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

The length and foot of the perpendicular from the point

(7, 14, 5) to the plane 2x+4y−z=2,2x + 4y - z = 2, are

A

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

B

321,(3,2,8)3\sqrt{21},(3,2,8)

C

213,(1,2,8)21\sqrt{3},(1,2,8)

D

321,(1,2,8)3\sqrt{21},(1,2,8)

Answer

321,(1,2,8)3\sqrt{21},(1,2,8)

Explanation

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)P ( 7,14,5 ) and has d.r.'s 2, 4, –1.

Therefore, its equation is x−72=y−144=z−5−1=r\frac { x - 7 } { 2 } = \frac { y - 14 } { 4 } = \frac { z - 5 } { - 1 } = r

(Say)

⇒ x=2r+7x = 2 r + 7 , y=4r+14y = 4 r + 14, z=−r+5z = - r + 5

Let co-ordinates of M be (2r+7,4r+14,−r+5)( 2 r + 7,4 r + 14 , - r + 5 )

Since M lies on the plane 2x+4y−z=22 x + 4 y - z = 2, therefore 2(2r+7)+4(4r+14)−(−r+5)=22 ( 2 r + 7 ) + 4 ( 4 r + 14 ) - ( - r + 5 ) = 2 ⇒ r=−3r = - 3

So, co-ordinates of foot of perpependicular are M(1,2,8)M ( 1,2,8 )

Now, PM = Length of perpendicular from P

= (1−7)2+(2−14)2+(8−5)2=321\sqrt { ( 1 - 7 ) ^ { 2 } + ( 2 - 14 ) ^ { 2 } + ( 8 - 5 ) ^ { 2 } } = 3 \sqrt { 21 }