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(C) The equation of the line passing through $B(0,4,1)$ and $C(-2,0,4)$ is given by $\frac{x-0}{-2-0} = \frac{y-4}{0-4} = \frac{z-1}{4-1} = \lambda$.

Vedclass · Accepted Answer

$18: 11$

Vedclass · Answer

(C) The equation of the line passing through $B(0,4,1)$ and $C(-2,0,4)$ is given by $\frac{x-0}{-2-0} = \frac{y-4}{0-4} = \frac{z-1}{4-1} = \lambda$. This simplifies to $\frac{x}{-2} = \frac{y-4}{-4} = \frac{z-1}{3} = \lambda$. Thus, any point $D$ on the line is $(-2\lambda, 4-4\lambda, 3\lambda+1)$. The vector $\vec{AD} = (-2\lambda-2, 4-4\lambda, 3\lambda-2)$. The direction vector of line $BC$ is $\vec{v} = (-2, -4, 3)$. Since $AD \perp BC$, their dot product is zero: $(-2)(-2\lambda-2) + (-4)(4-4\lambda) + (3)(3\lambda-2) = 0$. $4\lambda + 4 - 16 + 16\lambda + 9\lambda - 6 = 0$. $29\lambda - 18 = 0$, so $\lambda = \frac{18}{29}$. Using the section formula, if $D$ divides $BC$ in ratio $m:n$, then $x_D = \frac{m(-2) + n(0)}{m+n} = -2\lambda$. $\frac{-2m}{m+n} = -2(\frac{18}{29})$. $\frac{m}{m+n} = \frac{18}{29}$. $29m = 18m + 18n \implies 11m = 18n$. Therefore, $\frac{m}{n} = \frac{18}{11}$.

Let $D$ be the foot of the perpendicular drawn from the point $A(2,0,3)$ to the line joining the points $B(0,4,1)$ and $C(-2,0,4)$. Then the ratio in which $D$ divides $BC$ is

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