Pure Mathematics Practice: Vectors — Answers

Questions: Vectors Drills.

Question 1

(a)

$$\mathbf{a}\cdot \mathbf{b}=2(1)+(-1)(4)+3(-2)=-8.$$

(b)

$$\mathbf{a}\times \mathbf{b}=\left|\begin{array}{ccc}\mathbf{i}& \mathbf{j}& \mathbf{k}\\ 2& -1& 3\\ 1& 4& -2\end{array}\right|=\left(\begin{array}{c}-10\\ 7\\ 9\end{array}\right).$$

(c)

$$\cos \theta =\frac{\mathbf{a}\cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|}=\frac{-8}{\sqrt{14}\sqrt{21}}.$$

Hence

$$\theta =117.8{\dots }^{\circ }\approx 118^{\circ }.$$

Question 2

(a)

$$\overrightarrow{AB}=\left(\begin{array}{c}3\\ -2\\ 6\end{array}\right),$$

so one equation is

$$\mathbf{r}=\left(\begin{array}{c}1\\ 2\\ -1\end{array}\right)+\lambda \left(\begin{array}{c}3\\ -2\\ 6\end{array}\right).$$

(b) The $x$-coordinate gives $1+3\lambda =7$, so $\lambda =2$. For this same value,

$$2-2(2)=-2,-1+6(2)=11.$$

All three coordinates agree, so $P$ lies on $l$.

Question 3

(a)

$$2(x-1)-y+3(z-2)=0,$$

so

$$2x-y+3z=8.$$

(b)

$$d=\frac{|2(3)-4+3(1)-8|}{\sqrt{2^2+(-1{)}^2+3^2}}=\frac{3}{\sqrt{14}}.$$

Question 4

(a) Substituting the line coordinates into the plane gives

$$(1+2\lambda )+(2-\lambda )+\lambda =8.$$

Hence $\lambda =5/2$, and the intersection point is

$$\left(6,-\frac{1}{2},\frac{5}{2}\right).$$

(b) A direction vector of $l$ is

$$\mathbf{d}=\left(\begin{array}{c}2\\ -1\\ 1\end{array}\right),$$

and a normal to $\Pi$ is

$$\mathbf{n}=\left(\begin{array}{c}1\\ 1\\ 1\end{array}\right).$$

If $\alpha$ is the acute angle between the line and plane,

$$\sin \alpha =\frac{|\mathbf{d}\cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}=\frac{2}{\sqrt{6}\sqrt{3}}=\frac{\sqrt{2}}{3}.$$

Thus

$$\alpha =28.1^{\circ }$$

to 1 decimal place.