Find the image of the point (1, 2, 3) in the | vedclass

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(B) Let the given point be $P(1, 2, 3)$ and let $Q(x_1, y_1, z_1)$ be its image in the line. Let $L$ be the foot of the perpendicular from $P$ to the

Vedclass · Accepted Answer

$(5, 8, 15)$

Vedclass · Answer

(B) Let the given point be $P(1, 2, 3)$ and let $Q(x_1, y_1, z_1)$ be its image in the line. Let $L$ be the foot of the perpendicular from $P$ to the line.The general point on the line is given by $L = (6 + 3\lambda, 7 + 2\lambda, 7 - 2\lambda)$.The vector $\vec{PL} = (6 + 3\lambda - 1)\hat{i} + (7 + 2\lambda - 2)\hat{j} + (7 - 2\lambda - 3)\hat{k} = (3\lambda + 5)\hat{i} + (2\lambda + 5)\hat{j} + (4 - 2\lambda)\hat{k}$.Since $\vec{PL}$ is perpendicular to the line vector $\vec{b} = 3\hat{i} + 2\hat{j} - 2\hat{k}$,their dot product is zero:$(3\lambda + 5)(3) + (2\lambda + 5)(2) + (4 - 2\lambda)(-2) = 0$$9\lambda + 15 + 4\lambda + 10 - 8 + 4\lambda = 0$$17\lambda + 17 = 0 \implies \lambda = -1$.Substituting $\lambda = -1$ into the expression for $L$,we get $L = (6 - 3, 7 - 2, 7 + 2) = (3, 5, 9)$.Since $L$ is the midpoint of $PQ$,we have:$\frac{1 + x_1}{2} = 3 \implies x_1 = 5$$\frac{2 + y_1}{2} = 5 \implies y_1 = 8$$\frac{3 + z_1}{2} = 9 \implies z_1 = 15$.Thus,the image of the point $P(1, 2, 3)$ is $(5, 8, 15)$.

Find the image of the point $(1, 2, 3)$ in the line $\vec{r} = (6\hat{i} + 7\hat{j} + 7\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 2\hat{k})$.

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