The image point of (5, 4, 6) in the plane x + | vedclass

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(A) Let $Q$ be the image of the point $P(5, 4, 6)$ in the given plane. Then $PQ$ is normal to the plane. The direction ratios of the normal to the pla

Vedclass · Accepted Answer

$(3, 2, 2)$

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(A) Let $Q$ be the image of the point $P(5, 4, 6)$ in the given plane. Then $PQ$ is normal to the plane. The direction ratios of the normal to the plane $x + y + 2z - 15 = 0$ are $1, 1, 2$.Since $PQ$ passes through $P(5, 4, 6)$ and has direction ratios $1, 1, 2$,the equation of line $PQ$ is:$\frac{x - 5}{1} = \frac{y - 4}{1} = \frac{z - 6}{2} = r$Thus,any point on the line $PQ$ is $(r + 5, r + 4, 2r + 6)$.Let $Q$ be $(r + 5, r + 4, 2r + 6)$. The midpoint $R$ of $PQ$ is:$R = \left( \frac{r + 5 + 5}{2}, \frac{r + 4 + 4}{2}, \frac{2r + 6 + 6}{2} \right) = \left( \frac{r + 10}{2}, \frac{r + 8}{2}, r + 6 \right)$Since $R$ lies on the plane $x + y + 2z - 15 = 0$:$\frac{r + 10}{2} + \frac{r + 8}{2} + 2(r + 6) - 15 = 0$$r + 10 + r + 8 + 4r + 24 - 30 = 0$$6r + 12 = 0 \Rightarrow r = -2$Substituting $r = -2$ into the coordinates of $Q$:$Q = (-2 + 5, -2 + 4, 2(-2) + 6) = (3, 2, 2)$Therefore,the image point is $(3, 2, 2)$.

The image point of $(5, 4, 6)$ in the plane $x + y + 2z - 15 = 0$ is

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