }\)
\(\\[15pt] \hspace{1.8em} \ {\small = \ 2(2) \ + \ {\large \frac{3}{2}}(-1) \ + \ 0(2)}\)
\(\\[15pt] \hspace{1.8em} \ {\small = \ 4 \ – \ {\large \frac{3}{2}} \ + \ 0 }\)
\(\\[17pt] \hspace{1.8em} \ {\small = \ {\large \frac{5}{2}} }\)
\(\\[5pt]\) Therefore, the plane equation in cartesian form is,
\( \hspace{3em} {\small 2x \ – \ y \ + \ 2z \ = \ {\large \frac{5}{2}} }\)
\(\hspace{2.8em}\rule{4cm}{1pt} {\small \times 2}\)
\(\hspace{3em} {\small 4x \ – \ 2y \ + \ 4z \ = \ 5 }\)
\(\\[1pt]\)
Now for the second question, to find the point of intersection between the line l and plane m, we substitute the position vector of the line l (\({\small \vec{{r}_{l}}}\)) into the position vector (\({\small \vec{r}}\)) of the plane m vector equation.
\(\\[1pt]\)
\(\\[15pt] { \small \big( \vec{r} \ – \ 2\textbf{i} \ – \ \frac{3}{2}\textbf{j} \big) \cdot (2\textbf{i} \ – \ \textbf{j} \ + \ 2\textbf{k}) = 0 }\)
\(\\[15pt] {\small \vec{r} \ = \ \vec{{r}_{l}} }\),
\(\\[27pt] \ {\small \begin{pmatrix}
2 \ + \ \mu \ – \ 2 \\[1pt]
1 \ + \ \mu \ – \ {\large \frac{3}{2}} \\[1pt]
1 \ + \ 2\mu
\end{pmatrix} \cdot \begin{pmatrix}
2 \\[1pt]
-1 \\[1pt]
2 \\[1pt]
\end{pmatrix} } \ = \ 0 \)
\(\\[27pt] \ {\small \begin{pmatrix}
\mu \\[1pt]
{\large -\frac{1}{2}} \ + \ \mu \\[1pt]
1 \ + \ 2\mu
\end{pmatrix} \cdot \begin{pmatrix}
2 \\[1pt]
-1 \\[1pt]
2 \\[1pt]
\end{pmatrix} } \ = \ 0 \)
\(\\[15pt] {\small \mu(2) \ + \ ({\large -\frac{1}{2}} \ + \ \mu)(-1) \ + \ (1 \ + \ 2\mu)(2) = 0}\)
\(\\[15pt] {\small 2\mu \ + \ {\large \frac{1}{2}} \ – \ \mu \ + \ 2 \ + \ 4\mu = 0}\)
\(\\[15pt] \hspace{6em} {\small 5\mu \ = \ {\large -\frac{5}{2}} }\)
\(\\[15pt] \hspace{6.3em} {\small \mu \ = \ {\large -\frac{1}{2}} }\)
Then, we substitute \( {\small \mu \ = \ {\large -\frac{1}{2}} }\) into the position vector of line l to finally find point P,
\(\\[1pt]\)
\(\\[15pt] {\small P: \hspace{1em} 2\textbf{i} + \textbf{j} + \textbf{k} + {\large -\frac{1}{2}} (\textbf{i} + \textbf{j} + 2\textbf{k}) }\)
\(\\[15pt]{\hspace{2.2em} \small 2\textbf{i} \ {\large -\frac{1}{2}}\textbf{i} \ + \textbf{j} \ {\large -\frac{1}{2}}\textbf{j} \ + \textbf{k} \ – \textbf{k} }\)
\({\hspace{4em} \small {\large \frac{3}{2}}\textbf{i} \ + \ {\large \frac{1}{2}}\textbf{j} }\)
\(\\[1pt]\)
\({\small 4.\enspace}\)
\(\\[1pt]\)
The diagram shows a set of rectangular axes
Ox,
Oy and
Oz, and four points
A,
B,
C and
D with position vectors \({\small \overrightarrow{OA} \ = \ 3\textbf{i} }\), \({\small \overrightarrow{OB} \ = \ 3\textbf{i} \ + \ 4\textbf{j} \ }\), \({\small \overrightarrow{OC} \ = \ \textbf{i} \ + \ 3\textbf{j} \ }\) and \({\small \overrightarrow{OD} \ = \ 2\textbf{i} \ + \ 3\textbf{j} \ + \ 5\textbf{k} \ }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\) Find the equation of the plane
BCD, giving your answer in the form \({\small ax + by + cz = d}\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}\) Calculate the acute angle between the planes
BCD and
OABC .
\(\\[7pt]{\small \hspace{1.2em}\textrm{(i)}. \enspace }\) We need to find
the cartesian form of the plane
BCD given 3 points on it,
\(\\[7pt]{\small \hspace{1.2em} \overrightarrow{OB} \ = \ 3\textbf{i} \ + \ 4\textbf{j} \ }\),
\(\\[7pt]{\small \hspace{1.2em} \overrightarrow{OC} \ = \ \textbf{i} \ + \ 3\textbf{j} \ }\) and
\(\\[12pt]{\small \hspace{1.2em} \overrightarrow{OD} \ = \ 2\textbf{i} \ + \ 3\textbf{j} \ + \ 5\textbf{k} \ }\).
Step 1:\(\enspace\) Construct 2 vectors from any 2 points from the given 3 points. Note that, the choice is arbitrary as long each of the points are on the plane. \(\\[1pt]\)
\(\\[7pt]{\small \hspace{1.2em} \overrightarrow{CB} \ = \ \overrightarrow{OB} \ – \ \overrightarrow{OC} }\)
\(\\[7pt]{\small \hspace{2.9em} = \ 3\textbf{i} \ + \ 4\textbf{j} \ – \ (\textbf{i} \ + \ 3\textbf{j}) }\)
\(\\[7pt]{\small \hspace{2.9em} = \ 3\textbf{i} \ – \ \textbf{i} \ + \ 4\textbf{j} \ – \ 3\textbf{j} }\)
\(\\[15pt]{\small \hspace{2.9em} = \ 2\textbf{i} \ + \ \textbf{j} }\)
\(\\[7pt]{\small \hspace{1.2em} \overrightarrow{CD} \ = \ \overrightarrow{OD} \ – \ \overrightarrow{OC} }\)
\(\\[7pt]{\small \hspace{2.9em} = \ 2\textbf{i} \ + \ 3\textbf{j} \ + \ 5\textbf{k} \ – \ (\textbf{i} \ + \ 3\textbf{j}) }\)
\(\\[7pt]{\small \hspace{2.9em} = \ 2\textbf{i} \ – \ \textbf{i} \ + \ 3\textbf{j} \ – \ 3\textbf{j} \ + \ 5\textbf{k} }\)
\(\\[15pt]{\small \hspace{2.9em} = \ \textbf{i} \ + \ 5\textbf{k} }\)
Step 2:\(\enspace\) Cross product the 2 vectors found from Step 1. The resulting vector from the cross product is a vector that is perpendicular to both vectors and thus it is the normal vector of plane BCD. \(\\[1pt]\)
\(\\[27pt]{\small \hspace{1.2em} \overrightarrow{CB} \ \times \overrightarrow{CD} \ = \ \begin{pmatrix}
2 \\[1pt]
1 \\[1pt]
0
\end{pmatrix} \times \begin{pmatrix}
1 \\[1pt]
0 \\[1pt]
5 \\[1pt]
\end{pmatrix} }\)
\(\\[7pt]{\small \hspace{2em} = \ (2)(5)(-\textbf{j}) \ + \ (1)(1)(-\textbf{k}) \ + \ (1)(5)(\textbf{i}) }\)
\(\\[7pt]{\small \hspace{2em} = \ -10\textbf{j} \ – \ \textbf{k} \ + \ 5\textbf{i} }\)
\(\\[15pt]{\small \hspace{2em} = \ 5\textbf{i} \ – \ 10\textbf{j} \ – \ \textbf{k} }\)
Step 3:\(\enspace\) Plug in the cartesian components of the normal vector found in Step 2 into the plane equation and found the unknown constant d. \(\\[1pt]\)
\(\\[10pt] \hspace{3em} {\small ax \ + \ by \ + \ cz \ = \ d }\)
with \(\\[27pt] \ {\small \begin{pmatrix}
a \\[1pt]
b \\[1pt]
c
\end{pmatrix} = \begin{pmatrix}
5 \\[1pt]
-10 \\[1pt]
-1 \\[1pt]
\end{pmatrix} }\) \(\\[7pt]\) and
\(\\[10pt] \hspace{1.2em} {\small d \ = \ \vec{a} \ \cdot \vec{n} \ }\)
\(\\[30pt]\hspace{1.8em} \ {\small = \begin{pmatrix}
3 \\[1pt]
4 \\[1pt]
0
\end{pmatrix} \cdot \begin{pmatrix}
5 \\[1pt]
-10 \\[1pt]
-1
\end{pmatrix} }\)
\(\\[15pt] \hspace{1.8em} \ {\small = \ 3(5) \ + \ 4(-10) \ + \ 0(-1)}\)
\(\\[15pt] \hspace{1.8em} \ {\small = \ 15 \ – \ 40 \ + \ 0 }\)
\(\\[15pt] \hspace{1.8em} \ {\small = \ -25 }\)
Note that i am using \({\small \overrightarrow{OB}}\) as \({\small \vec{a}}\) in the above equation. However, we can also use either \({\small \overrightarrow{OC}}\) or \({\small \overrightarrow{OD}}\) since each of them lies on the plane
BCD.
