Question

The mirror image of $P(2, 4, -1)$ in the plane $x - y + 2z - 2 = 0$ is $(a, b, c)$, then the value of $a+b+c$ is

• A. 4
• B. 5
• C. 7
• D. 9

Hint:

To find the mirror image of a point in a plane, first find the foot of the perpendicular from the point to the plane. This foot of the perpendicular will be the midpoint of the line segment connecting the original point and its image.

Answer 1

The Correct Option is A

Step 1: The direction ratios (d.r.s.) of the normal to the plane $x - y + 2z - 2 = 0$ are $(1, -1, 2)$. The equation of the line $PM$ passing through $P(2,4,-1)$ and perpendicular to the plane can be written as:\n\[\frac{x-2}{1} = \frac{y-4}{-1} = \frac{z+1}{2} = \lambda \text{ (say)}\]
Step 2: From the line equation, any point $M$ on the line can be represented as $M(\lambda+2, -\lambda+4, 2\lambda-1)$. This point $M$ is the foot of the perpendicular from $P$ to the plane.\n
Step 3: Since $M$ lies on the plane, its coordinates must satisfy the plane's equation:\n\[(\lambda+2) - (-\lambda+4) + 2(2\lambda-1) - 2 = 0\] \[\lambda+2 + \lambda-4 + 4\lambda-2 - 2 = 0\] \[6\lambda - 6 = 0 \implies \lambda = 1\]
Step 4: Substitute $\lambda=1$ back into the coordinates of $M$ to find its exact location:\n\[M = (1+2, -1+4, 2(1)-1) = (3, 3, 1)\]
Step 5: Let $Q(a,b,c)$ be the mirror image of $P(2,4,-1)$. The point $M$ is the midpoint of the line segment $PQ$. Using the midpoint formula:\n\[\frac{2+a}{2} = 3 \quad ; \quad \frac{4+b}{2} = 3 \quad ; \quad \frac{-1+c}{2} = 1\]
Step 6: Solve these equations for $a, b, c$ and then find their sum.\n\[2+a = 6 \implies a = 4\] \[4+b = 6 \implies b = 2\] \[-1+c = 2 \implies c = 3\] Therefore, the value of $a+b+c = 4+2+3 = 9$.