Let the image of the point P(0, -5, 0) in the | vedclass

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(A) $1$. For point $P(0, -5, 0)$ and line $L_1: \frac{x-1}{2} = \frac{y}{1} = \frac{z+1}{-2} = \lambda$, the foot of perpendicular $F_1$ is $(2\lambda

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$162$

Vedclass · Answer

(A) $1$. For point $P(0, -5, 0)$ and line $L_1: \frac{x-1}{2} = \frac{y}{1} = \frac{z+1}{-2} = \lambda$, the foot of perpendicular $F_1$ is $(2\lambda+1, \lambda, -2\lambda-1)$. The vector $\vec{PF_1} = (2\lambda+1, \lambda+5, -2\lambda-1)$. Since $\vec{PF_1} \cdot (2, 1, -2) = 0$, we get $2(2\lambda+1) + 1(\lambda+5) - 2(-2\lambda-1) = 0$, which simplifies to $9\lambda + 9 = 0$, so $\lambda = -1$. Thus $F_1 = (-1, -1, 1)$. The image $R = 2F_1 - P = 2(-1, -1, 1) - (0, -5, 0) = (-2, 3, 2)$.
$2$. For point $Q(0, -1/2, 0)$ and line $L_2: \frac{x-1}{-1} = \frac{y+9}{4} = \frac{z+1}{1} = \mu$, the foot of perpendicular $F_2$ is $(-\mu+1, 4\mu-9, \mu-1)$. The vector $\vec{QF_2} = (-\mu+1, 4\mu-8.5, \mu-1)$. Since $\vec{QF_2} \cdot (-1, 4, 1) = 0$, we get $1(\mu-1) + 16\mu - 34 + \mu - 1 = 0$, which simplifies to $18\mu - 36 = 0$, so $\mu = 2$. Thus $F_2 = (-1, -1, 1)$. The image $S = 2F_2 - Q = 2(-1, -1, 1) - (0, -0.5, 0) = (-2, -1.5, 2)$.
$3$. Vectors: $\vec{PQ} = (0, 4.5, 0)$ and $\vec{PS} = (-2, 3.5, 2)$.
$4$. Area of parallelogram $PQRS = |\vec{PQ} 	imes \vec{PS}| = |(0, 4.5, 0) 	imes (-2, 3.5, 2)| = |(9, 0, 9)| = \sqrt{81 + 0 + 81} = \sqrt{162}$.
$5$. Square of the area = $162$.

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