\(\\[1pt]\)
\(\\[7pt]\) Hence, the equation of plane
BCD is,
\(\\[20pt] \hspace{3em} 5x \ – \ 10y \ – \ z \ = \ -25 \)
\({\small \hspace{1.2em}\textrm{(ii)}. \enspace }\) The angle between the normal vectors of two planes is also the angle between the planes. The illustration of the angle between two planes can be seen below.
\(\\[1pt]\)
We can use
the dot product of the normal vectors between plane
BCD and
OABC to find the angle,
\(\\[1pt]\)
\(\\[12pt]{\small \hspace{1.2em} \overrightarrow{{n}_{OABC}} \ = \ \textbf{k} \ }\) since the plane
OABC lies on the xy-plane,
\(\\[15pt]{\small \hspace{1.2em} |\overrightarrow{{n}_{OABC}}| \ = \ 1 }\),
\(\\[12pt]{\small \hspace{1.2em} \overrightarrow{{n}_{BCD}} \ = \ 5\textbf{i} \ – \ 10\textbf{j} \ – \ \textbf{k} }\),
\(\\[12pt]{\small \hspace{1.2em} |\overrightarrow{{n}_{BCD}}| \ = \ \sqrt{(5)^2 \ + \ (-10)^2 \ + \ (-1)^2} }\)
\(\\[12pt]{\small \hspace{4.2em} = \ \sqrt{25 \ + \ 100 \ + \ 1} }\)
\(\\[15pt]{\small \hspace{4.2em} = \ \sqrt{126} }\),
\(\\[7pt]\) Therefore,
\(\\[15pt]{\small \overrightarrow{{n}_{OABC}} \ \cdot \ \overrightarrow{{n}_{BCD}} \ = \ |\overrightarrow{{n}_{OABC}} | | \overrightarrow{{n}_{BCD}} | \ \cos \theta }\)
\(\\[20pt]{\small \hspace{3.7em} \cos \theta \ = \ {\large\frac{\overrightarrow{{n}_{OABC}} \ \cdot \ \overrightarrow{{n}_{BCD}}}{ |\overrightarrow{{n}_{OABC}} | | \overrightarrow{{n}_{BCD}} | }} }\)
\(\\[20pt]{\small \hspace{5em} \theta \ = \ \cos^{-1} \bigg({\large\frac{\overrightarrow{{n}_{OABC}} \ \cdot \ \overrightarrow{{n}_{BCD}}}{ |\overrightarrow{{n}_{OABC}} | | \overrightarrow{{n}_{BCD}} | }} \bigg) }\)
\(\\[20pt]{\small \hspace{5em} \theta \ = \ \cos^{-1} \bigg( {\large\frac{ (\textbf{k}) \ \cdot \ (5\textbf{i} \ – \ 10\textbf{j} \ – \ \textbf{k}) }{ 1 \times \sqrt{126} } } \bigg) }\)
\(\\[20pt]{\small \hspace{5em} \theta \ = \ \cos^{-1} \bigg( {\large\frac{ -1 }{ \sqrt{126} } } \bigg) }\)
\(\\[7pt]{\small \hspace{5em} \theta \ = \ 95.11^{\large{\circ}} }\)
\(\\[10pt]\) since we want the acute angle,
\(\\[10pt]{\small \hspace{5em} \theta \ = \ 180^{\large{\circ}} \ – \ 95.11^{\large{\circ}} }\)
\({\small \hspace{5em} \theta \ = \ 84.9^{\large{\circ}} }\)
\(\\[1pt]\)
\({\small 5.\enspace}\) The line
l has equation
r =
i + 2
j + 3
k + \({\small \mu}\)(2
i –
j – 2
k).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\) The point
P has position vector 4
i + 2
j – 3
k. Find the length of the perpendicular from
P to
l.
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}\) It is given that
l lies in the plane \({\small ax + by + 2z = 13}\), where
a and
b are constants. Find the values of
a and
b.
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\) Find the distance from a point to a line.
\(\\[1pt]\)
Step 1:\(\enspace\) Find the point of intersection between a perpendicular line from point P to l.
\(\\[1pt]\)
Let the foot of the perpendicular line from P to l as Q,
\(\\[1pt]\)
\(\\[10pt]{\small \hspace{1.2em} \overrightarrow{OQ} \ = \ \textbf{i} \ + \ 2\textbf{j} \ + \ 3\textbf{k} \ + \ \mu (2\textbf{i} \ – \ \textbf{j} \ – \ 2\textbf{k}) }\)
\(\\[33pt]{\small \hspace{3em} = \ \begin{pmatrix}
1 \ + \ 2\mu \\[1pt]
2 \ – \ \mu \\[1pt]
3 \ – \ 2\mu
\end{pmatrix} }\)
Step 2:\(\enspace\) Construct the vector joining point P and Q, \( {\small \overrightarrow{PQ} }\).
\(\\[1pt]\)
\(\\[10pt]{\small \hspace{2.2em} \overrightarrow{PQ} \ = \ \overrightarrow{OQ} \ – \ \overrightarrow{OP} }\)
\(\\[30pt]{\small \hspace{4em} = \ \begin{pmatrix}
1 \ + \ 2\mu \ – \ 4 \\[1pt]
2 \ – \ \mu \ – \ 2 \\[1pt]
3 \ – \ 2\mu \ – \ (-3)
\end{pmatrix} }\)
\(\\[33pt]{\small \hspace{4em} = \ \begin{pmatrix}
-3 \ + \ 2\mu \\[1pt]
– \ \mu \\[1pt]
6 \ – \ 2\mu
\end{pmatrix} }\)
Step 3:\(\enspace\) Find the \({\small \mu }\) value by calculating the dot product between \( {\small \overrightarrow{PQ} }\) and the direction vector of line l equal to zero. The zero result of dot product enforces the perpendicularity of both vectors.
\(\\[1pt]\)
\(\\[10pt]{\small \hspace{2.3em} \overrightarrow{PQ} \ \cdot \ (2\textbf{i} \ – \ \textbf{j} \ – \ 2\textbf{k}) \ = \ 0 }\)
\(\\[30pt] \ {\small \begin{pmatrix}
-3 \ + \ 2\mu \\[1pt]
– \ \mu \\[1pt]
6 \ – \ 2\mu
\end{pmatrix} \cdot \begin{pmatrix}
2 \\[1pt]
-1 \\[1pt]
-2 \\[1pt]
\end{pmatrix} } \ = \ 0 \)
\(\\[10pt] {\small (-3 \ + \ 2\mu)(2) \ + \ (-\mu)(-1) \ + \ (6 \ – \ 2\mu)(-2) = 0}\)
\(\\[10pt] {\small -6 \ + \ 4\mu \ + \ \mu \ – \ 12 \ + \ 4\mu = 0}\)
\(\\[10pt] \hspace{8.6em} {\small 9\mu \ = \ 18 }\)
\(\\[10pt] \hspace{9em} {\small \mu \ = \ 2 }\)
Then, we substitute \( {\small \mu \ = \ 2 }\) to find point \( {\small \overrightarrow{PQ} }\),
\(\\[1pt]\)
\(\\[27pt]{\small \hspace{1.2em} \overrightarrow{PQ} \ = \ \begin{pmatrix}
-3 \ + \ 2(2) \\[1pt]
– \ (2) \\[1pt]
6 \ – \ 2(2)
\end{pmatrix} }\)
\(\\[27pt]{\small \hspace{3em} = \ \begin{pmatrix}
-3 \ + \ 4 \\[1pt]
-2 \\[1pt]
6 \ – \ 4
\end{pmatrix} }\)
\(\\[35pt]{\small \hspace{3em} = \ \begin{pmatrix}
1 \\[1pt]
-2 \\[1pt]
2
\end{pmatrix} }\)
Step 4:\(\enspace\) Find the length of \( {\small \overrightarrow{PQ} }\), that is the length of the perpendicular from P to l.
