# 问题及解答

## 点到某直线在某平面上的投影的距离

Posted by haifeng on 2015-08-31 18:22:34 last update 2015-08-31 22:12:52 | Edit | Answers (1)

\Gamma:\ \left\{ \begin{aligned} x^2+y^2+z^2 &=6,\\ x+y+z &=0 \end{aligned} \right.

1

Posted by haifeng on 2015-08-31 23:17:34

$z+\frac{5}{4}=(x-1)^2+(y-\frac{1}{2})^2$

$n_1=(2x,2y,2z)\bigr|_{Q}=(2,2,-4),\quad n_2=(1,1,1).$

$n_1\times n_2= \begin{vmatrix} i & j & k\\ 2 & 2 & -4\\ 1 & 1 & 1 \end{vmatrix}=6(1,-1,0).$

$\frac{x-1}{1}=\frac{y-1}{-1}=\frac{z+2}{0},$

\left\{ \begin{aligned} x+y-2&=0,\\ z+2&=0. \end{aligned} \right.

$\ell'$ 是 $\ell$ 在平面 $\pi$ 上的投影, 于是 $\ell'$ 和 $\ell$ 所在的平面与 $\pi$ 垂直. 可以先求出 $v=n_{\pi}\times n_{\ell}$, 然后再求 $v\times n_{\pi}$, 这就是 $\ell'$ 的方向向量.

$v=n_{\pi}\times n_{\ell}= \begin{vmatrix} i & j & k\\ 0 & 0 & 1\\ 1 & -1 & 0 \end{vmatrix}=(1,1,0).$

$v_{\ell'}=v\times n_{\pi}= \begin{vmatrix} i & j & k\\ 1 & 1 & 0\\ 0 & 0 & 1 \end{vmatrix}=(1,-1,0).$

$\frac{x-1}{1}=\frac{y-1}{-1}=\frac{z+\frac{5}{4}}{0},$

\left\{ \begin{aligned} x+y-2&=0,\\ z+\frac{5}{4}&=0. \end{aligned} \right.

$\mathrm{dist}(P,\ell')=\frac{|1+\frac{1}{2}-2|}{\sqrt{1+1}}=\frac{1}{2\sqrt{2}}.$

$\Sigma_{\lambda}:\ (x+y-2)+\lambda(z+2)=0.$

\left\{ \begin{aligned} x+y-2&=0,\\ z+\frac{5}{4}&=0. \end{aligned} \right.