Vector Equations of Lines & The Scalar Product (Cambridge (CIE) A Level Maths: Pure 3): Exam Questions

Find the equation of each of these lines in vector form.

The line joining $\left(4,1\right)$ to $\left(7,6\right)$.

Find, in vector form, the equation of the line passing through $\left(2,5\right)$ in the same direction as $2\mathbf{i}+\mathbf{j}$.

Find the equation of each of these lines in vector form.

(i) Through $\left(3,1,-2\right)$ in the direction $\left(\begin{array}{c}0\\ 1\\ 5\end{array}\right)$

(ii) Through $\left(2,0,7\right)$ and $\left(5,2,-3\right)$.

Show that the point $\left(-4,-4,27\right)$ lies on the line found in part (a)(ii).

$A$ and $B$ are the points on the line $\mathbf{r}=\left(\begin{array}{c}3\\ -2\\ 5\end{array}\right)+t\left(\begin{array}{c}1\\ 3\\ -1\end{array}\right)$ with $t=4$ and $t=7$ respectively.

(i) Find the position vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$.

(ii) Find $\left|\overrightarrow{AB}\right|$.

The coordinates of three points are $A\left(1,0,-4\right)$, $B\left(3,-1,6\right)$ and $C\left(-2,5,7\right)$.

Find $\overrightarrow{AB}$ and $\overrightarrow{AC}$.

Calculate $\left|\overrightarrow{AB}\right|$ and $\left|\overrightarrow{AC}\right|$, and the scalar product $\overrightarrow{AB}·\overrightarrow{AC}$.

Hence, find the angle between the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$.

Give your answer in degrees correct to 1 decimal place.

The vertices of triangle $ABC$ are the points with coordinates $A\left(2,4,-3\right)$, $B\left(0,2,1\right)$ and $C\left(-4,-2,-3\right)$.

(i) Show that $\overrightarrow{BA}·\overrightarrow{BC}=0$.

(ii) What does the result of part (i) tell you about the triangle $ABC$?

Two lines $l$ and $m$ have equations $\mathbf{r}=-3\mathbf{i}-6\mathbf{j}+8\mathbf{k}+s\left(\mathbf{i}+\mathbf{j}-\mathbf{k}\right)$ and $\mathbf{r}=8\mathbf{i}-10\mathbf{j}+6\mathbf{k}+t\left(2\mathbf{i}-3\mathbf{j}+\mathbf{k}\right)$ respectively.

Show that the two lines meet and find the position vector of the point of intersection.

Two lines $l$ and $m$ have equations $\mathbf{r}=8\mathbf{i}-5\mathbf{j}-6\mathbf{k}+s\left(-\mathbf{i}+2\mathbf{j}+\mathbf{k}\right)$ and $\mathbf{r}=2\mathbf{i}+2\mathbf{j}-2\mathbf{k}+t\left(2\mathbf{i}+\mathbf{j}+\mathbf{k}\right)$ respectively.

(i) Show that the two lines do not intersect.

(ii) Are the two lines skew? Be sure to justify your answer.

The line $l$ has equation $\mathbf{r}=\left(\begin{array}{c}1\\ -4\\ 10\end{array}\right)+s\left(\begin{array}{c}1\\ -1\\ 3\end{array}\right)$.

Point $N$ is the point on line $l$ such that the line connecting $N$ and the point $P\left(3,4,1\right)$ is perpendicular to $l$.

Find the position vector, $\overrightarrow{ON}$, of point $N$.

Hence find the shortest distance from point $P$ to the line $l$.

In the triangle $ABC$, the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are given by

$\overrightarrow{AB}=-2\mathbf{i}+3\mathbf{j}-\mathbf{k}$

$\overrightarrow{AC}=-5\mathbf{i}-4\mathbf{j}-7\mathbf{k}$

Show that angle $BAC=81.9$°, correct to 1 decimal place.

With respect to the origin $O$, the points $R$ and $S$ have position vectors given by

$\overrightarrow{OR}=-\mathbf{i}+5\mathbf{j}+14\mathbf{k}$

$\overrightarrow{OS}=7\mathbf{i}-2\mathbf{j}+12\mathbf{k}$

Find

(i) the vector $\overrightarrow{RS}$,

(ii) a unit vector parallel to $\overrightarrow{RS}$.

Find the angle that $\overrightarrow{RS}$ makes with the positive $y$-axis.

The vector $\overrightarrow{TU}$ is given by $\overrightarrow{TU}=-24\mathbf{i}+21\mathbf{j}+6\mathbf{k}$.

