Question

Find the image of the point P (3,5,7) in the plane 2x+y+z = 0 ?

• A. (-3, 2,4)
• B. (-9, -1,1)
• C. (-1, -9,1)
• D. None of these

Answer 1

The correct option is B (-9, -1,1)

Given plane is 2x+y+z = 0.
Direction ratios of normal to the plane are 2,1,1
So equation of the normal to plane passing through P will be -
$\frac{x-3}{2}=\frac{y-5}{1}=\frac{z-7}{1}=r$
Let Point R lie on the plane and line.
Then,
R (2r+3, r+5, r+7) ……..(1) (any point on the line)
Point R lies on the plane, so we can satisfy its coordinates in the equation of plane -
2(2r+3)+ (r+5)+ (r+7) = 0
6r+18 = 0
Or r = - 3
And the coordinates of R will be (-3, 2, 4) (from equation 1)
Let the coordinates of point Image Q be - $\alpha ,\beta ,\gamma$
As R is the midpoint of PQ -
$-3=\frac{\alpha +3}{2}$
Or $\alpha =-9$
Similarly,
$2=\frac{\beta +5}{2}$
Or $\beta =-1$
Also, $4=\frac{\gamma +7}{2}$
Or $\gamma =1$
So the coordinates of Point Q or the image of point P will be (-9, -1,1).