Q.

If the foot of the perpendicular from the point A(1,4,3) on the plane P:2x+my+nz=4 is 2,72,32then the distance of the point A from the plane P, measured parallel to a line with direction ratios 3, -1, -4, is equal to :

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a

1

b

22

c

26

d

14

answer is B.

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Detailed Solution

Let foot of perpendicular be B2,72,32
D.R.'s of line A B are<-1+2,4-72,3-32>
i.e.,<2,1,3>
AB is r to plane, so D.R.'s of normal of plane will be <2,1,3>
 m=1,n=3
Now, equation of AC is
x+13=y41=z34=λ( say )
Any point that is C on AC is (3λ1,λ+4,4λ+3)
C lies on plane 2x+y+3z=4
 6λ2λ+412λ+9=47λ=7λ=1 C(2,3,1)
Question Image
Thus, AC=(2+1)2+(34)2+(13)2
=9+1+16=26 units  

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If the foot of the perpendicular from the point A(−1,4,3) on the plane P:2x+my+nz=4 is −2,72,32then the distance of the point A from the plane P, measured parallel to a line with direction ratios 3, -1, -4, is equal to :