Explain, giving a reason for your answer, whether the vectors $\overrightarrow{RS}$ and $\overrightarrow{TU}$ are parallel.

Find the equation of each of these lines in vector form.

The line joining $\left(3,-2\right)$ to $\left(-1,5\right)$.

Find, in vector form, the equation of the line passing through $\left(3,1\right)$ parallel to $4\mathbf{i}-12\mathbf{j}$.

Find the equation of each of these lines in vector form.

(i) Through $\left(2,-7,9\right)$ in the same direction as $6\mathbf{i}+3\mathbf{j}-9\mathbf{k}$.

(ii) Through $\left(4,-3,1\right)$ and $\left(-8,3,-5\right)$.

Show that the point $\left(10,-6,5\right)$ does not lie on the line found in part (a)(ii).

The coordinates of three points are $A\left(2,-1,5\right)$, $B\left(4,1,1\right)$ and $C\left(-2,-5,9\right)$.

Find $\overrightarrow{BA}$ and $\overrightarrow{BC}$.

By considering the scalar product $\overrightarrow{BA}·\overrightarrow{BC}$, or otherwise, calculate the angle between $\overrightarrow{BA}$ and $\overrightarrow{BC}$. Give your answer in degrees, accurate to 1 decimal place.

In the triangle $ABC$, the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are given by

$\overrightarrow{AB}=7\mathbf{i}+\mathbf{j}-\mathbf{k}$

$\overrightarrow{AC}=-2\mathbf{i}+5\mathbf{k}$

Show that angle $BAC=119.6$°, correct to 1 decimal place.

Hence find the area of the triangle $ABC$, giving your answer correct to 3 significant figures.

The vertices of triangle $ABC$ are the points with coordinates $A\left(5,-3,0\right)$, $B\left(-2,0,-1\right)$ and $C\left(1,2,1\right)$.

(i) Calculate the scalar products $\overrightarrow{AB}·\overrightarrow{AC}$, $\overrightarrow{BA}·\overrightarrow{BC}$ and $\overrightarrow{CA}·\overrightarrow{CB}$.

(ii) What does the result of part (i) tell you about the triangle $ABC$?

Given that the coordinates of $A$ and $B$ are $\left(-1,7\right)$ and $\left(-7,-5\right)$ respectively, find the equation of each of the following lines in vector form.

(i) The line joining $\left(5,-6\right)$ to the midpoint of $AB$.

(ii) The line passing through $B$, parallel to the line in part (i).

$A$ and $B$ are the points on the line $\mathbf{r}=\left(\begin{array}{c}1\\ -4\\ 3\end{array}\right)+t\left(\begin{array}{c}2\\ a\\ -1\end{array}\right)$ with $t=-1$ and $t=2$ respectively.

Given that the distance from $A$ to $B$ is 9 units, find the possible values of $a$.

With respect to the origin $O$, the point $R$ has position vector $\mathbf{i}+6\mathbf{j}-2\mathbf{k}$ and the point $S$ has position vector $10\mathbf{i}+13\mathbf{k}$.

Find a unit vector parallel to $\overrightarrow{RS}$.

Find the angle that $\overrightarrow{RS}$ makes with the negative $z$-axis.

The vector $\overrightarrow{TU}$ is given by $\overrightarrow{TU}=-12\mathbf{i}+8\mathbf{j}-20\mathbf{k}$.

Explain, giving a reason for your answer, whether the vectors $\overrightarrow{RS}$ and $\overrightarrow{TU}$ are parallel.

Determine whether each of the following pairs of lines intersect, are parallel, or are skew. If the lines intersect, find the coordinates of the point of intersection.

(i) $\mathbf{r}=-7\mathbf{i}-\mathbf{j}+6\mathbf{k}+s\left(\mathbf{i}+\mathbf{j}-\mathbf{k}\right)$ and $\mathbf{r}=2\mathbf{i}+2\mathbf{j}+11\mathbf{k}+t\left(2\mathbf{i}-\mathbf{j}+5\mathbf{k}\right)$

(ii) $\mathbf{r}=3\mathbf{i}-3\mathbf{j}+\mathbf{k}+s\left(\mathbf{i}-2\mathbf{j}+\mathbf{k}\right)$ and $\mathbf{r}=-\mathbf{i}-\mathbf{j}+5\mathbf{k}+t\left(2\mathbf{i}+2\mathbf{j}-3\mathbf{k}\right)$

Find the perpendicular distance of the point $P\left(6,0,3\right)$ to the line $\mathbf{r}=-\mathbf{i}+5\mathbf{j}+2\mathbf{k}+s\left(3\mathbf{i}-6\mathbf{j}-2\mathbf{k}\right)$.

