MHTCET 26 Apr 2025 Shift 1 Paper

Concept:The foot of the perpendicular from a point to a plane is the midpoint of the point and its mirror image in that plane.Explanation:The normal vector to the plane $x - y - 2z + 1 = 0$ is $\vec n = (1, -1, -2)$.So the line through $P(-1,2,-4)$ perpendicular to the plane has parametric equations:$x = -1 + t$, $y = 2 - t$, $z = -4 - 2t$.Let the foot of the perpendicular be $M(-1+t,\ 2-t,\ -4-2t)$.Since $M$ lies on the plane, substitute into $x - y - 2z + 1 = 0$:$(-1+t) - (2-t) - 2(-4-2t) + 1 = 0$$-1 + t - 2 + t + 8 + 4t + 1 = 0$$6t + 6 = 0 \Rightarrow t = -1$.Therefore $M = (-2, 3, -2)$.Let the image point be $Q(a, b, c)$. Since $M$ is the midpoint of $PQ$:$\frac{-1+a}{2} = -2 \Rightarrow a = -3$$\frac{2+b}{2} = 3 \Rightarrow b = 4$$\frac{-4+c}{2} = -2 \Rightarrow c = 0$Thus $Q = (-3, 4, 0)$.Answer:The mirror image is $(-3, 4, 0)$, i.e., Option B.