Vector Line Equation

Vector Line and Plane Equation

Vector Line and Plane Equation

We have learned the cartesian form of a line equation: \( \boxed{ \ y \ = \ mx \ + \ c \ }\).

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.

Using the vector form of a line equation and a plane equation helps us to solve 3D problems much easier than using its cartesian form.

Given any two points, A and B, we can draw the vector \({\small \vec{a}}\) and \({\small \vec{b}}\) from the origin. Then, the line equation of line AB in the vector form can be written as follows:

\( \hspace{3em} \vec{r} \ = \ \vec{a} \ + \ \lambda(\vec{b} \ – \ \vec{a}) \)

\(\\[5pt] {\small \vec{r} \ = \ }\) position vector
: it represents any points along the line AB

\(\\[5pt] {\small \vec{a} \ \ \textrm{or} \ \ \vec{b} \ = \ }\) location vector
: it shows the location vector of any one point along the line AB which can be represented by \({\small \vec{a}}\) or \({\small \vec{b}}\)

\(\\[5pt] {\small \vec{b} \ – \ \vec{a} \ = \ }\) direction vector
: it gives the direction vector of the line AB

\(\\[5pt] {\small \lambda \ }\) is a constant value.

Vector Line Equation

For the 2D shape, the vector form of a plane equation is shown below:

\( \hspace{3em} (\vec{r} \ – \ \vec{a}) \cdot \vec{n} \ = \ 0 \)

\(\\[5pt] {\small \vec{r} \ = \ }\) position vector
: it represents any points on the plane

\(\\[5pt] {\small \vec{a} \ = \ }\) location vector
: it shows the location of a point on the plane which is represented by \({\small \vec{a}}\)

\(\\[5pt] {\small \vec{n} \ = \ }\) normal vector
: it is the vector that gives perpendicular direction to the plane.

Vector Plane Equation

Note that we can find the cartesian form of a plane equation from its vector form,

\(\\[8pt] \hspace{3em} (\vec{r} \ – \ \vec{a}) \cdot \vec{n} \ \hspace{0.7em} \ = \ 0 \)
\(\\[8pt] \hspace{3em} \vec{r} \ \cdot \vec{n} \ – \ \vec{a} \ \cdot \vec{n} \ = \ 0 \)
\(\\[8pt] \hspace{3em} \vec{r} \ \cdot \vec{n} \hspace{3.3em} \ = \ \vec{a} \ \cdot \vec{n} \)

Let \( \ {\small \vec{r}}=\begin{pmatrix}
x \\[1pt]
y \\[1pt]
z
\end{pmatrix}
\ \) and \( \ {\small \vec{n}}=\begin{pmatrix}
a \\[1pt]
b \\[1pt]
c
\end{pmatrix} \)

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:

\( \hspace{3em} ax \ + \ by \ + \ cz \ = \ d \)

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 ! =) .
\(\\[1pt]\)


EXAMPLE:

\({\small 1.\enspace}\) 9709/03/SP/20 – Specimen Paper 2020 Pure Maths 3 No 8
\(\\[1pt]\)
9709/03/SP/20 – Specimen Paper 2020 Pure Maths 3 No 8
\(\\[1pt]\)
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 \( {\small \ {\large \frac{3}{5} } \sqrt 5 }\).

\(\\[1pt]\)
\({\small 2.\enspace}\) 9709/11/O/N/16 – Paper 11 Oct Nov 2020 Pure Maths 1 No 9
\(\\[1pt]\)
9709/11/O/N/16 – Paper 11 Oct Nov 2020 Pure Maths 1 No 9
\(\\[1pt]\)
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 \({\small \ \overrightarrow{XQ} }\).

\(\\[1pt]\)
\({\small 3.\enspace}\) The points A and B have position vectors i + 2jk and 3i + j + k respectively. The line l has equation r = 2i + j + k + \({\small \mu}\)(i + j + 2k).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\) 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.

\(\\[1pt]\)
\({\small 4.\enspace}\)
Vector Line and Plane Equation Example 2
\(\\[1pt]\)
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 .

\(\\[1pt]\)
\({\small 5.\enspace}\) The line l has equation r = i + 2j + 3k + \({\small \mu}\)(2ij – 2k).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\) The point P has position vector 4i + 2j – 3k. 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.

\(\\[1pt]\)
\({\small 6.\enspace}\) 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.

\(\\[1pt]\)
\({\small 7.\enspace}\) 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 = 4i + 2j + k + \({\small \lambda}\)(i + j + 2k).
\(\\[1pt]\)
\({\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}\) 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.

\(\\[1pt]\)
\({\small 8.\enspace}\) The line l has equation r = 5i – 3jk + \({\small \lambda}\)(i – 2j + k). The plane p has equation (ri – 2j) . (3i + 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.

\(\\[1pt]\)


PRACTICE MORE WITH THESE QUESTIONS BELOW!

\({\small 1.\enspace}\) The points A and B have position vectors, relative to the origin O, given by \({\small \overrightarrow{OA} \ = \ \textbf{i} \ + \ \textbf{j} \ + \ \textbf{k} }\) and \({\small \overrightarrow{OB} \ = \ 2\textbf{i} \ + \ 3\ \textbf{k} }\). The line l has vector equation \( {\small \vec{r} = 2\textbf{i} \ – \ 2\textbf{j} \ – \ \textbf{k} + \mu(-\textbf{i} + 2\textbf{j} + \textbf{k} ) }\).
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}\) Show that the line passing through A and B does not intersect l.
\(\\[1pt]\)
\({\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}\) Show that the length of the perpendicular from A to l is \({\small {\large\frac{1}{\sqrt{2}} }}\).
\(\\[1pt]\)

\({\small 2. \enspace}\) 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
4jk. 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 =) .