question_answer

The distance of the z-axis from the image of the point \[M(2,-3,3)\]in the plane \[x-2y-z+1=0,\] is

A)  1                                

B)  \[\sqrt{2}\]

C)  2                                

D)  4

Correct Answer: A

Solution :

We have \[n=\vec{i}-2\hat{j}-\hat{k}=\] normal vector of plane \[\therefore \] Equation of line through A and perpendicular to plane is \[\frac{x-2}{1}=\frac{y+3}{-2}\,=\frac{z-3}{-1}\,=\lambda \] \[\therefore \,\,x=\lambda +2,\,\,y=-(2\lambda +3)\,,\,\,z=3-\lambda \] \[\lambda +2+2(2\lambda +3)\,-(3-\lambda )\,+1=0\] \[\therefore \,\,6\lambda +6=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\lambda =-1\] \[\therefore \,\,Q(1,\,-1,\,4)\] Now, \[{{a}_{1}}+2=2\,\,\Rightarrow \,{{a}_{1}}=0\] \[{{a}_{2}}-3=-2\,\,\,\,\,\,\,\,\,\Rightarrow \,\,{{a}_{2}}=1\] \[{{a}_{3}}+3=8\,\,\,\,\Rightarrow \,\,{{a}_{3}}=5\] \[\Rightarrow \,\]Image (0, 1, 5) Obviously distance of image of the point from z-axis is 1.