\(\\[1pt]\)
\(\\[12pt]{\small \hspace{1.7em} \overrightarrow{PQ} \ = \ \textbf{i} \ – \ 2\textbf{j} \ + \ 2\textbf{k} }\),
\(\\[12pt]{\small \hspace{1.2em} |\overrightarrow{PQ}| \ = \ \sqrt{(1)^2 \ + \ (-2)^2 \ + \ (2)^2} }\)
\(\\[12pt]{\small \hspace{3.4em} = \ \sqrt{1 \ + \ 4 \ + \ 4} }\)
\(\\[12pt]{\small \hspace{3.4em} = \ \sqrt{9} }\)
\(\\[15pt]{\small \hspace{3.4em} = \ 3 }\).
\({\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}\) Find the plane equation given a vector line equation lies on the plane.
\(\\[1pt]\)
Step 1:\(\enspace\) Find any two points on the line l. We can do so by choosing arbitrary values of \({\small \mu}\).
\(\\[1pt]\)
\(\\[7pt]\) Choose \({\small \mu \ = \ 0, }\)
\(\\[7pt]\) Let’s name it \( {\small \overrightarrow{OA} }\),
\(\\[15pt]{\small \hspace{1.2em} \overrightarrow{OA} \ = \ \textbf{i} \ + \ 2\textbf{j} \ + \ 3\textbf{k} }\)
\(\\[7pt]\) Choose \({\small \mu \ = \ 1, }\)
\(\\[7pt]\) Let’s name it \( {\small \overrightarrow{OB} }\),
\(\\[10pt]{\small \hspace{1.2em} \overrightarrow{OB} \ = \ \textbf{i} \ + \ 2\textbf{j} \ + \ 3\textbf{k} \ + \ (1) (2\textbf{i} \ – \ \textbf{j} \ – \ 2\textbf{k}) }\)
\(\\[10pt]{\small \hspace{1.2em} \overrightarrow{OB} \ = \ 3\textbf{i} \ + \ \textbf{j} \ + \ \textbf{k} }\)
\(\\[12pt]\) The two points are A(1, 2, 3) and B(3, 1, 1) .
Step 2:\(\enspace\) Substitute both points into the cartesian plane equation to find the values of a and b.
\(\\[1pt]\)
\(\\[10pt] \hspace{3.1em} {\small ax \ + \ by \ + \ 2z \ = \ 13 }\)
\(\\[10pt] \hspace{1.2em} {\small a(1) \ + \ b(2) \ + \ 2(3) \ = \ 13 }\)
\(\\[10pt] \hspace{4.1em} {\small a \ + \ 2b \ + \ 6 \ = \ 13 }\)
\(\\[15pt] \hspace{6em} {\small a \ + \ 2b \ = \ 7 \quad \ldots \quad (1) }\)
\(\\[10pt] \hspace{1.2em} {\small a(3) \ + \ b(1) \ + \ 2(1) \ = \ 13 }\)
\(\\[10pt] \hspace{4.1em} {\small 3a \ + \ b \ + \ 2 \ = \ 13 }\)
\(\\[15pt] \hspace{6em} {\small 3a \ + \ b \ = \ 11 \quad \ldots \quad (2) }\)
\(\\[7pt]\) Using elimination or substitution method for (1) and (2),
\(\hspace{1.2em} a \ = \ 3 \ \) and \( \ b \ = \ 2 \).
\(\\[1pt]\)
\({\small 6.\enspace}\) Two planes have equations \({\small 2x + 3y \ – \ z \ = \ 1 \ }\) and \( \ {\small x \ – \ 2y + z \ = \ 3}\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\) Find the acute angle between the planes.
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}\) Find a vector equation for the line of intersection of the planes.
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\) As has been described in detail in Example 2, the acute angle can be found from the dot product of the normal vectors of both planes.
\(\\[1pt]\)
\(\\[7pt]{\small \hspace{1.2em} \textrm{Plane 1}: \quad 2x \ + \ 3y \ – \ z \ = \ 1}\)
\(\\[12pt]{\small \hspace{1.4em} \overrightarrow{{n}_{1}} \ = \ 2\textbf{i} \ + \ 3\textbf{j} \ – \ \textbf{k} \ }\)
\(\\[15pt]{\small \hspace{1.2em} |\overrightarrow{{n}_{1}}| \ = \ \sqrt{(2)^2 \ + \ (3)^2 \ + \ (-1)^2} }\)
\(\\[12pt]{\small \hspace{3.3em} = \ \sqrt{4 \ + \ 9 \ + \ 1} }\)
\(\\[15pt]{\small \hspace{3.3em} = \ \sqrt{14} }\)
\(\\[7pt]{\small \hspace{1.2em} \textrm{Plane 2}: \quad x \ – \ 2y \ + \ z \ = \ 3}\)
\(\\[12pt]{\small \hspace{1.4em} \overrightarrow{{n}_{2}} \ = \ \textbf{i} \ – \ 2\textbf{j} \ + \ \textbf{k} \ }\)
\(\\[15pt]{\small \hspace{1.2em} |\overrightarrow{{n}_{2}}| \ = \ \sqrt{(1)^2 \ + \ (-2)^2 \ + \ (1)^2} }\)
\(\\[12pt]{\small \hspace{3.3em} = \ \sqrt{1 \ + \ 4 \ + \ 1} }\)
\(\\[15pt]{\small \hspace{3.3em} = \ \sqrt{6} }\)
\(\\[7pt]\) Therefore,
\(\\[15pt]{\small \hspace{1.2em} \overrightarrow{{n}_{1}} \ \cdot \ \overrightarrow{{n}_{2}} \ = \ |\overrightarrow{{n}_{1}} | | \overrightarrow{{n}_{2}} | \ \cos \theta }\)
\(\\[20pt]{\small \hspace{1.7em} \cos \theta \ = \ {\large\frac{\overrightarrow{{n}_{1}} \ \cdot \ \overrightarrow{{n}_{2}}}{ |\overrightarrow{{n}_{1}} | | \overrightarrow{{n}_{2}} | }} }\)
\(\\[20pt]{\small \hspace{3em} \theta \ = \ \cos^{-1} \bigg({\large\frac{\overrightarrow{{n}_{1}} \ \cdot \ \overrightarrow{{n}_{2}}}{ |\overrightarrow{{n}_{1}} | | \overrightarrow{{n}_{2}} | }} \bigg) }\)
\(\\[20pt]{\small \hspace{3em} \theta \ = \ \cos^{-1} \bigg( {\large\frac{ (2\textbf{i} \ + \ 3\textbf{j} \ – \ \textbf{k}) \ \cdot \ (\textbf{i} \ – \ 2\textbf{j} \ + \ \textbf{k}) }{ \sqrt{14} \times \sqrt{6} } } \bigg) }\)
\(\\[20pt]{\small \hspace{3em} \theta \ = \ \cos^{-1} \bigg( {\large\frac{ 2(1) \ + \ 3(-2) \ + \ -1(1) }{ 14 } } \bigg) }\)
\(\\[20pt]{\small \hspace{3em} \theta \ = \ \cos^{-1} \bigg( {\large\frac{ 2 \ – \ 6 \ – \ 1 }{ \sqrt{84} } } \bigg) }\)
\(\\[20pt]{\small \hspace{3em} \theta \ = \ \cos^{-1} \bigg( {\large\frac{-5 }{ 2\sqrt{21} } } \bigg) }\)
\(\\[7pt]{\small \hspace{3em} \theta \ = \ 123.06^{\large{\circ}} }\)
\(\\[10pt]\) since we want the acute angle,
\(\\[10pt]{\small \hspace{3em} \theta \ = \ 180^{\large{\circ}} \ – \ 123.06^{\large{\circ}} }\)
\({\small \hspace{3em} \theta \ = \ 56.9^{\large{\circ}} }\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}\) Find the line of intersection between two planes.
