# 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.

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.

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 ! =) .
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EXAMPLE:

$${\small 1.\enspace}$$ 9709/03/SP/20 – Specimen Paper 2020 Pure Maths 3 No 8
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$$\\[1pt]$$
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.
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$${\small\hspace{1.2em}(\textrm{a}).\hspace{0.8em}}$$ Express the vectors $${\small \ \overrightarrow{ON} \ }$$ and $${\small \ \overrightarrow{CM} \ }$$ in terms of i, j and k.
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$${\small\hspace{1.2em}(\textrm{b}).\hspace{0.8em}}$$ Calculate the angle between the directions of $${\small \ \overrightarrow{ON} \ }$$ and $${\small \ \overrightarrow{CM} \ }$$.
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$${\small\hspace{1.2em}(\textrm{c}).\hspace{0.8em}}$$ Show that the length of the perpendicular from M to ON is $${\small \ {\large \frac{3}{5} } \sqrt 5 }$$.

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$${\small 2.\enspace}$$ 9709/11/O/N/16 – Paper 11 Oct Nov 2020 Pure Maths 1 No 9
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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.
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$${\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}$$ Find the possible values of p such that angle $${\small \ OPX \ = \ 90^{\large{\circ}}}$$.
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$${\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}$$ For the case where p = 9, find the unit vector in the direction of $${\small \ \overrightarrow{XP} }$$.
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$${\small\hspace{1.2em}(\textrm{iii}).\hspace{0.5em}}$$ A point Q lies on the face CBFG and is such that XQ is parallel to AG. Find $${\small \ \overrightarrow{XQ} }$$.

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$${\small 3.\enspace}$$ 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).
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$${\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}$$ Show that l does not intersect the line passing through A and B.
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$${\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}$$ 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.

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$${\small 4.\enspace}$$

$$\\[1pt]$$
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 $${\small \overrightarrow{OD} \ = \ 2\textbf{i} \ + \ 3\textbf{j} \ + \ 5\textbf{k} \ }$$.
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$${\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}$$ Find the equation of the plane BCD, giving your answer in the form $${\small ax + by + cz = d}$$.
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$${\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}$$ Calculate the acute angle between the planes BCD and OABC .

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$${\small 5.\enspace}$$ The line l has equation r = i + 2j + 3k + $${\small \mu}$$(2ij – 2k).
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$${\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}$$ The point P has position vector 4i + 2j – 3k. Find the length of the perpendicular from P to l.
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$${\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}$$ 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.

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$${\small 6.\enspace}$$ Two planes have equations $${\small 2x + 3y \ – \ z \ = \ 1 \ }$$ and $$\ {\small x \ – \ 2y + z \ = \ 3}$$.
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$${\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}$$ Find the acute angle between the planes.
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$${\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}$$ Find a vector equation for the line of intersection of the planes.

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$${\small 7.\enspace}$$ 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).
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$${\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}$$ Show that l is parallel to m.
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$${\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}$$ Calculate the acute angle between the planes m and n.
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$${\small\hspace{1.2em}\textrm{(iii)}.\hspace{0.8em}}$$ 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.

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$${\small 8.\enspace}$$ 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.
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$${\small\hspace{1.2em}\textrm{(i)}.\hspace{0.8em}}$$ Find the position vector of A.
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$${\small\hspace{1.2em}\textrm{(ii)}.\hspace{0.8em}}$$ Calculate the acute angle between l and p.
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$${\small\hspace{1.2em}\textrm{(iii)}.\hspace{0.8em}}$$ Find the equation of the line which lies in p and intersects l at right angles.

