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(B) The line $L$ is given by $\frac{x-1}{1} = \frac{y+1}{-1} = \frac{z-2}{2} = \mu$. Any point on $L$ is $P(\mu+1, -\mu-1, 2\mu+2)$.Since $AP \perp L$

Vedclass · Accepted Answer

$27$

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(B) The line $L$ is given by $\frac{x-1}{1} = \frac{y+1}{-1} = \frac{z-2}{2} = \mu$. Any point on $L$ is $P(\mu+1, -\mu-1, 2\mu+2)$.Since $AP \perp L$,the vector $\overrightarrow{AP} = (\mu+1-1, -\mu-1-2, 2\mu+2-2) = (\mu, -\mu-3, 2\mu)$ is perpendicular to the direction vector of $L$,$\vec{d} = (1, -1, 2)$.$\overrightarrow{AP} \cdot \vec{d} = 0 \Rightarrow (\mu)(1) + (-\mu-3)(-1) + (2\mu)(2) = 0$.$\mu + \mu + 3 + 4\mu = 0 \Rightarrow 6\mu = -3 \Rightarrow \mu = -\frac{1}{2}$.Thus,$P = (-\frac{1}{2}+1, -(-\frac{1}{2})-1, 2(-\frac{1}{2})+2) = (\frac{1}{2}, -\frac{1}{2}, 1)$.The line $L_2$ is given by $\vec{r} = (-1, 1, -2) + \lambda(1, -1, 1)$. Any point on $L_2$ is $Q(-1+\lambda, 1-\lambda, -2+\lambda)$.Since $Q$ lies on $L$,it must satisfy the equation of $L$: $\frac{(-1+\lambda)-1}{1} = \frac{(1-\lambda)+1}{-1} = \frac{(-2+\lambda)-2}{2} = \mu$.$\frac{\lambda-2}{1} = \frac{2-\lambda}{-1} = \frac{\lambda-4}{2}$.From $\lambda-2 = \lambda-2$ (always true) and $\lambda-2 = \frac{\lambda-4}{2} \Rightarrow 2\lambda-4 = \lambda-4 \Rightarrow \lambda = 0$.For $\lambda = 0$,$Q = (-1, 1, -2)$.Now,$PQ^2 = (\frac{1}{2} - (-1))^2 + (-\frac{1}{2} - 1)^2 + (1 - (-2))^2 = (\frac{3}{2})^2 + (-\frac{3}{2})^2 + (3)^2 = \frac{9}{4} + \frac{9}{4} + 9 = \frac{18}{4} + 9 = 4.5 + 9 = 13.5$.Therefore,$2(PQ)^2 = 2(13.5) = 27$.

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