\(\\[1pt]\)
Step 1:\(\enspace\) Find the direction vector of the line of intersection. It will be in perpendicular direction to both normal vectors of the planes. We can make use of the cross product between the two normal vectors.
\(\\[1pt]\)
\(\\[27pt]{\small \hspace{1.2em} \overrightarrow{{n}_{1}} \ \times \overrightarrow{{n}_{2}} \ = \ \begin{pmatrix}
2 \\[1pt]
3 \\[1pt]
-1
\end{pmatrix} \times \begin{pmatrix}
1 \\[1pt]
-2 \\[1pt]
1 \\[1pt]
\end{pmatrix} }\)
\(\\[10pt]{\small \hspace{4.8em} = \ -4\textbf{k} \ – \ 2\textbf{j} \ – \ 3\textbf{k} \ – \ \textbf{j} \ – \ 2\textbf{i} }\)
\(\\[18pt]{\small \hspace{4.8em} = \ \textbf{i} \ – \ 3\textbf{j} \ – \ 7\textbf{k} }\)
Step 2:\(\enspace\) Find one point on the line of intersection in order to construct our location vector.
\(\\[1pt]\)
\(\\[10pt]\) Set \({\small x \ = \ 0}\) into our plane equations,
\(\\[7pt]{\small \hspace{1.2em} 2(0) + 3y \ – \ z \ = \ 1}\)
\(\\[12pt]{\small \hspace{3.8em} 3y \ – \ z \ = \ 1 \quad \ldots \quad (1) }\)
\(\\[7pt]{\small \hspace{1.2em} (0) \ – \ 2y + z \ = \ 3}\)
\(\\[12pt]{\small \hspace{2.2em} -2y \ + \ z \ = \ 3 \quad \ldots \quad (2) }\)
\(\\[10pt]\hspace {1.2em}\) (1) + (2),
\(\\[15pt]\hspace{2 em} { \small y \ = \ 4 } \)
\(\\[10pt]\hspace{1.7 em} {\small 3y \ – \ z \ = \ 1} \quad \ldots \quad {\small (1)} \)
\(\\[10pt]\hspace{1.4 em} {\small 3(4) \ – \ z \ = \ 1} \)
\(\\[10pt]\hspace{3.8 em} {\small z \ = \ 12 \ – \ 1} \)
\(\\[15pt]\hspace{3.8 em} {\small z \ = \ 11 } \)
The location vector is \({\small 4\textbf{j} \ + \ 11\textbf{k} }\)
\(\\[1pt]\)
The vector equation for the line of intersection of the planes is
\(\\[1pt]\)
\( \hspace{3em} {\small \vec{r} \ = \ 4\textbf{j} \ + \ 11\textbf{k} \ + \ \lambda(\textbf{i} \ – \ 3\textbf{j} \ – \ 7\textbf{k} ) }\).
\(\\[1pt]\)
\({\small 7.\enspace}\) The planes
m and
n have equations \({\small 3x + y \ – \ 2z \ = \ 10}\) and \({\small x \ – \ 2y + 2z \ = \ 5}\) respectively. The line
l has equation
r = 4
i + 2
j +
k + \({\small \lambda}\)(
i +
j + 2
k).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\) Show that
l is parallel to
m.
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}\) Calculate the acute angle between the planes
m and
n.
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(iii)}.\hspace{0.8em}}\) A point
P lies on the line
l. The perpendicular distance
P from the plane
l is equal to 2. Find the position vectors of the two possible positions of
P.
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\) To show that a line is parallel to a plane.
\(\\[1pt]\)
We can find the dot product of the direction vector of line l and the normal vector of plane m.
\(\\[1pt]\)
If it is zero, it means they have the same direction (parallel) to each other.
\(\\[1pt]\)
Then, we can choose one point on the line and check whether the line lies on the plane.
\(\\[1pt]\)
\(\\[10pt]{\small \hspace{1.2em} \textrm{Plane m}: \quad 3x \ + \ y \ – \ 2z \ = \ 10}\)
\(\\[12pt]{\small \hspace{3.4em} \overrightarrow{{n}_{m}} \ = \ 3\textbf{i} \ + \ \textbf{j} \ – \ 2\textbf{k} }\)
\(\\[10pt]{\small \hspace{1.2em} \textrm{Line} \ l: \quad \vec{r} \ = \ 4\textbf{i} + 2\textbf{j} + \textbf{k} + \lambda(\textbf{i} + \textbf{j} + 2\textbf{k} ) }\)
\(\\[20pt] \hspace{2em} \) direction vector \({\small \ = \ \textbf{i} + \textbf{j} + 2\textbf{k} \ }\)
The dot product of \({\small \overrightarrow{{n}_{m}} }\) and the direction vector,
\(\\[1pt]\)
\(\\[12pt]{\small \hspace{1.2em} ( 3\textbf{i} + \textbf{j} \ – \ 2\textbf{k} ) \ \cdot \ ( \textbf{i} + \textbf{j} + 2\textbf{k} ) }\)
\(\\[10pt] {\small \hspace{2em} \ = \ 3(1) \ + \ 1(1) \ + \ -2(2) }\)
\(\\[7pt] {\small \hspace{2em} \ = \ 3 \ + \ 1 \ – \ 4 }\)
\(\\[15pt] {\small \hspace{2em} \ = \ 0 }\)
Then, we choose one point on the line l, \({\small 4\textbf{i} + 2\textbf{j} + \textbf{k} \ }\) or (4, 2, 1) and substitute the point onto the plane equation to check whether the line lies on the plane.
\(\\[1pt]\)
\(\\[10pt]{\small \hspace{1.2em} \textrm{Plane m}: \quad 3x \ + \ y \ – \ 2z \ = \ 10}\)
\(\\[7pt]{\small \hspace{1.2em} 3(4) + 2 \ – \ 2(1) \quad ? \quad 10}\)
\(\\[7pt]{\small \hspace{2.6em} 12 \ + \ 2 \ – \ 2 \quad ? \quad 10 }\)
\(\\[12pt]{\small \hspace{6.3em} 12 \ \ne \ 10 }\)
Therefore, line l is parallel to plane m and does not lie on it.
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}\) Find the angle between two planes.
