The coordinates of the foot of the perpendicu | vedclass

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(D) Let $A(0, 0, 0)$ be the origin and $B(-9, 4, 5)$ and $C(10, 0, -1)$ be the given points. Let $D$ be the foot of the perpendicular from $A$ to the

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(D) Let $A(0, 0, 0)$ be the origin and $B(-9, 4, 5)$ and $C(10, 0, -1)$ be the given points. Let $D$ be the foot of the perpendicular from $A$ to the line $BC$. Let $D$ divide $BC$ in the ratio $\lambda : 1$.The coordinates of $D$ are given by the section formula: $D = \left( \frac{10\lambda - 9}{\lambda + 1}, \frac{0\lambda + 4}{\lambda + 1}, \frac{-1\lambda + 5}{\lambda + 1} ight) = \left( \frac{10\lambda - 9}{\lambda + 1}, \frac{4}{\lambda + 1}, \frac{5 - \lambda}{\lambda + 1} ight)$.The direction ratios of the line $BC$ are $(10 - (-9), 0 - 4, -1 - 5) = (19, -4, -6)$.Since $AD \perp BC$,the dot product of vector $\vec{AD}$ and the direction vector of $BC$ must be zero.$vec{AD} = \left( \frac{10\lambda - 9}{\lambda + 1}, \frac{4}{\lambda + 1}, \frac{5 - \lambda}{\lambda + 1} ight)$.$19 \left( \frac{10\lambda - 9}{\lambda + 1} ight) - 4 \left( \frac{4}{\lambda + 1} ight) - 6 \left( \frac{5 - \lambda}{\lambda + 1} ight) = 0$.$19(10\lambda - 9) - 16 - 6(5 - \lambda) = 0$.$190\lambda - 171 - 16 - 30 + 6\lambda = 0$.$196\lambda - 217 = 0 \Rightarrow \lambda = \frac{217}{196} = \frac{31}{28}$.Substituting $\lambda = \frac{31}{28}$ into the coordinates of $D$:$x = \frac{10(\frac{31}{28}) - 9}{\frac{31}{28} + 1} = \frac{310 - 252}{31 + 28} = \frac{58}{59}$.$y = \frac{4}{\frac{31}{28} + 1} = \frac{4 	imes 28}{59} = \frac{112}{59}$.$z = \frac{5 - \frac{31}{28}}{\frac{31}{28} + 1} = \frac{140 - 31}{59} = \frac{109}{59}$.The coordinates are $(\frac{58}{59}, \frac{112}{59}, \frac{109}{59})$,which is not among the options $A, B, C$. Therefore,the correct option is $D$.

The coordinates of the foot of the perpendicular drawn from the origin to the line joining the points $(-9, 4, 5)$ and $(10, 0, -1)$ are

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