Mathematics Extension 2

Which of the following is a true statement about the lines
$\ell_1=\begin{pmatrix}-1\\2\\5\end{pmatrix}+\lambda\begin{pmatrix}-1\\3\\1\end{pmatrix},\qquad \ell_2=\begin{pmatrix}3\\-10\\1\end{pmatrix}+\mu\begin{pmatrix}1\\-3\\-1\end{pmatrix}?$

A. $\ell_1$ and $\ell_2$ are the same line.
B. $\ell_1$ and are not parallel and they intersect. C. and are parallel and they do not intersect. D. and are not parallel and they do not intersect.

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

Find a vector equation of the line through the points $A(-3,1,5)$ and $B(0,2,3)$.

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 .

Find the equations of a sphere whose centre is at $(1,0,1)$ and touches the line

Consider the line $\ell_1$ joining

Determine the vector equation of $\ell_1$.

Another line, $\ell_2$, is defined by the vector equation

Find the possible values of $a$ when the angle between $\ell_1$ and $\ell_2$ is $\frac{\pi}{4}$.

$\ell_1$ and $\ell_2$ have a unique point of intersection when $a\ne2$. Find the point of intersection in terms of $a$.

Let $A$, $B$, $P$ be three points in three-dimensional space with $A\ne B$.

If $P$ is on the line $AB$, then there exists a real number $\lambda$ such that $\overrightarrow{AP}=\lambda\overrightarrow{AB}$.

Which of the following is the contrapositive of this statement?

A. If for all real numbers $\lambda$, $\overrightarrow{AP}=\lambda\overrightarrow{AB}$, then is on the line . B. If for all real numbers , , then is not on the line . C. If there exists a real number such that , then is on the line . D. If there exists a real number such that , then is not on the line .

The straight line
$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\4\\3\end{pmatrix}+\lambda\begin{pmatrix}-1\\1\\2\end{pmatrix}$
has symmetric (or Cartesian) equation

A. $x-1=y-4=\frac{z-3}{2}$

B. $x-1=y-4=\frac{z+3}{2}$

C. $1-x=y-4=\frac{z-3}{2}$

D. $1-x=y-4=\frac{z+3}{2}$

Given
$\overrightarrow{OA}=\begin{bmatrix}1\\1\\0\end{bmatrix},\quad \overrightarrow{OB}=\begin{bmatrix}2\\-1\\1\end{bmatrix},\quad \overrightarrow{OC}=\begin{bmatrix}3\\\mu\\\lambda\end{bmatrix},$
where $O$ is the origin, what are the values of $\mu$ and $\lambda$ if $A$, $B$ and $C$ are collinear?

A. $\mu=3,\ \lambda=2$

B. $\mu=-3,\ \lambda=2$

C. $\mu=3,\ \lambda=-2$

D. $\mu=-3,\ \lambda=-2$

The equation of the curve in 3-dimensional space below is

A. $x=t^2$, $y=t$, $z=0$

B. $x=0$, $y=t^2$, $z=t$

C. $x=0$, $y=t$, $z=t^2$

D. $x=t^2$, $y=t^2$, $z=0$

Let $A$ and $B$ be two distinct points in three-dimensional space. Let $M$ be the midpoint of $AB$.

Let $S_1$ be the set of all points $P$ such that $\overrightarrow{AP}\cdot\overrightarrow{BP}=0$.

Let $S_2$ be the set of all points $N$ such that $|\overrightarrow{AN}|=|\overrightarrow{MN}|$.

The intersection of $S_1$ and $S_2$ is the circle $S$.

What is the radius of the circle $S$?

A. $\dfrac{|\overrightarrow{AB}|}{2}$
B. ${\displaystyle }$ C. D.

Find the vector equation for the sphere $x^2+4x+y^2-6y+z^2=12$.

Let $\ell_1$ be the line with equation
$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}-1\\7\end{pmatrix}+\lambda\begin{pmatrix}3\\2\end{pmatrix},\quad \lambda\in\mathbb R.$

The line $\ell_2$ passes through the point $A(-6,5)$ and is parallel to $\ell_1$. Find the equation of the line in the form .

The equation of line $l_2$ is
$\frac{x+3}{k}=\frac{4-y}{6}=z-1.$
Find the value of $k$ for which $l_1$ and $l_2$ intersect.

The point $P$ is defined by
$\overrightarrow{AP}=-6\overrightarrow{AB}-8\overrightarrow{AC}.$

Does $P$ lie on the line $AL$? Justify your answer.

Using part (ii), or otherwise, determine the position of $L$ by showing that
$\overrightarrow{BL}=\frac47\overrightarrow{BC}.$

Show by finding it that the only point of intersection between
$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\10\\5\end{pmatrix}+\lambda\begin{pmatrix}2\\-3\\1\end{pmatrix}$
and the sphere
$(x-5)^2+(y+2)^2+(z-3)^2=38$
is $\begin{pmatrix}7\\1\\8\end{pmatrix}$.

Show by finding it that the point of intersection between the lines
$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\10\\5\end{pmatrix}+\lambda\begin{pmatrix}2\\-3\\1\end{pmatrix}$
and
$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}15\\-1\\6\end{pmatrix}+\delta\begin{pmatrix}4\\-1\\-1\end{pmatrix}$
is $\begin{pmatrix}7\\1\\8\end{pmatrix}$.

Show that the line
$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}15\\-1\\6\end{pmatrix}+\delta\begin{pmatrix}4\\-1\\-1\end{pmatrix}$
is also tangent to the sphere
$(x-5)^2+(y+2)^2+(z-3)^2=38$
by showing that it is perpendicular to the radius of the sphere at $\begin{pmatrix}7\\1\\8\end{pmatrix}$.

The line has vector equation
$\vec r=\lambda\begin{bmatrix}1\\3\\1\end{bmatrix}.$
For $P(x_1,y_1,z_1)$, the point $Q$ on the line is such that $PQ$ is a minimum. Using vector projection methods, show that
$\overrightarrow{OQ}=\frac{x_1+3y_1+z_1}{11}\begin{bmatrix}1\\3\\1\end{bmatrix}.$

It is given that
$\vec c=\begin{bmatrix}1\\4\\0\end{bmatrix}+t\begin{bmatrix}-1\\1\\5\end{bmatrix}$
does not intersect $\vec r$. Point $T$ is on $\vec c$ and $S$ is a point on $\vec r$ such that $TS$ is a minimum. By first expressing $\vec c$ in parametric form, find an expression for $\overrightarrow{TS}$ in terms of $t$.

Which of the following is a vector equation of the line joining $A(4,2,5)$ and $B(-2,2,1)$?

A. $\vec r=\begin{pmatrix}4\\2\\5\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\3\end{pmatrix}$
B. $\vec r=\begin{pmatrix}4\\2\\5\end{pmatrix}+\lambda\begin{pmatrix}3\\0\\2\end{pmatrix}$
C. $\vec r=\begin{pmatrix}1\\2\\3\end{pmatrix}+\lambda\begin{pmatrix}4\\2\\5\end{pmatrix}$
D. $\vec r=\begin{pmatrix}3\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}4\\2\\5\end{pmatrix}$