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$${\small 9.\enspace}$$ 9709/33/M/J/20 – Paper 33 June 2020 Pure Maths 3 No 8
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Relative to the origin $$O$$, the points $$A$$, $$B$$ and $$D$$ have position vectors given by
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$${\small \ \overrightarrow{OA} = \textbf{i} \ + \ 2\textbf{j} \ + \ \textbf{k} \ }$$, $${\small \ \overrightarrow{OB} = 2\textbf{i} \ + \ 5\textbf{j} \ + \ 3\textbf{k} \ }$$ and $${\small \ \overrightarrow{OD} = 3\textbf{i} \ + \ 2\textbf{k} }$$.
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A fourth point $$C$$ is such that $$ABCD$$ is a parallelogram.
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$${\small (\textrm{a}).\hspace{0.8em}}$$ Find the position vector of $$C$$ and verify that the parallelogram is not a rhombus.
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$${\small (\textrm{b}).\hspace{0.8em}}$$ Find angle $$BAD$$, giving your answer in degrees.
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$${\small (\textrm{c}).\hspace{0.8em}}$$ Find the area of the parallelogram correct to 3 significant figures.

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$${\small 10.\enspace}$$ 9709/32/M/J/20 – Paper 32 June 2020 Pure Maths 3 No 10
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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 $$AN = 2NB$$.
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$${\small (\textrm{a}).\hspace{0.8em}}$$ Find a vector equation for the line through $$M$$ and $$N$$.
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The line through $$M$$ and $$N$$ intersects the line through $$O$$ and $$B$$ at the point $$P$$.
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$${\small (\textrm{b}).\hspace{0.8em}}$$ Find the position vector of $$P$$.
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$${\small (\textrm{c}).\hspace{0.8em}}$$ Calculate angle $$OPM$$, giving your answer in degrees.

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$${\small 11.\enspace}$$ 9709/31/M/J/20 – Paper 31 June 2020 Pure Maths 3 No 9
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With respect to the origin $$O$$, the vertices of a triangle $$ABC$$ have position vectors
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$${\small \ \overrightarrow{OA} = 2\textbf{i} \ + \ 5\textbf{k} \ }$$, $${\small \ \overrightarrow{OB} = 3\textbf{i} \ + \ 2\textbf{j} \ + \ 3\textbf{k} \ }$$ 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.
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$${\small (\textrm{b}).\hspace{0.8em}}$$ Show that triangle $$ABC$$ is isosceles.
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$${\small (\textrm{c}).\hspace{0.8em}}$$ Find the exact length of the perpendicular from $$O$$ to the line through $$B$$ and $$C$$.

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$${\small 12.\enspace}$$ 9709/32/F/M/21 – Paper 32 March 2021 Pure Maths 3 No 7
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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 $$\ {\small \vec{r}}=\begin{pmatrix} 2 \\[1pt] 1 \\[1pt] 4 \end{pmatrix} + t \begin{pmatrix} 1 \\[1pt] -1 \\[1pt] 4 \end{pmatrix}$$
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$${\small (\textrm{a}).\hspace{0.8em}}$$ Show that the lines are skew.
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$${\small (\textrm{b}).\hspace{0.8em}}$$ Find the acute angle between the directions of the two lines.

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$${\small 13.\enspace}$$ 9709/33/M/J/21 – Paper 33 June 2021 Pure Maths 3 No 9
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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
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$${\small \ \overrightarrow{OA} = -\textbf{i} \ + \ 2\textbf{j} \ + \ 3\textbf{k} \ }$$, $${\small \ \overrightarrow{OB} = \textbf{i} \ + \ 3\textbf{j} \ + \ \textbf{k} \ }$$ and $${\small \ \overrightarrow{OC} = 2\textbf{i} \ + \ 2\textbf{j} \ – \ 3\textbf{k} }$$.
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$${\small (\textrm{a}).\hspace{0.8em}}$$ Given that $${\small \ \overrightarrow{DC} = 3 \overrightarrow{AB} }$$, find the position vector of $$D$$.
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$${\small (\textrm{b}).\hspace{0.8em}}$$ State a vector equation for the line through $$A$$ and $$B$$.
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$${\small (\textrm{c}).\hspace{0.8em}}$$ Find the distance between the parallel sides and hence find the area of the trapezium.

