Let the plane P contain the line 2 x+y-z-3=0=5 x-3 y+4 z+9 and be parallel to the line (x+…

Question

Let the plane $P$ contain the line $2 x+y-z-3=0=5 x-3 y+4 z+9$ and be parallel to the line $\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}$. Then the distance of the point $\mathrm{A}(8,-1,-19)$ from the plane $\mathrm{P}$ measured parallel to ...

Answer: 26

Accepted Answer

Plane $P\equiv {P}_{1}+\lambda {P}_{2}=0$

$\begin{gathered} (2x+y-z-3)+\lambda (5x-3y)+4z+9)=0 \ (5\lambda +2)x+(1-3\lambda )y+(4\lambda -1)z+9\lambda -3=0 \ \overset{	o }{n}\cdot \overset{	o }{b}=0\overset{	o }{b}(2,4,5) \ 2(5\lambda +2)+4(1-3\lambda )+5(4\lambda -1)=0 \ \lambda =-\frac{1}{6} \end{gathered}$

Plane $7x+9y-10z-27=0$

Equation of line $AB$ is

$\frac{x-8}{-3}=\frac{y+1}{4}=\frac{z+19}{12}=\lambda $

Let $B=(8-3\lambda ,-1+4\lambda ,-19+12\lambda )$ lies on plane $P$

$\begin{gathered} 	herefore 7(8-3\lambda )+9(4\lambda -1)-10(12\lambda -19)=27 \ \lambda =2 \ 	herefore B=(2,7,5) \ AB=\sqrt{{6}^{2}+{8}^{2}+{24}^{2}}=26 \end{gathered}$

Solutions · 1