In the parallelogram $ABCD$, $AB$ is parallel to $CD$ and $BC$ is parallel to $AD$.

Given that

$\overrightarrow{AB}=3\mathbf{i}-\mathbf{j}-2\mathbf{k}$

$\overrightarrow{AD}=7\mathbf{i}-\mathbf{j}+4\mathbf{k}$

find the area of the parallelogram, giving your answer correct to 3 significant figures.

Given that the coordinates of $A$, $B$ and $C$ are $\left(6,1,-3\right)$, $\left(-2,-7,5\right)$ and $\left(3,12,-9\right)$ respectively, find the equation of each of the following lines in vector form.

(i) The line through $A$ and $B$.

(ii) The line through $B$, parallel to $\overrightarrow{OC}$.

Determine whether the point $\left(1,5,-4\right)$ lies on either of the lines found in part (a).

The coordinates of three points are $A\left(-3,4,0\right)$, $B\left(2,-2,-3\right)$ and $C\left(-1,-1,-4\right)$.

Calculate the angle between $\overrightarrow{CA}$ and $\overrightarrow{CB}$. Give your answer in degrees, accurate to 1 decimal place.

The vertices of triangle $ABC$ are the points with coordinates $A\left(-2,5,4\right)$, $B\left(3,1,0\right)$ and $C\left(-1,-3,-1\right)$.

Use a vector method to show that $ABC$ is a right-angled triangle.

Determine whether each of the following pairs of lines intersect, are parallel, or are skew. For any lines that intersect, determine the point of intersection.

(i) $\mathbf{r}=3\mathbf{i}-5\mathbf{j}+\mathbf{k}+s\left(\mathbf{i}+3\mathbf{j}-2\mathbf{k}\right)$ and $\mathbf{r}=\mathbf{i}+2\mathbf{j}+2\mathbf{k}+t\left(\mathbf{i}-\mathbf{j}-\mathbf{k}\right)$

(ii) $\mathbf{r}=\mathbf{i}-2\mathbf{j}+3\mathbf{k}+s\left(2\mathbf{i}+6\mathbf{j}-2\mathbf{k}\right)$ and $\mathbf{r}=4\mathbf{i}+7\mathbf{j}+t\left(-5\mathbf{i}-15\mathbf{j}+5\mathbf{k}\right)$

(iii) $\mathbf{r}=-\mathbf{i}-4\mathbf{j}-4\mathbf{k}+s\left(2\mathbf{i}-2\mathbf{j}-4\mathbf{k}\right)$ and $\mathbf{r}=-4\mathbf{i}-\mathbf{j}-2\mathbf{k}+t\left(3\mathbf{i}-3\mathbf{j}+6\mathbf{k}\right)$

Find the coordinates of the point on the line $\mathbf{r}=2\mathbf{i}-12\mathbf{j}+3\mathbf{k}+s\left(\mathbf{i}-6\mathbf{j}+4\mathbf{k}\right)$ that is closest to the point $P\left(2,3,-1\right)$, and hence determine the minimum distance from point $P$ to the line.

The following diagram depicts a large bank vault, the walls, floor and ceiling of which are all perfectly rectangular.

The corner $O$ is taken to be the origin of the coordinate system, and the units on the diagram are all given in metres.

$CE$ is the diagonal running across the room from corner $C$ to corner $E$.

By first finding the vector $\overrightarrow{CE}$, write a vector equation of the line that passes through $C$ and $E$.

A motion detector is located at point $O$, and is set to sound an alarm if anything moves within 5 metres of it.

If a small insect flies in a straight line from point $C$ to point $E$, will the alarm be triggered? You must provide clear mathematical workings to justify your answer.

The vector $\overrightarrow{RS}=x\mathbf{i}-9\mathbf{j}+3\mathbf{k}$ makes an angle $\theta$ with the positive $x$-axis.

Show that $x^2=\frac{90}{{\text{tan}}^2\theta }$.

Given that $\theta$ is acute and that $\text{cos}\theta =\frac{4}{5}$, find a unit vector parallel to $\overrightarrow{RS}$.

The coordinates of three points are $A\left(-2,3,-5\right)$, $B\left(2,-5,-9\right)$ and $C\left(-10,-1,-1\right)$.

The point $M$ is the midpoint of $AB$, and the point $N$ lies on $BC$.

Given that $\left|\overrightarrow{BN}\right|=3\left|\overrightarrow{NC}\right|$, find the equation of the line through points $M$ and $N$ in vector form.