Find the reflection of the point (1, 6, 3) in | vedclass

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(D) Let $P(1, 6, 3)$ be the given point and $L$ be the foot of the perpendicular drawn from $P$ to the given line.The general point on the given line

Vedclass · Accepted Answer

$(1, 0, 7)$

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(D) Let $P(1, 6, 3)$ be the given point and $L$ be the foot of the perpendicular drawn from $P$ to the given line.The general point on the given line is given by $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = \lambda$.Thus,the coordinates of $L$ are $(\lambda, 2\lambda + 1, 3\lambda + 2)$.The direction ratios of $PL$ are $(\lambda - 1, 2\lambda + 1 - 6, 3\lambda + 2 - 3)$,which simplifies to $(\lambda - 1, 2\lambda - 5, 3\lambda - 1)$.Since $PL$ is perpendicular to the line with direction ratios $(1, 2, 3)$,we have:$1(\lambda - 1) + 2(2\lambda - 5) + 3(3\lambda - 1) = 0$$\lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0$$14\lambda - 14 = 0 \implies \lambda = 1$.Substituting $\lambda = 1$ into the coordinates of $L$,we get $L(1, 3, 5)$.Let $Q(x_1, y_1, z_1)$ be the reflection of $P(1, 6, 3)$ in the line. Then $L$ is the midpoint of $PQ$.$\frac{x_1 + 1}{2} = 1 \implies x_1 = 1$$\frac{y_1 + 6}{2} = 3 \implies y_1 = 0$$\frac{z_1 + 3}{2} = 5 \implies z_1 = 7$Thus,the reflection of the point $P(1, 6, 3)$ in the given line is $(1, 0, 7)$.

Find the reflection of the point $(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$.

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