\(\\[1pt]\)
\(\\[7pt]{\small \hspace{1.2em} \textrm{Plane m}: \quad 3x \ + \ y \ – \ 2z \ = \ 10}\)
\(\\[12pt]{\small \hspace{1.5em} \overrightarrow{{n}_{m}} \ = \ 3\textbf{i} \ + \ \textbf{j} \ – \ 2\textbf{k} }\)
\(\\[15pt]{\small \hspace{1.2em} |\overrightarrow{{n}_{m}}| \ = \ \sqrt{(3)^2 \ + \ (1)^2 \ + \ (-2)^2} }\)
\(\\[12pt]{\small \hspace{3.3em} = \ \sqrt{9 \ + \ 1 \ + \ 4} }\)
\(\\[15pt]{\small \hspace{3.3em} = \ \sqrt{14} }\)
\(\\[7pt]{\small \hspace{1.2em} \textrm{Plane n}: \quad x \ – \ 2y \ + \ 2z \ = \ 5}\)
\(\\[12pt]{\small \hspace{1.5em} \overrightarrow{{n}_{n}} \ = \ \textbf{i} \ – \ 2\textbf{j} \ + \ 2\textbf{k} }\)
\(\\[15pt]{\small \hspace{1.2em} |\overrightarrow{{n}_{n}}| \ = \ \sqrt{(1)^2 \ + \ (-2)^2 \ + \ (2)^2} }\)
\(\\[12pt]{\small \hspace{3.3em} = \ \sqrt{1 \ + \ 4 \ + \ 4} }\)
\(\\[12pt]{\small \hspace{3.3em} = \ \sqrt{9} }\)
\(\\[15pt]{\small \hspace{3.3em} = \ 3 }\)
\(\\[7pt]\) Therefore,
\(\\[15pt]{\small \hspace{1.2em} \overrightarrow{{n}_{m}} \ \cdot \ \overrightarrow{{n}_{n}} \ = \ |\overrightarrow{{n}_{m}} | | \overrightarrow{{n}_{n}} | \ \cos \theta }\)
\(\\[20pt]{\small \hspace{1.7em} \cos \theta \ = \ {\large\frac{\overrightarrow{{n}_{m}} \ \cdot \ \overrightarrow{{n}_{n}}}{ |\overrightarrow{{n}_{m}} | | \overrightarrow{{n}_{n}} | }} }\)
\(\\[20pt]{\small \hspace{3em} \theta \ = \ \cos^{-1} \bigg({\large\frac{\overrightarrow{{n}_{m}} \ \cdot \ \overrightarrow{{n}_{n}}}{ |\overrightarrow{{n}_{m}} | | \overrightarrow{{n}_{n}} | }} \bigg) }\)
\(\\[20pt]{\small \hspace{3em} \theta \ = \ \cos^{-1} \bigg( {\large\frac{ (3\textbf{i} \ + \ \textbf{j} \ – \ 2\textbf{k} ) \ \cdot \ (\textbf{i} \ – \ 2\textbf{j} \ + \ 2\textbf{k}) }{ \sqrt{14} \times 3 } } \bigg) }\)
\(\\[20pt]{\small \hspace{3em} \theta \ = \ \cos^{-1} \bigg( {\large\frac{ 3(1) \ + \ 1(-2) \ + \ -2(2) }{ 3\sqrt{14} } } \bigg) }\)
\(\\[20pt]{\small \hspace{3em} \theta \ = \ \cos^{-1} \bigg( {\large\frac{ 3 \ – \ 2 \ – \ 4 }{ 3\sqrt{14} } } \bigg) }\)
\(\\[20pt]{\small \hspace{3em} \theta \ = \ \cos^{-1} \bigg( {\large\frac{-3 }{ 3\sqrt{14} } } \bigg) }\)
\(\\[7pt]{\small \hspace{3em} \theta \ = \ 105.50^{\large{\circ}} }\)
\(\\[10pt]\) since we want the acute angle,
\(\\[10pt]{\small \hspace{3em} \theta \ = \ 180^{\large{\circ}} \ – \ 105.50^{\large{\circ}} }\)
\({\small \hspace{3em} \theta \ = \ 74.5^{\large{\circ}} }\)
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(iii)}.\hspace{0.8em}}\) Find the distance of a point lies on a line to a plane.
\(\\[1pt]\)
Since point P lies on the line l, it can be represented by the position vector of line l,
\(\\[1pt]\)
\(\\[10pt]{\small \hspace{1.2em} \textrm{P:} \quad \vec{r} \ = \ 4\textbf{i} + 2\textbf{j} + \textbf{k} + \lambda(\textbf{i} + \textbf{j} + 2\textbf{k} ) }\)
\(\\[33pt]{\small \hspace{4em} = \ \begin{pmatrix}
4 \ + \ \lambda \\[1pt]
2 \ + \ \lambda \\[1pt]
1 \ + \ 2\lambda
\end{pmatrix} }\)
We can use the general formula of the distance from a point \({\small A(x_{1}, y_{1}, z_{1}) \ }\) to a plane \({\small ax \ + \ by \ + \ cz \ = \ d }\),
\(\\[1pt]\)
\(\\[30pt] \hspace{3em} s \ = \ {\large \frac{ |ax_{1} \ + \ by_{1} \ + \ cz_{1} \ – \ d| }{ \sqrt{a^2 \ + \ b^2 \ + \ c^2 \ } } } \).
\(\\[7pt]\) Then,
\(\\[25pt] \hspace{3em} 2 \ = \ \frac{ |(4 \ + \ \lambda) \ – \ 2(2 \ + \ \lambda) \ + \ 2(1 \ + \ 2\lambda) \ – \ 5| }{ \sqrt{1^2 \ + \ (-2)^2 \ + \ 2^2 \ } } \)
\(\\[25pt] \hspace{3em} 2 \ = \ \frac{ |4 \ + \ \lambda \ – \ 4 \ – \ 2\lambda \ + \ 2 \ + \ 4\lambda \ – \ 5| }{ \sqrt{1 \ + \ 4 \ + \ 4 \ } } \)
\(\\[15pt] \hspace{3em} 2 \ = \ \frac{ |3\lambda \ – \ 3| }{ 3 } \)
\(\\[12pt] \hspace{3.2em} {\small 6 \ = \ |3\lambda \ – \ 3| } \)
\(\\[10pt]\) The first value of \({\small \lambda }\),
\(\\[10pt] \hspace{3em}{\small 6 \ = \ 3\lambda \ – \ 3 }\)
\(\\[10pt] \hspace{3em}{\small 9 \ = \ 3\lambda }\)
\(\\[10pt] \hspace{3em}{\small \lambda \ = \ 3 }\)
\(\\[10pt]\) Substitute to the position vector of line l,
\(\\[10pt]{\small \hspace{1.2em} \textrm{P:} \quad \vec{r} \ = \ 4\textbf{i} + 2\textbf{j} + \textbf{k} + 3(\textbf{i} + \textbf{j} + 2\textbf{k} ) }\)
\(\\[17pt]{\small \hspace{4em} = \ 7\textbf{i} + 5\textbf{j} + 7\textbf{k} }\)
\(\\[10pt]\) The second value of \({\small \lambda }\),
\(\\[10pt] \hspace{3em}{\small 6 \ = \ -(3\lambda \ – \ 3) }\)
\(\\[10pt] \hspace{3em}{\small 6 \ = \ -3\lambda \ + \ 3) }\)
\(\\[10pt] \hspace{3em}{\small 3 \ = \ -3\lambda }\)
\(\\[10pt] \hspace{3em}{\small \lambda \ = \ -1 }\)
\(\\[10pt]\) Substitute to the position vector of line l,
\(\\[10pt]{\small \hspace{1.2em} \textrm{P:} \quad \vec{r} \ = \ 4\textbf{i} + 2\textbf{j} + \textbf{k} – 1(\textbf{i} + \textbf{j} + 2\textbf{k} ) }\)
\(\\[10pt]{\small \hspace{4em} = \ 3\textbf{i} + \textbf{j} \ – \textbf{k} }\).
\(\\[1pt]\)
\({\small 8.\enspace}\) The line
l has equation
r = 5
i – 3
j –
k + \({\small \lambda}\)(
i – 2
j +
k). The plane
p has equation (
r –
i – 2
j) . (3
i +
j +
k) = 0. The line
l intersects the plane
p at the point
A.
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\) Find the position vector of
A.
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}\) Calculate the acute angle between
l and
p.
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(iii)}.\hspace{0.8em}}\) Find the equation of the line which lies in
p and intersects
l at right angles.
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\)
Find the point of intersection between a line and a plane.
\(\\[1pt]\)
We substitute the position vector of the line
l (\({\small \vec{{r}_{l}}}\)) into the position vector (\({\small \vec{r}}\)) of the plane
p vector equation.
