Discussion questions · Section 1 of 2

Find the vector $\mathbf p$ given that $|\mathbf p|=6$ and $\mathbf p$ is in the direction of $2\mathbf i-3\mathbf j+6\mathbf k$.

Referred to an origin $O$, the position vectors of four non-collinear points $A$, $B$, $C$ and $D$ are $\mathbf a$, $\mathbf b$, $\mathbf c$ and $\mathbf d$ respectively. Given that $\mathbf a-\mathbf b=\mathbf d-\mathbf c$, show that $ABCD$ is a parallelogram.

Given also that $|\mathbf a-\mathbf c|=|\mathbf b-\mathbf d|$, identify the shape of the parallelogram $ABCD$, justify your answer.

The position vectors of vertices $A$, $B$ and $C$, relative to the origin $O$, are $-3\mathbf i+4\mathbf k$, $-2\mathbf i+4\mathbf j-3\mathbf k$ and $11\mathbf i-4\mathbf j-9\mathbf k$ respectively.

(i) Find a unit vector parallel to $\overrightarrow{OA}$.

(ii) A point $P$ divides $AC$ in the ratio $1:3$. Find the position vector of $P$.

(iii) Show that the points $O$, $B$ and $P$ are collinear.

(iv) $OBDC$ forms a parallelogram. Find the position vector of $D$.

Referred to the origin $O$, points $A$ and $B$ have position vectors $\mathbf a=-\mathbf i-6\mathbf j-5\mathbf k$ and $\mathbf b=4\mathbf i+4\mathbf j+10\mathbf k$ respectively. Point $P$ lies on $AB$, between $A$ and $B$, such that $AP:PB=2:3$.

(i) Find the position vector $\overrightarrow{OP}$.

(ii) Show that $\overrightarrow{AB}$ and $\overrightarrow{OP}$ are perpendicular.

The vector $\mathbf c$ is a unit vector in the direction of $\overrightarrow{OP}$.

(iii) Find the value of $|\mathbf a\cdot\mathbf c|$ and give the geometrical interpretation of $|\mathbf a\cdot\mathbf c|$. Hence, or otherwise, find the area of triangle $OAB$.

Given $|\mathbf a|=4$, $|\mathbf b|=5$ and the angle between $\mathbf a$ and $\mathbf b$ is $45^\circ$, find the exact values of the following:

(i) $\mathbf a\cdot\mathbf b$

(ii) $(4\mathbf a+2\mathbf b)\cdot(3\mathbf a-\mathbf b)$

(iii) $|\mathbf a-\mathbf b|^2$

The angle between two unit vectors $\mathbf a$ and $\mathbf b$ is $\theta$.

By expanding the scalar product, show that $(3\mathbf a-\mathbf b)\cdot(\mathbf a+3\mathbf b)=8\cos\theta$.

Given that $\theta=60^\circ$, show that the length of projection of $(3\mathbf a-\mathbf b)$ onto $(\mathbf a+3\mathbf b)$ is $\dfrac4{\sqrt{13}}$.