题目

Problem

Relative to a fixed origin $O$

the point $A$ has coordinates $(-10,5,-4)$
the point $B$ has coordinates $(-6,4,-1)$

The straight line $l_1$ passes through $A$ and $B$ .

(a) Find a vector equation for $l_1$ .

(2)

The line $l_2$ has equation

\mathbf{r}= \begin{pmatrix} 3\\ p\\ q \end{pmatrix} +\mu \begin{pmatrix} 3\\ -4\\ 1 \end{pmatrix}

where $p$ and $q$ are constants and $\mu$ is a scalar parameter.

Given that $l_1$ and $l_2$ intersect at $B$ ,

(b) find the value of $p$ and the value of $q$ .

(3)

The acute angle between $l_1$ and $l_2$ is $\theta$ .

(3)

Given that the point $C$ lies on $l_2$ such that $AC$ is perpendicular to $l_2$ ,

(d) find the exact length of $AC$ , giving your answer as a surd.

(2)

题目中文翻译

相对于固定原点 $O$ ，

点 $A$ 的坐标为 $(-10,5,-4)$ ；
点 $B$ 的坐标为 $(-6,4,-1)$ 。

直线 $l_1$ 经过点 $A$ 和点 $B$ 。

(a) 求 $l_1$ 的一个向量方程。

直线 $l_2$ 的方程为

\mathbf{r}= \begin{pmatrix} 3\\ p\\ q \end{pmatrix} +\mu \begin{pmatrix} 3\\ -4\\ 1 \end{pmatrix}

其中 $p,q$ 为常数， $\mu$ 为标量参数。

已知 $l_1$ 与 $l_2$ 交于点 $B$ ，

(b) 求 $p$ 与 $q$ 的值。

$l_1$ 与 $l_2$ 的锐角为 $\theta$ 。

已知点 $C$ 在 $l_2$ 上，且 $AC$ 垂直于 $l_2$ ，

(d) 求 $AC$ 的精确长度，答案写成根式。

解答

(a)

The direction vector of $l_1$ is

\overrightarrow{AB} = \begin{pmatrix} -6\\ 4\\ -1 \end{pmatrix} - \begin{pmatrix} -10\\ 5\\ -4 \end{pmatrix} = \begin{pmatrix} 4\\ -1\\ 3 \end{pmatrix}

Therefore a vector equation for $l_1$ is

\boxed{ \mathbf r= \begin{pmatrix} -10\\ 5\\ -4 \end{pmatrix} +\lambda \begin{pmatrix} 4\\ -1\\ 3 \end{pmatrix} }

where $\lambda$ is a scalar parameter.

(b)

Since $l_1$ and $l_2$ intersect at $B$ , the point $B(-6,4,-1)$ lies on $l_2$ .

For $l_2$ ,

\mathbf{r}= \begin{pmatrix} 3\\ p\\ q \end{pmatrix} +\mu \begin{pmatrix} 3\\ -4\\ 1 \end{pmatrix}

Using the $x$ -coordinate at $B$ :

3+3\mu=-6

3\mu=-9

and hence

\mu=-3

Now use the $y$ -coordinate:

p-4(-3)=4

Thus

p+12=4

p=-8

Use the $z$ -coordinate:

q+(-3)=-1

q=2

Therefore

\boxed{p=-8,\qquad q=2}

(c)

A direction vector for $l_1$ is

\mathbf a= \begin{pmatrix} 4\\ -1\\ 3 \end{pmatrix}

A direction vector for $l_2$ is

\mathbf b= \begin{pmatrix} 3\\ -4\\ 1 \end{pmatrix}

For the acute angle $\theta$ between the two lines,

\cos\theta=\frac{|\mathbf a\cdot\mathbf b|}{|\mathbf a||\mathbf b|}

Now

\mathbf a\cdot\mathbf b =4(3)+(-1)(-4)+3(1) =12+4+3 =19

Also,

|\mathbf a|=\sqrt{4^2+(-1)^2+3^2}=\sqrt{26}

and

|\mathbf b|=\sqrt{3^2+(-4)^2+1^2}=\sqrt{26}

Therefore

\cos\theta =\frac{19}{\sqrt{26}\sqrt{26}} =\frac{19}{26}

Hence

\boxed{\cos\theta=\frac{19}{26}}

(d)

Since $C$ lies on $l_2$ , and $B$ also lies on $l_2$ , we can write

\overrightarrow{BC}=s \begin{pmatrix} 3\\ -4\\ 1 \end{pmatrix}

C=B+s \begin{pmatrix} 3\\ -4\\ 1 \end{pmatrix}

and therefore

\overrightarrow{AC} =\overrightarrow{AB}+s \begin{pmatrix} 3\\ -4\\ 1 \end{pmatrix}

Now

\overrightarrow{AB}= \begin{pmatrix} 4\\ -1\\ 3 \end{pmatrix}

Thus

\overrightarrow{AC} = \begin{pmatrix} 4+3s\\ -1-4s\\ 3+s \end{pmatrix}

Since $AC$ is perpendicular to $l_2$ ,

\overrightarrow{AC}\cdot \begin{pmatrix} 3\\ -4\\ 1 \end{pmatrix} =0

3(4+3s)-4(-1-4s)+(3+s)=0

Hence

12+9s+4+16s+3+s=0

Therefore

19+26s=0

s=-\frac{19}{26}

Substitute this into $\overrightarrow{AC}$ :

\overrightarrow{AC} = \begin{pmatrix} 4-\frac{57}{26}\\ -1+\frac{76}{26}\\ 3-\frac{19}{26} \end{pmatrix} = \begin{pmatrix} \frac{47}{26}\\ \frac{25}{13}\\ \frac{59}{26} \end{pmatrix}

Thus

\begin{aligned} AC^2 &=\left(\frac{47}{26}\right)^2+\left(\frac{25}{13}\right)^2+\left(\frac{59}{26}\right)^2 \\ &=\frac{47^2+50^2+59^2}{26^2} \\ &=\frac{8190}{676} \\ &=\frac{315}{26} \end{aligned}

AC=\sqrt{\frac{315}{26}}

Rationalising this into the form used for a surd length,

AC=\frac{\sqrt{8190}}{26} =\frac{3\sqrt{910}}{26}

Therefore

\boxed{AC=\frac{3\sqrt{910}}{26}}

IAL 2024 Oct Q8

题目

解答

(a)

(b)

(c)

(d)