\(\\[1pt]\)
\(\\[15pt] { \small \big( \vec{r} \ – \ \textbf{i} \ – \ 2\textbf{j} \big) \cdot (3\textbf{i} \ + \ \textbf{j} \ + \ \textbf{k}) = 0 }\)
\(\\[15pt] {\small \vec{r} \ = \ \vec{{r}_{l}} }\),
\(\\[27pt] \ {\small \begin{pmatrix}
5 \ + \ \lambda \ – \ 1 \\[1pt]
-3 \ – \ 2\lambda \ – \ 2 \\[1pt]
-1 \ + \ \lambda \
\end{pmatrix} \cdot \begin{pmatrix}
3 \\[1pt]
1 \\[1pt]
1 \\[1pt]
\end{pmatrix} } \ = \ 0 \)
\(\\[27pt] \ {\small \begin{pmatrix}
4 \ + \ \lambda \\[1pt]
-5 \ – \ 2\lambda \\[1pt]
-1 \ + \ \lambda \
\end{pmatrix} \cdot \begin{pmatrix}
3 \\[1pt]
1 \\[1pt]
1 \\[1pt]
\end{pmatrix} } \ = \ 0 \)
\(\\[15pt] {\small (4 \ + \ \lambda)(3) \ – \ 5 \ – \ 2\lambda \ – \ 1 \ + \ \lambda = 0}\)
\(\\[15pt] {\small 12 \ + \ 3\lambda \ – \ 5 \ – \ 2\lambda \ – \ 1 \ + \ \lambda = 0}\)
\(\\[15pt] \hspace{6em} {\small 2\lambda \ = \ -6 }\)
\(\\[15pt] \hspace{6.3em} {\small \lambda \ = \ -3 }\)
Then, we substitute \( {\small \lambda \ = \ -3 }\) into the position vector of line
l to finally find the position vector of
A,
\(\\[1pt]\)
\(\\[10pt] {\small \overrightarrow{OA}: \hspace{1em} 5\textbf{i} \ – 3\textbf{j} \ – \textbf{k} + (-3)(\textbf{i} \ – 2\textbf{j} + \textbf{k}) }\)
\(\\[10pt]{\hspace{3em} \small 5\textbf{i} \ – 3\textbf{i} \ – 3\textbf{j} \ + \ 6 \ \textbf{j} – \textbf{k} \ – 3\textbf{k} }\)
\({\hspace{3em} \small 2\textbf{i} \ + \ 3\textbf{j} \ – \ 4\textbf{k} }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}\)
Find the angle between a line and a plane.
\(\\[1pt]\)
The angle between a line and a plane can be found using
the dot product of the normal vector of the plane and
the direction vector of the line.
\(\\[1pt]\)
The complementary angle of the angle found from the dot product is the angle between the line and the plane.
\(\\[1pt]\)
The illustration of the concept can be seen below.
\(\\[1pt]\)
The dot product of the normal vector of plane
p and the direction vector of line
l,
\(\\[1pt]\)
\(\\[12pt]{\small \hspace{1.2em} \overrightarrow{{n}_{p}} \ = \ 3\textbf{i} \ + \ \textbf{j} \ + \ \textbf{k} }\),
\(\\[15pt]{\small \hspace{1.2em} |\overrightarrow{{n}_{p}}| \ = \ \sqrt{(3)^2 \ + \ (1)^2 \ + \ (1)^2} }\)
\(\\[12pt]{\small \hspace{3em} = \ \sqrt{9 \ + \ 1 \ + \ 1} }\)
\(\\[15pt]{\small \hspace{3em} = \ \sqrt{11} }\),
\(\\[15pt]\) Let the direction vector of line
l is \( \ {\small \vec{d_{k}}}\),
\(\\[12pt]{\small \hspace{1.2em} \vec{d_{k}} \ = \ \textbf{i} \ – \ 2\textbf{j} \ + \ \textbf{k} }\),
\(\\[12pt]{\small \hspace{1.2em} |\vec{d_{k}}| \ = \ \sqrt{(1)^2 \ + \ (-2)^2 \ + \ (1)^2} }\)
\(\\[12pt]{\small \hspace{3em} = \ \sqrt{1 \ + \ 4 \ + \ 1} }\)
\(\\[15pt]{\small \hspace{3em} = \ \sqrt{6} }\),
\(\\[7pt]\) Therefore,
\(\\[15pt]{\small \overrightarrow{{n}_{p}} \ \cdot \ \vec{d_{k}} \ = \ |\overrightarrow{{n}_{p}} | | \vec{d_{k}} | \ \cos \theta }\)
\(\\[20pt]{\small \hspace{1.7em} \cos \theta \ = \ {\large\frac{\overrightarrow{{n}_{p}} \ \cdot \ \vec{d_{k}} }{ |\overrightarrow{{n}_{p}} | | \vec{d_{k}} | }} }\)
\(\\[20pt]{\small \hspace{2em} \theta \ = \ \cos^{-1} \bigg({\large\frac{\overrightarrow{{n}_{p}} \ \cdot \ \vec{d_{k}} }{ |\overrightarrow{{n}_{p}} | | \vec{d_{k}} | }} \bigg) }\)
\(\\[20pt]{\small \hspace{2em} \theta \ = \ \cos^{-1} \bigg( {\large\frac{ (3\textbf{i} \ + \ \textbf{j} \ + \ \textbf{k}) \ \cdot \ (\textbf{i} \ – \ 2\textbf{j} \ + \ \textbf{k}) }{ \sqrt{11} \times \sqrt{6} } } \bigg) }\)
\(\\[20pt]{\small \hspace{2em} \theta \ = \ \cos^{-1} \bigg( {\large\frac{ 3 \ – \ 2 \ + \ 1 }{ \sqrt{66} } } \bigg) }\)
\(\\[20pt]{\small \hspace{2em} \theta \ = \ \cos^{-1} \bigg( {\large\frac{ 2 }{ \sqrt{66} } } \bigg) }\)
\(\\[7pt]{\small \hspace{2em} \theta \ = \ 75.75^{\large{\circ}} }\)
\(\\[10pt]\) since we want
the complementary angle,
\(\\[10pt]{\small \hspace{2em} 90^{\large{\circ}} \ – \ \theta \ = \ 90^{\large{\circ}} \ – \ 75.75^{\large{\circ}} }\)
\(\\[17pt]{\small \hspace{5.3em} = \ 14.3^{\large{\circ}} }\)
\({\small\hspace{1.2em}\textrm{(iii)}.\hspace{0.8em}}\)
Find a line equation that lies in a plane and also intersect another line at \({\small 90^{\large{\circ}} }\) angle.
\(\\[1pt]\)
The direction vector of the new line is
perpendicular to both
the normal vector of
p and
the direction vector of
l.
\(\\[1pt]\)
We can use of
the cross product between the normal vector of
p and the direction vector of line
l to get
the direction vector of our new line,
\(\\[1pt]\)
\(\\[27pt]{\small \hspace{1.2em} \overrightarrow{{n}_{p}} \ \times \vec{d_{k}} \ = \ \begin{pmatrix}
3 \\[1pt]
1 \\[1pt]
1
\end{pmatrix} \times \begin{pmatrix}
1 \\[1pt]
-2 \\[1pt]
1 \\[1pt]
\end{pmatrix} }\)
\(\\[10pt]{\small \hspace{4.8em} = \ -6\textbf{k} \ – \ 3\textbf{j} \ – \ \textbf{k} \ + \ \textbf{i} \ + \ \textbf{j} \ + \ 2\textbf{i} }\)
\(\\[18pt]{\small \hspace{4.8em} = \ 3\textbf{i} \ – \ 2\textbf{j} \ – \ 7\textbf{k} }\)
We can use point
A found in part (i) as our
location vector.
\(\\[1pt]\)
The location vector is \({\small 2\textbf{i} \ + \ 3\textbf{j} \ – \ 4\textbf{k} }\)
\(\\[1pt]\)
Thus, the vector equation of the line is
\(\\[1pt]\)
\( \hspace{3em} {\small \vec{r} \ = \ 2\textbf{i} \ + \ 3\textbf{j} \ – \ 4\textbf{k} \ + \ \mu(3\textbf{i} \ – \ 2\textbf{j} \ – \ 7\textbf{k} ) }\).
\(\\[1pt]\)
\({\small 9.\enspace}\)
9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 8 \(\\[1pt]\)
Relative to the origin \(O\), the points \(A\), \(B\) and \(D\) have position vectors given by
\(\\[1pt]\)
\({\small \ \overrightarrow{OA} = \textbf{i} \ + \ 2\textbf{j} \ + \ \textbf{k} \ }\), \({\small \ \overrightarrow{OB} = 2\textbf{i} \ + \ 5\textbf{j} \ + \ 3\textbf{k} \ }\) and \({\small \ \overrightarrow{OD} = 3\textbf{i} \ + \ 2\textbf{k} }\).
\(\\[1pt]\)
A fourth point \(C\) is such that \(ABCD\) is a parallelogram.
\(\\[1pt]\)
\({\small (\textrm{a}).\hspace{0.8em}}\) Find the position vector of \(C\) and verify that the parallelogram is not a rhombus.
