The foot of the perpendicular from the point | vedclass

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(A) Let the coordinates of any point $P$ on the line be $(3\lambda + 6, 2\lambda + 7, -2\lambda + 7)$.Given point is $A(1, 2, 3)$.The direction ratios

Vedclass · Accepted Answer

$(3, 5, 9)$

Vedclass · Answer

(A) Let the coordinates of any point $P$ on the line be $(3\lambda + 6, 2\lambda + 7, -2\lambda + 7)$.Given point is $A(1, 2, 3)$.The direction ratios of the line $AP$ are $(3\lambda + 6 - 1, 2\lambda + 7 - 2, -2\lambda + 7 - 3) = (3\lambda + 5, 2\lambda + 5, -2\lambda + 4)$.Since $AP$ is perpendicular to the given line with direction ratios $(3, 2, -2)$,the dot product of their direction ratios must be zero:$3(3\lambda + 5) + 2(2\lambda + 5) - 2(-2\lambda + 4) = 0$$9\lambda + 15 + 4\lambda + 10 + 4\lambda - 8 = 0$$17\lambda + 17 = 0$$\lambda = -1$Substituting $\lambda = -1$ into the coordinates of $P$:$P = (3(-1) + 6, 2(-1) + 7, -2(-1) + 7) = (3, 5, 9)$.

The foot of the perpendicular from the point $(1, 2, 3)$ on the line $\vec{r} = (6 \hat{i} + 7 \hat{j} + 7 \hat{k}) + \lambda(3 \hat{i} + 2 \hat{j} - 2 \hat{k})$ has the coordinates:

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