The mirror image of P(2, 4, -1) in the plane | vedclass

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Question

(D) The normal vector to the plane $x - y + 2z - 2 = 0$ is $\vec{n} = (1, -1, 2)$.Let $PQ$ be the line passing through $P(2, 4, -1)$ and perpendicular

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) The normal vector to the plane $x - y + 2z - 2 = 0$ is $\vec{n} = (1, -1, 2)$.Let $PQ$ be the line passing through $P(2, 4, -1)$ and perpendicular to the plane. The equation of line $PQ$ is:$\frac{x - 2}{1} = \frac{y - 4}{-1} = \frac{z + 1}{2} = \lambda$So,any point on the line is $M(\lambda + 2, 4 - \lambda, 2\lambda - 1)$.Since $M$ lies on the plane,it satisfies the plane equation:$(\lambda + 2) - (4 - \lambda) + 2(2\lambda - 1) - 2 = 0$$\lambda + 2 - 4 + \lambda + 4\lambda - 2 - 2 = 0$$6\lambda - 6 = 0 \Rightarrow \lambda = 1$.Substituting $\lambda = 1$,we get the coordinates of $M$ as $(3, 3, 1)$.Since $M$ is the midpoint of $PQ$ and $Q = (a, b, c)$:$\frac{2 + a}{2} = 3 \Rightarrow a = 4$$\frac{4 + b}{2} = 3 \Rightarrow b = 2$$\frac{-1 + c}{2} = 1 \Rightarrow c = 3$Thus,$a + b + c = 4 + 2 + 3 = 9$.

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

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