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$${\small 14.\enspace}$$ 9709/32/M/J/21 – Paper 32 June 2021 Pure Maths 3 No 11
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With respect to the origin $$O$$, the points $$A$$ and $$B$$ have position vectors given by
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$${\small \ \overrightarrow{OA} = 2\textbf{i} \ – \ \textbf{j} \ }$$ 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.
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The midpoint of $$AB$$ is $$M$$. The point $$P$$ on the line through $$O$$ and $$M$$ is such that $$PA : OA = \sqrt{7} : 1$$.
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$${\small (\textrm{b}).\hspace{0.8em}}$$ Find the possible position vectors of $$P$$.

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$${\small 15.\enspace}$$ 9709/31/M/J/21 – Paper 31 June 2021 Pure Maths 3 No 8
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With respect to the origin $$O$$, the points $$A$$ and $$B$$ have position vectors given by
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$$\ {\small \overrightarrow{OA}}=\begin{pmatrix} 1 \\[1pt] 2 \\[1pt] 1 \end{pmatrix}$$ 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 $$l$$.
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$${\small (\textrm{b}).\hspace{0.8em}}$$ Find the position vector of the point $$P$$ on $$l$$ such that $$AP = BP$$.

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$${\small 16.\enspace}$$ 9709/32/F/M/22 – Paper 32 March 2022 Pure Maths 3 No 10
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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.
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The line $$l$$ has vector equation $${\small \ \vec{r} \ = \ \textbf{i} \ + \ 2\textbf{j} \ – \ 3\textbf{k} \ + \ \mu ( \textbf{i} \ – \ 3\textbf{j} \ – \ 2\textbf{k} ) }$$.
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$${\small (\textrm{a}).\hspace{0.8em}}$$ Find a vector equation for the line through $$A$$ and $$B$$.
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$${\small (\textrm{b}).\hspace{0.8em}}$$ Find the acute angle between the directions of $$AB$$ and $$l$$, giving your answer in degrees.
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$${\small (\textrm{c}).\hspace{0.8em}}$$ Show that the line through $$A$$ and $$B$$ does not intersect the line $$l$$.

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$${\small 17.\enspace}$$ 9709/12/O/N/19 – Paper 12 Oct Nov 2019 Pure Maths 1 No 7
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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 $${\small \ \overrightarrow{OD} \ = \ 2\textbf{i} \ + \ 10\textbf{k} }$$.
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$${\small\hspace{1.2em}(\textrm{i}).\hspace{0.7em}}$$ Express each of the vectors $${\small \ \overrightarrow{PB} \ }$$ and $${\small \ \overrightarrow{PQ} \ }$$ in terms of i, j and k.
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$${\small\hspace{1.2em}(\textrm{ii}).\hspace{0.7em}}$$ Determine whether P is nearer to Q or to B.
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$${\small\hspace{1.2em}(\textrm{iii}).\hspace{0.5em}}$$ Use a scalar product to find angle BPQ.

