Question

The length and foot of the perpendicular from the point $(7,14,5)$ to the plane $2x+4y-z=2$, are

• A. $\sqrt{21},(1,2,8)$
• B. $3\sqrt{21},(3,2,8)$
• C. $21\sqrt{3},(1,2,8)$
• D. $3\sqrt{21},(1,2,8)$

Answer 1

Answer: (D)

$3\sqrt{21},(1,2,8)$

Since the line passing through $(7,14,5)$ is perpendicular to the plane $2x+4y-z=2$, its direction ratios are the direction ratios of the normal to the plane.
Thus, the direction ratios of the perpendicular are $2,4$and$-1$.
So the equation of the perpendicular is $\frac{x-7}{2}=\frac{y-14}{4}=\frac{z-5}{-1}$
Any point on this line is $(2\alpha +7,4\alpha +14,-\alpha +5)$
Since the foot of the perpendicular is on the plane, we have $2(2\alpha +7)+4(4\alpha +14)-(-\alpha +5)=2$
This equation yeilds $21\alpha +63=0\Rightarrow \alpha =-3$
$\therefore$ the foot of the perpendicular is $(2\alpha +7,4\alpha +14,-\alpha +5)=(1,2,8)$
Now, the length of the perpendicular is the distance between $(97,14,5)$ and $(1,2,8)$, which is $P=\sqrt{{(7-1)}^2+{(14-2)}^2+{(5-8)}^2}=\sqrt{189}=3\sqrt{21}$