question_answer

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

Correct Answer: D

Solution :

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  \[\frac{x-7}{2}=\frac{y-14}{4}=\frac{z-5}{-1}=r,\]

\[\Rightarrow \,\,x=2r+7,\] \[y=4r+14,\] \[z=-r+5\]

Co-ordinates of M be \[(2r+7,\,4r+14,\,-r+5)\]

Since M lies on the plane \[2x+4y-z=2,\]therefore \[2(2r+7)+4(4r+14)-(-r+5)=2\Rightarrow r=-3\]

Co-ordinates of foot of perpendicular are \[M(1,2,8)\].

PM= Length of perpendicular from P \[=\sqrt{{{(1-7)}^{2}}+{{(2-14)}^{2}}+{{(8-5)}^{2}}}=3\sqrt{21}\].