\(\\[1pt]\)
\({\small (\textrm{b}).\hspace{0.8em}}\) Find angle \(BAD\), giving your answer in degrees.
\(\\[1pt]\)
\({\small (\textrm{c}).\hspace{0.8em}}\) Find the area of the parallelogram correct to 3 significant figures.
Check out my solution here:
\(\\[1pt]\)
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\({\small 10.\enspace}\)
9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 10 \(\\[1pt]\)
With respect to the origin \(O\), the points \(A\) and \(B\) have position vectors given by \({\small \ \overrightarrow{OA} = 6\textbf{i} \ + \ 2\textbf{j} \ }\) and \({\small \ \overrightarrow{OB} = 2\textbf{i} \ + \ 2\textbf{j} \ + \ 3\textbf{k} }\). The midpoint of \(OA\) is \(M\). The point \(N\) lying on \(AB\), between \(A\) and \(B\), is such that \(AN = 2NB\).
\(\\[1pt]\)
\({\small (\textrm{a}).\hspace{0.8em}}\) Find a vector equation for the line through \(M\) and \(N\).
\(\\[1pt]\)
The line through \(M\) and \(N\) intersects the line through \(O\) and \(B\) at the point \(P\).
\(\\[1pt]\)
\({\small (\textrm{b}).\hspace{0.8em}}\) Find the position vector of \(P\).
\(\\[1pt]\)
\({\small (\textrm{c}).\hspace{0.8em}}\) Calculate angle \(OPM\), giving your answer in degrees.
Check out my solution here:
\(\\[1pt]\)
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\({\small 11.\enspace}\)
9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 9 \(\\[1pt]\)
With respect to the origin \(O\), the vertices of a triangle \(ABC\) have position vectors
\(\\[1pt]\)
\({\small \ \overrightarrow{OA} = 2\textbf{i} \ + \ 5\textbf{k} \ }\), \({\small \ \overrightarrow{OB} = 3\textbf{i} \ + \ 2\textbf{j} \ + \ 3\textbf{k} \ }\) and \({\small \ \overrightarrow{OC} = \textbf{i} \ + \ \textbf{j} \ + \ \textbf{k} }\).
\(\\[1pt]\)
\({\small (\textrm{a}).\hspace{0.8em}}\) Using a scalar product, show that angle \(ABC\) is a right angle.
\(\\[1pt]\)
\({\small (\textrm{b}).\hspace{0.8em}}\) Show that triangle \(ABC\) is isosceles.
\(\\[1pt]\)
\({\small (\textrm{c}).\hspace{0.8em}}\) Find the exact length of the perpendicular from \(O\) to the line through \(B\) and \(C\).
Check out my solution here:
\(\\[1pt]\)
https://www.instagram.com/p/Cbv8Weostsv/ \(\\[1pt]\)
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\({\small 12.\enspace}\)
9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 7 \(\\[1pt]\)
Two lines have equations \( \ {\small \vec{r}}=\begin{pmatrix}
1 \\[1pt]
3 \\[1pt]
2
\end{pmatrix}
+ s \begin{pmatrix}
2 \\[1pt]
-1 \\[1pt]
3
\end{pmatrix} \) and \( \ {\small \vec{r}}=\begin{pmatrix}
2 \\[1pt]
1 \\[1pt]
4
\end{pmatrix}
+ t \begin{pmatrix}
1 \\[1pt]
-1 \\[1pt]
4
\end{pmatrix} \)
\(\\[1pt]\)
\({\small (\textrm{a}).\hspace{0.8em}}\) Show that the lines are skew.
\(\\[1pt]\)
\({\small (\textrm{b}).\hspace{0.8em}}\) Find the acute angle between the directions of the two lines.
Check out my solution here:
\(\\[1pt]\)
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\({\small 13.\enspace}\)
9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 9 \(\\[1pt]\)
The quadrilateral \(ABCD\) is a trapezium in which \(AB\) and \(DC\) are parallel. With respect to the origin \(O\), the position vectors of \(A\), \(B\) and \(C\) are given by
\(\\[1pt]\)
\({\small \ \overrightarrow{OA} = -\textbf{i} \ + \ 2\textbf{j} \ + \ 3\textbf{k} \ }\), \({\small \ \overrightarrow{OB} = \textbf{i} \ + \ 3\textbf{j} \ + \ \textbf{k} \ }\) and \({\small \ \overrightarrow{OC} = 2\textbf{i} \ + \ 2\textbf{j} \ – \ 3\textbf{k} }\).
\(\\[1pt]\)
\({\small (\textrm{a}).\hspace{0.8em}}\) Given that \({\small \ \overrightarrow{DC} = 3 \overrightarrow{AB} }\), find the position vector of \(D\).
\(\\[1pt]\)
\({\small (\textrm{b}).\hspace{0.8em}}\) State a vector equation for the line through \(A\) and \(B\).
\(\\[1pt]\)
\({\small (\textrm{c}).\hspace{0.8em}}\) Find the distance between the parallel sides and hence find the area of the trapezium.
Check out my solution here:
\(\\[1pt]\)
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\({\small 14.\enspace}\)
9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 11 \(\\[1pt]\)
With respect to the origin \(O\), the points \(A\) and \(B\) have position vectors given by
\(\\[1pt]\)
\({\small \ \overrightarrow{OA} = 2\textbf{i} \ – \ \textbf{j} \ }\) and \({\small \ \overrightarrow{OB} = \textbf{j} \ – \ 2\textbf{k} }\).
\(\\[1pt]\)
\({\small (\textrm{a}).\hspace{0.8em}}\) Show that \(OA = OB\) and use a scalar product to calculate angle \(AOB\) in degrees.
\(\\[1pt]\)
The midpoint of \(AB\) is \(M\). The point \(P\) on the line through \(O\) and \(M\) is such that \( PA : OA = \sqrt{7} : 1\).
\(\\[1pt]\)
\({\small (\textrm{b}).\hspace{0.8em}}\) Find the possible position vectors of \(P\).
Check out my solution here:
\(\\[1pt]\)
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\({\small 15.\enspace}\)
9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 8 \(\\[1pt]\)
With respect to the origin \(O\), the points \(A\) and \(B\) have position vectors given by
\(\\[1pt]\)
\( \ {\small \overrightarrow{OA}}=\begin{pmatrix}
1 \\[1pt]
2 \\[1pt]
1
\end{pmatrix} \) and \( \ {\small \overrightarrow{OB}}=\begin{pmatrix}
3 \\[1pt]
1 \\[1pt]
-2
\end{pmatrix} \). The line \(l\) has equation \( \ {\small \vec{r}}=\begin{pmatrix}
2 \\[1pt]
3 \\[1pt]
1
\end{pmatrix}
+ \lambda \begin{pmatrix}
1 \\[1pt]
-2 \\[1pt]
1
\end{pmatrix} \).
\(\\[1pt]\)
\({\small (\textrm{a}).\hspace{0.8em}}\) Find the acute angle between the directions of \(AB\) and \(l\).
\(\\[1pt]\)
\({\small (\textrm{b}).\hspace{0.8em}}\) Find the position vector of the point \(P\) on \(l\) such that \(AP = BP\).
Check out my solution here:
\(\\[1pt]\)
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\({\small 16.\enspace}\)
9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 10 \(\\[1pt]\)
The points \(A\) and \(B\) have position vectors \({\small \ 2\textbf{i} \ + \ \textbf{j} \ + \ \textbf{k} \ }\) and \({\small \ \textbf{i} \ – \ 2\textbf{j} \ + \ 2\textbf{k} \ }\) respectively.
\(\\[1pt]\)
The line \(l\) has vector equation \({\small \ \vec{r} \ = \ \textbf{i} \ + \ 2\textbf{j} \ – \ 3\textbf{k} \ + \ \mu ( \textbf{i} \ – \ 3\textbf{j} \ – \ 2\textbf{k} ) }\).
\(\\[1pt]\)
\({\small (\textrm{a}).\hspace{0.8em}}\) Find a vector equation for the line through \(A\) and \(B\).
\(\\[1pt]\)
\({\small (\textrm{b}).\hspace{0.8em}}\) Find the acute angle between the directions of \(AB\) and \(l\), giving your answer in degrees.
\(\\[1pt]\)
\({\small (\textrm{c}).\hspace{0.8em}}\) Show that the line through \(A\) and \(B\) does not intersect the line \(l\).
