The foot of the perpendicular from the point | vedclass

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(C) Let the line be $L: \frac{x+1}{2} = \frac{y-1}{5} = \frac{z+1}{-1} = \lambda$.Any point on the line is $P(2\lambda - 1, 5\lambda + 1, -\lambda - 1

Vedclass · Accepted Answer

$\frac{\beta}{\gamma} = -5$

Vedclass · Answer

(C) Let the line be $L: \frac{x+1}{2} = \frac{y-1}{5} = \frac{z+1}{-1} = \lambda$.Any point on the line is $P(2\lambda - 1, 5\lambda + 1, -\lambda - 1)$.Since $P$ is the foot of the perpendicular from $A(2, 0, 5)$ to the line,the vector $\vec{AP}$ must be perpendicular to the direction vector of the line $\vec{b} = 2\hat{i} + 5\hat{j} - \hat{k}$.$\vec{AP} = (2\lambda - 1 - 2)\hat{i} + (5\lambda + 1 - 0)\hat{j} + (-\lambda - 1 - 5)\hat{k} = (2\lambda - 3)\hat{i} + (5\lambda + 1)\hat{j} + (-\lambda - 6)\hat{k}$.Since $\vec{AP} \cdot \vec{b} = 0$,we have:$2(2\lambda - 3) + 5(5\lambda + 1) - 1(-\lambda - 6) = 0$$4\lambda - 6 + 25\lambda + 5 + \lambda + 6 = 0$$30\lambda + 5 = 0 \Rightarrow \lambda = -\frac{1}{6}$.Substituting $\lambda = -\frac{1}{6}$ into the coordinates of $P$:$\alpha = 2(-\frac{1}{6}) - 1 = -\frac{1}{3} - 1 = -\frac{4}{3}$$\beta = 5(-\frac{1}{6}) + 1 = -\frac{5}{6} + 1 = \frac{1}{6}$$\gamma = -(-\frac{1}{6}) - 1 = \frac{1}{6} - 1 = -\frac{5}{6}$Now check the options:$A) \frac{\alpha \beta}{\gamma} = \frac{(-\frac{4}{3})(\frac{1}{6})}{-\frac{5}{6}} = \frac{-\frac{4}{18}}{-\frac{5}{6}} = \frac{4}{18} 	imes \frac{6}{5} = \frac{4}{15}$ (Correct)$B) \frac{\alpha}{\beta} = \frac{-\frac{4}{3}}{\frac{1}{6}} = -\frac{4}{3} 	imes 6 = -8$ (Correct)$C) \frac{\beta}{\gamma} = \frac{\frac{1}{6}}{-\frac{5}{6}} = -\frac{1}{5}$ (Incorrect,as it is given as $-5$)$D) \frac{\gamma}{\alpha} = \frac{-\frac{5}{6}}{-\frac{4}{3}} = \frac{5}{6} 	imes \frac{3}{4} = \frac{5}{8}$ (Correct)Thus,option $C$ is not correct.

The foot of the perpendicular from the point $A(2, 0, 5)$ to the line $\frac{x+1}{2} = \frac{y-1}{5} = \frac{z+1}{-1}$ is $P(\alpha, \beta, \gamma)$. Then,which of the following is $NOT$ correct?

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