If the lines frac{x-1}{1}=frac{y-2}{2}=frac{z | vedclass

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(B) Let the point on the first line $L_1$ be $(\lambda+1, 2\lambda+2, \lambda-3)$.Let the point on the second line $L_2$ be $(2\mu+a, 3\mu-2, \mu+3)$.

Vedclass · Accepted Answer

$28$

Vedclass · Answer

(B) Let the point on the first line $L_1$ be $(\lambda+1, 2\lambda+2, \lambda-3)$.Let the point on the second line $L_2$ be $(2\mu+a, 3\mu-2, \mu+3)$.Since the lines intersect at point $P$,the coordinates must be equal:$1) \lambda+1 = 2\mu+a$$2) 2\lambda+2 = 3\mu-2 \Rightarrow 2\lambda = 3\mu-4$$3) \lambda-3 = \mu+3 \Rightarrow \lambda = \mu+6$Substitute $\lambda = \mu+6$ into the second equation:$2(\mu+6) = 3\mu-4 \Rightarrow 2\mu+12 = 3\mu-4 \Rightarrow \mu = 16$.Then $\lambda = 16+6 = 22$.Now find $a$ using the first equation:$22+1 = 2(16)+a \Rightarrow 23 = 32+a \Rightarrow a = -9$.The point $P$ is $(\lambda+1, 2\lambda+2, \lambda-3) = (22+1, 2(22)+2, 22-3) = (23, 46, 19)$.The distance of point $P(23, 46, 19)$ from the plane $z = -9$ is given by $|z_P - (-9)| = |19 + 9| = 28$.

If the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1}$ and $\frac{x-a}{2}=\frac{y+2}{3}=\frac{z-3}{1}$ intersect at the point $P$,then the distance of the point $P$ from the plane $z = a$ is:

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