Check out my solution here:
\(\\[1pt]\)
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\({\small 17.\enspace}\)
9709/12/O/N/19 – Paper 12 Oct Nov 2019 Pure Maths 1 No 7 \(\\[1pt]\)
\(\\[1pt]\)
The base
OABC and the upper surface
DEFG are identical horizontal rectangles. The parallelograms
OAED and
CBFG both lie in vertical planes. Points
P and
Q are the mid-points of
OD and
GF respectively. Unit vectors
i and
j are parallel to \({\small \ \overrightarrow{OA} \ }\) and \({\small \ \overrightarrow{OC} \ }\) respectively and the unit vector
k is vertically upwards. The position vectors of
A,
C and
D are given by \({\small \ \overrightarrow{OA} \ = \ 6\textbf{i} }\), \({\small \ \overrightarrow{OC} \ = \ 8\textbf{j} }\) and \({\small \ \overrightarrow{OD} \ = \ 2\textbf{i} \ + \ 10\textbf{k} }\).
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}\) Express each of the vectors \({\small \ \overrightarrow{PB} \ }\) and \({\small \ \overrightarrow{PQ} \ }\) in terms of
i,
j and
k.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}\) Determine whether
P is nearer to
Q or to
B.
\(\\[1pt]\)
\({\small\hspace{1.2em}(\textrm{iii}).\hspace{0.5em}}\) Use a scalar product to find angle
BPQ.
https://www.instagram.com/p/Cjg_jFiMxBx/ \(\\[1pt]\)
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\({\small 2. \enspace}\) The point P has position vector \({\small 3\textbf{i} \ – \ 2\textbf{j} \ + \ \textbf{k} }\). The line l has equation \( {\small \vec{r} = 4\textbf{i} + 2\textbf{j} + 5\textbf{k} + \mu(\textbf{i} + 2\textbf{j} + 3\textbf{k} ) }\).
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Find the length of the perpendicular from P to l, giving your answer correct to 3 significant figures.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) Find the equation of the plane containing l and P, giving your answer in the form \({\small ax + by + cz = d}\).
\(\\[1pt]\)
\({\small 3. \enspace}\) Two lines l and m have equations \( {\small \vec{r} = 2\textbf{i} \ – \textbf{j} + \textbf{k} + s(2\textbf{i} + 3\textbf{j} \ – \textbf{k} ) }\) and \( {\small \vec{r} = \textbf{i} + 3\textbf{j} + 4\textbf{k} + t(\textbf{i} + 2\textbf{j} + \textbf{k} ) }\) respectively.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Show that the lines are skew.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) A plane p is parallel to the lines l and m. Find a vector that is normal to p.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(iii)}.\hspace{0.5em}}\) Given that p is equidistant from the lines l and m, find the equation of p. Give your answer in
the form \({\small ax + by + cz = d}\).
\(\\[1pt]\)
\({\small 4. \enspace}\) The line l has equation \( {\small \vec{r} = 4\textbf{i} + 3\textbf{j} \ – \textbf{k} + \mu(\textbf{i} + 2\textbf{j} \ – 2\textbf{k} ) }\). The plane p has equation \({\small 2x \ – 3y \ – z = 4}\).
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Find the position vector of the point of intersection of l and p.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) Find the acute angle between l and p.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(iii)}.\hspace{0.5em}}\) A second plane q is parallel to l, perpendicular to p and contains the point with position vector
4j − k. Find the equation of q, giving your answer in the form \({\small ax + by + cz = d}\).
\(\\[1pt]\)
\({\small 5. \enspace}\) Two planes p and q have equations \({\small x + y + 3z = 8}\) and \({\small 2x \ – 2y + z = 3}\) respectively.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Calculate the acute angle between the planes p and q.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) The point A on the line of intersection of p and q has y-coordinate equal to 2. Find the equation of the plane which contains the point A and is perpendicular to both the planes p and q. Give your answer in the form \({\small ax + by + cz = d}\).
\(\\[1pt]\)
\({\small 6. \enspace}\) The equations of two lines l and m are \( {\small \vec{r} = 3\textbf{i} \ – \textbf{j} \ – 2\textbf{k} + \lambda(\ -\textbf{i} + \textbf{j} + 4\textbf{k} ) }\) and \( {\small \vec{r} = 4\textbf{i} + 4\textbf{j} \ – 3\textbf{k} + \mu(2\textbf{i} + \textbf{j} \ – 2\textbf{k} ) }\) respectively.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Show that the lines do not intersect.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) ) Calculate the acute angle between the directions of the lines.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(iii)}.\hspace{0.5em}}\) Find the equation of the plane which passes through the point (3, −2, −1) and which is parallel to both l and m. Give your answer in the form \({\small ax + by + cz = d}\).
\(\\[1pt]\)
\({\small 7. \enspace}\) The points A and B have position vectors, relative to the origin O, given by \({\small \overrightarrow{OA} \ = \ \textbf{i} \ + \ 2\textbf{j} \ + \ 3\textbf{k} }\) and \({\small \overrightarrow{OB} \ = \ 2\textbf{i} \ + \ \textbf{j} \ + \ 3\ \textbf{k} }\). The line l has vector equation \( {\small \vec{r} = (1 \ – \ 2t)\textbf{i} + (5 \ + \ t)\textbf{j} \ + (2 \ – \ t)\textbf{k} }\).
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Show that l does not intersect the line passing through A and B.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) ) The point P lies on l and is such that angle PAB is equal to \({\small 60^{\large{\circ}} }\). Given that the position vector of P is \({\small (1 \ – \ 2t)\textbf{i} + (5 \ + \ t)\textbf{j} \ + (2 \ – \ t)\textbf{k} }\), show that \({\small 3t^2 \ + \ 7t \ + \ 2 \ = \ 0 }\). Hence find the only possible position vector of P.
\(\\[1pt]\)
\({\small 8. \enspace}\) The plane p has equation \({\small 3x + 2y + 4z = 13}\). A second plane q is perpendicular to p and has equation \({\small ax + y + z = 4}\), where a is a constant.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Find the value of a.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) The line with equation \( {\small \vec{r} = \textbf{j} \ – \textbf{k} + \lambda( \textbf{i} + 2\textbf{j} + 2\textbf{k} ) }\) meets the plane p at the point A and the plane q at the point B. Find the length of AB.
\(\\[1pt]\)
\({\small 9. \enspace}\) The lines \({\small l_{1}}\) and \({\small l_{2}}\) have equations \( {\small \vec{r} = \textbf{i} + 2\textbf{j} + 3\textbf{k} + \lambda(a\textbf{i} + 4\textbf{j} + 3 \textbf{k} ) }\) and \( {\small \vec{r} = 4\textbf{i} \ – \textbf{k} + \mu(2\textbf{i} + 4\textbf{j} + b\textbf{k} ) }\) respectively. Given that \({\small l_{1}}\) and \({\small l_{2}}\) are parallel:
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Write down the values of a and b.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) Find the shortest distance d between \({\small l_{1}}\) and \({\small l_{2}}\).
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(iii)}.\hspace{0.5em}}\) Find a vector equation of the plane p containing \({\small l_{1}}\) and \({\small l_{2}}\).
\(\\[1pt]\)
\({\small 10.\enspace}\) Two lines \({\small l_{1}}\) and \({\small l_{2}}\) have equations \( {\small \vec{r} = \ -8\textbf{i} + 12\textbf{j} + 16\textbf{k} + \lambda(\textbf{i} + 7\textbf{j} \ – 2 \textbf{k} ) }\) and \( {\small \vec{r} = 4\textbf{i} + 6 \textbf{j} \ – 8\textbf{k} + \mu(2\textbf{i} \ – \textbf{j} + 2\textbf{k} ) }\) respectively.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Show that \({\small l_{1}}\) and \({\small l_{2}}\) are skew lines.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) The points P and Q lie on \({\small l_{1}}\) and \({\small l_{2}}\) respectively such that PQ is perpendicular to both \({\small l_{1}}\) and \({\small l_{2}}\). Show that \({\small \overrightarrow{PQ} \ = \ 16\textbf{i} \ – \ 8\textbf{j} \ – \ 20\textbf{k} }\).
\(\\[1pt]\)
As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .
