点 $A$ 和 $B$ 的位置向量分别为 $\mathbf{i}+2\mathbf{j}-2\mathbf{k}$ 和 $2\mathbf{i}-\mathbf{j}+\mathbf{k}$。直线 $l$ 的方程为

(a) 证明 $l$ 与经过 $A$ 和 $B$ 的直线不相交。

(b) 求从 $A$ 向 $l$ 作垂线所得垂足的位置向量。

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

先用 $A,B$ 写出直线方程。若有交点，三个分量必须由同一组参数同时满足；用两个分量求参数后检查第三个。

The vector $\overrightarrow{AB}$ is

Hence the line through $A$ and $B$ has equation

At an intersection with $l$,

The first gives $\lambda=2\mu$. Substituting into the second,

The third components are then $-2+3(2)=4$ and $3+4(1)=7$, which are unequal.

Therefore no parameter pair satisfies all three equations, and $\boxed{l\text{ does not intersect }AB}$.

设垂足 $P$ 在线 $l$ 上，用 $\overrightarrow{AP}$ 与直线方向向量的点积为零求参数。

A general point on $l$ is

Since $A=(1,2,-2)$,

For $P$ to be the foot of the perpendicular,

Thus

giving $\mu=-1$. Hence

也可将 $AP^2$ 写成参数的二次式，配方后求使距离最短的参数，再代回直线。

For $P=(1+2\mu,-1-3\mu,3+4\mu)$,

This is minimised at $\mu=-1$. Equivalently, $\dfrac{\mathrm{d}(AP^2)}{\mathrm{d}\mu}=58\mu+58=0$ gives $\mu=-1$, and the positive quadratic coefficient confirms a minimum. Thus this point is the perpendicular foot. Substituting into $l$,