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

The correct option is D $3\sqrt{21},(1,2,8)$


Let M be the foot of perpendicular from (7,14,5) to the given plane, then PM is normal to the plane. So, its d.r.'s are 2, 4, -1. Since PM passes through P(7,14,5) and has d.r.'s 2,4,-1.
Therefore, its equation is $\frac{x-7}{2}=\frac{y-14}{4}=\frac{z-5}{-1}=r$,
$\Rightarrow x=2r+7,y=4r+14,z=-r+5$
Let co-ordinates of M be (2r+7, 4r+14, -r+5)
Since M lies on the plane 2x+4y-z=2, therefore 2(2r+7)+4(4r+14)-(-r+5)=2 $\Rightarrow r=-3$
So, co-ordinates of foot of perpendicular are M(1,2,8)
Noe, PM = Length of perpendicular from P
$=\sqrt{(1-7{)}^2+(2-14{)}^2+(8-5{)}^2}=3\sqrt{21}$