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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 $${\small \vec{r} = 2\textbf{i} \ – \ 2\textbf{j} \ – \ \textbf{k} + \mu(-\textbf{i} + 2\textbf{j} + \textbf{k} ) }$$.
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$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Show that the line passing through A and B does not intersect l.
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$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ Show that the length of the perpendicular from A to l is $${\small {\large\frac{1}{\sqrt{2}} }}$$.
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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 $${\small \vec{r} = 4\textbf{i} + 2\textbf{j} + 5\textbf{k} + \mu(\textbf{i} + 2\textbf{j} + 3\textbf{k} ) }$$.
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$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Find the length of the perpendicular from P to l, giving your answer correct to 3 significant figures.
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$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ Find the equation of the plane containing l and P, giving your answer in the form $${\small ax + by + cz = d}$$.
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$${\small 3. \enspace}$$ 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.
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$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Show that the lines are skew.
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$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ A plane p is parallel to the lines l and m. Find a vector that is normal to p.
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$${\small\hspace{2.8em}\textrm{(iii)}.\hspace{0.5em}}$$ Given that p is equidistant from the lines l and m, find the equation of p. Give your answer in
the form $${\small ax + by + cz = d}$$.
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$${\small 4. \enspace}$$ 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 $${\small 2x \ – 3y \ – z = 4}$$.
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$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Find the position vector of the point of intersection of l and p.
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$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ Find the acute angle between l and p.
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$${\small\hspace{2.8em}\textrm{(iii)}.\hspace{0.5em}}$$ 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 $${\small ax + by + cz = d}$$.
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$${\small 5. \enspace}$$ Two planes p and q have equations $${\small x + y + 3z = 8}$$ and $${\small 2x \ – 2y + z = 3}$$ respectively.
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$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Calculate the acute angle between the planes p and q.
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$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ 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 $${\small ax + by + cz = d}$$.
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$${\small 6. \enspace}$$ 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.
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$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Show that the lines do not intersect.
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$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ ) Calculate the acute angle between the directions of the lines.
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$${\small\hspace{2.8em}\textrm{(iii)}.\hspace{0.5em}}$$ 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 $${\small ax + by + cz = d}$$.
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$${\small 7. \enspace}$$ 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 $${\small \vec{r} = (1 \ – \ 2t)\textbf{i} + (5 \ + \ t)\textbf{j} \ + (2 \ – \ t)\textbf{k} }$$.
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$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Show that l does not intersect the line passing through A and B.
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$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ ) 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.
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$${\small 8. \enspace}$$ 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.
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$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Find the value of a.
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$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ 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.
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$${\small 9. \enspace}$$ 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:
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$${\small\hspace{2.8em}\textrm{(i)}.\hspace{0.7em}}$$ Write down the values of a and b.
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$${\small\hspace{2.8em}\textrm{(ii)}.\hspace{0.7em}}$$ Find the shortest distance d between $${\small l_{1}}$$ and $${\small l_{2}}$$.
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$${\small\hspace{2.8em}\textrm{(iii)}.\hspace{0.5em}}$$ 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 =) .

1. Two lines l and m have equations r⃗ =2i –j+k+s(2i+3j –k)
and r⃗ =i+3j+4k+t(i+2j+k)
respectively.

(i).
Show that the lines are skew.

(ii).
A plane p is parallel to the lines l and m. Find a vector that is normal to p.

(iii).
Given that p is equidistant from the lines l and m, find the equation of p. Give your answer in
the form ax+by+cz=d
.

3. Are these answers for question 4. right?
Question 4. :
4. The line l has equation r⃗ =4i+3j –k+μ(i+2j –2k). The plane p has equation 2x –3y –z=4.

(i). Find the position vector of the point of intersection of l and p.

(ii). Find the acute angle between l and p.

(iii). 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 ax+by+cz=d.

(i) Point of intersection = 2i -1j+ 3k
(ii) acute angle rounded off to: 10.5 degrees
(iii) Equation of plane q = 8x -3y- 7z= -5

1. Hi Estra,
part (i) is correct,
part (ii) i’m getting $${\small{10.3}^{\large{\circ}} }$$ after round off
part (iii) $${ \small q: \ 8x + 3y + 7z = 5 }$$

1. Thanks a great deal for your help Mr. Will,
I’ve rechecked my answers, and now they match up with yours.
Was a silly mistake of reading off the wrong value from the calculator and skewing up signs for part (iii) 🙂

1. Hi Estra,
Glad you’ve understood the topic well.
Keep up the effort and stay healthy

Will

4. are there answers for the part with practice more with these questions?

1. Nope, but you can try and maybe post your answers here, i could check ’em.
Cheers,
Will