Question

Find the equation of the plane passing through the point $A(3,-2,1)$and perpendicular to the vector $4i+7j-4k,$ if PM be the perpendicular from the point P $(1,2,-1)$ to this plane, find its length.

• A. $\frac{28}{9}$ units.
• B. $\frac{14}{9}$ units.
• C. $\frac{14}{3}$ units.
• D. None

Answer 1

The correct option is A $\frac{28}{9}$ units.

Let $A=a=3i-2j+k$ be the point through which the plane passes. Let us choose $L=r=(xi+yj+zk)$ any point $(x,y,z)$ on this plane.

Therefore,

$\overrightarrow{AL}=r-a$or $\overrightarrow{AL}(x-3)i+(y+2)j+(2-1)k.$ Since $4i+7j-4k=n,$ say, is normal to the plane, therefore $(r-a).n=0$

or $(x-3).4+(y+2).7+(z-1)(-4)=0$

or $4x+7y-4z+6=0.$ Again PM is perpendicular from $p(1,2-1)=i+2j-k$ to the plane so that PM is projection of AP along the vector a which is normal to the plane. $\therefore \frac{\overrightarrow{PA}.\overrightarrow{n}}{\left|\overrightarrow{n}\right|}=\frac{(2i-4j+2k).(4i+7j-4k)}{\sqrt{(16+49+16)}}$

$=\frac{8-28-8}{9}=-\frac{28}{9}=\frac{28}{9}$ units.