Question

The distance between $(0,1,-1)$ and the point of intersection of the line $\frac{x}{2}=\frac{y-1}{3}=\frac{z+1}{4}$ and the plane $x+y+z=9$ is

• A. $\sqrt{29}$
• B. $\frac{1}{2}\sqrt{29}$
• C. $\frac{4}{9}$
• D. $\frac{2}{9}\sqrt{29}$

Answer 1

The correct option is A $\sqrt{29}$

$\begin{array}{c}\frac{x}{2}=\frac{y-1}{3}=\frac{z+1}{4}\\ x=2t,y=1+3t,z=-1+4t\\ x+y+z=9\\ 2t+1+3t-1+4t=9\\ 9t=9\\ t=1\\ \therefore \left(x,y,z\right)=\left(2,4,3\right)\\ dis\tan ce=\sqrt{4+9+16}\\ =\sqrt{29}\\ Hence,\\ optionAiscorrectanswer.\end{array}$