Mathematics Extension 2

Which of the equations best represent the curve below?

[Diagram: A three-dimensional spiral around the $z$-axis, with radius increasing as $|t|$ increases.]

(A) $\displaystyle \vec r(t)=(\cos t)\vec i+(\sin t)\vec j+t\vec k$

(B) $\displaystyle \vec r(t)=(t\cos t)\vec i+(t\sin t)\vec j+t\vec k$

(C) $\displaystyle \vec r(t)=(t\cos t)\vec i+(t\sin t)\vec j+\frac{1}{t}\vec k$

(D) $\displaystyle \vec r(t)=(\cos t)\vec i+(\sin t)\vec j+\frac{1}{t}\vec k$

In the diagram below, $OABCD$ is a solid figure where $|\overrightarrow{OA}|=|\overrightarrow{OB}|=4$ units and $|\overrightarrow{OD}|=3$ units. The edge $\overrightarrow{OD}$ is vertical, $\overrightarrow{DC}$ is parallel to $\overrightarrow{OB}$ and $|\overrightarrow{DC}|=1$ unit. The base, $OAB$, is horizontal and $\angle AOB=90^\circ$. Unit vectors $\vec i$, $\vec j$ and $\vec k$ are parallel to $\overrightarrow{OA}$, $\overrightarrow{OB}$ and $\overrightarrow{OD}$ respectively. The midpoint of $\overline{AB}$ is $M$ and the point $N$ on $\overline{BC}$ is such that $\overrightarrow{NC}=2\overrightarrow{BN}$.

Express vectors $\overrightarrow{MD}$ and $\overrightarrow{ON}$ in terms of , and .

Consider the line $\ell_1$ joining

Determine the vector equation of $\ell_1$.

What is the unit vector in the same direction as $\vec a=2\mathbf i+\mathbf j-3\mathbf k$?

A. $-\frac1{14}\begin{bmatrix}2\\1\\-3\end{bmatrix}$ B. C. D.

Point $A$ has position vector $8\mathbf i+13\mathbf j-2\mathbf k$, point $B$ has position vector $10\mathbf i+14\mathbf j-4\mathbf k$, and point $C$ has position vector $9\mathbf i+9\mathbf j+6\mathbf k$. Find $\overrightarrow{AB}$.

The points $X$, $Y$ and $Z$ bisect $AB$, $BC$ and $CA$ respectively. Show that
$\overrightarrow{AY}+\overrightarrow{BZ}+\overrightarrow{CX}=\vec0.$

$ABCD$ is a parallelogram. $N$ is on $AD$ so that $\overrightarrow{AN}=\frac13\overrightarrow{AD}$. $M$ is on the diagonal $AC$. Letting $\overrightarrow{AB}=\vec p$ and $\overrightarrow{AD}=\vec q$, express $\overrightarrow{NB}$ in terms of $\vec p$ and $\vec q$.

$ABCD$ is a parallelogram. $N$ is on $AD$ so that $\overrightarrow{AN}=\frac13\overrightarrow{AD}$. $M$ is on the diagonal $AC$. By letting $\overrightarrow{AM}=\lambda\overrightarrow{AC}$ and $\overrightarrow{MB}=\delta\overrightarrow{NB}$, prove that
$\overrightarrow{AM}=\frac14\overrightarrow{AC}.$

Four cubes are placed in a line as shown in the diagram. Which vector is equal to $\overrightarrow{AB}+\overrightarrow{CQ}$?

A. $\overrightarrow{AQ}$
B. $\overrightarrow{CP}$ C. D.

Find the value of $\lambda$ such that $\overrightarrow{HG}=\lambda\overrightarrow{HS}$.

Using part (i), or otherwise, show that
$4\overrightarrow{GH}+\overrightarrow{GS}=\vec0.$

The diagram shows pyramid $ABCDS$, where $ABCD$ is a square. The diagonals of the square bisect each other at $H$. Show that
$\overrightarrow{HA}+\overrightarrow{HB}+\overrightarrow{HC}+\overrightarrow{HD}=\vec0.$

What is the length of the vector $-\mathbf{i}+18\mathbf{j}-6\mathbf{k}$?

A. $5$
B. $19$
C. $25$
D. $361$

Using parts (ii) and (iii), or otherwise, prove that $T$ is the point that divides the interval $PR$ in the ratio $2:1$.

Using the same definitions as in part (i), prove that
$\overrightarrow{OC}=\frac m{m+n}\mathbf a+\frac n{m+n}\mathbf b.$

The point $C$ divides the interval $AB$ so that
$\frac{CB}{AC}=\frac mn.$
The position vectors of $A$ and $B$ are $\mathbf a$ and $\mathbf b$, respectively, as shown in the diagram.

(i) Show that
$\overrightarrow{AC}=\frac n{m+n}(\mathbf b-\mathbf a).$