Vectors: NJC 2024 Promo Q7

National Junior College2024 PromoPromo Q7●●○ Standard8 marks

What this question tests

Vectors, lines and planes.

Question

1. Given that $\mathbf{p}\cdot\mathbf{q} = 0$ , what can be deduced about the vectors $\mathbf{p}$ and $\mathbf{q}$ ? [2]
2. With respect to the origin on a $xy$ -plane, a fixed point $G$ has position vector $\mathbf{g}$ and a variable point $R$ has position vector $\mathbf{r}$ . Describe the locus of $R$ if $\mathbf{r}\cdot(\mathbf{r} - \mathbf{g}) = 0$ . [2]
With respect to the origin, plane $p$ has an equation given by $\mathbf{r} = \mathbf{i} + \mathbf{j} + \mathbf{k} + \lambda(\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}) + \mu(2\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}), \qquad \lambda, \mu \in \mathbb{R}.$ Points $Q$ and $R$ have position vectors $3\mathbf{i} + 9\mathbf{j} + 5\mathbf{k}$ and $4\mathbf{i} + 11\mathbf{j} + 7\mathbf{k}$ respectively. Find the length of the projection of $\overrightarrow{QR}$ onto $p$ . [4]

Show full worked solution▾

(Vectors I and II)

Suggested Solution

(a) (i) $\mathbf p$ is perpendicular to $\mathbf q$ or $\mathbf p$ is a zero vector or $\mathbf q$ is a zero vector.

(a) (ii) The locus of $R$ is a circle with diameter $OG$ .

(b) $\begin{pmatrix}1\\-2\\2\end{pmatrix}\times \begin{pmatrix}2\\3\\-2\end{pmatrix} =\begin{pmatrix}-2\\6\\7\end{pmatrix}.$ $\overrightarrow{QR}=\overrightarrow{OR}-\overrightarrow{OQ} =\begin{pmatrix}1\\2\\2\end{pmatrix}.$ Projection of $\overrightarrow{QR}$ onto $p$ $\begin{aligned} &=\begin{pmatrix}1\\2\\2\end{pmatrix}\times \frac{\begin{pmatrix}-2\\6\\7\end{pmatrix}}{\sqrt{(-2)^2+6^2+7^2}}\\ &=\frac{\begin{pmatrix}2\\-11\\10\end{pmatrix}}{\sqrt{89}}\\ &=\frac{15}{\sqrt{89}}. \end{aligned}$

Answers

(a)(i) $\mathbf{p}$ is perpendicular to $\mathbf{q}$ , or $\mathbf{p} = \mathbf{0}$ , or $\mathbf{q} = \mathbf{0}$
(a)(ii)a circle with diameter $OG$ (b) $\dfrac{15}{\sqrt{89}}$

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