The lines frac{x-2}{2}=frac{y}{-2}=frac{z-7}{ | vedclass

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Question

(A) Let the first line be $\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16} = \lambda$. Then any point on this line is $(2\lambda+2, -2\lambda, 16\lambda+7)$

Vedclass · Accepted Answer

$108$

Vedclass · Answer

(A) Let the first line be $\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16} = \lambda$. Then any point on this line is $(2\lambda+2, -2\lambda, 16\lambda+7)$.Let the second line be $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1} = k$. Then any point on this line is $(4k-3, 3k-2, k-2)$.Since they intersect at $P$,we equate the coordinates:$2\lambda+2 = 4k-3 \Rightarrow 2\lambda - 4k = -5$$-2\lambda = 3k-2 \Rightarrow 2\lambda + 3k = 2$Adding these equations: $-k = -7 \Rightarrow k=1$. Substituting $k=1$ in $2\lambda+3(1)=2$,we get $2\lambda = -1 \Rightarrow \lambda = -1/2$.Checking the $z$-coordinate: $16(-1/2)+7 = -8+7 = -1$ and $k-2 = 1-2 = -1$. Since they match,the intersection point $P$ is $(4(1)-3, 3(1)-2, 1-2) = (1, 1, -1)$.Now,we find the distance $l$ of $P(1, 1, -1)$ from the line $\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$.Let $A = (-1, 1, 1)$ be a point on the line and $\vec{v} = 2\hat{i}+3\hat{j}+\hat{k}$ be the direction vector.Vector $\vec{AP} = (1 - (-1))\hat{i} + (1-1)\hat{j} + (-1-1)\hat{k} = 2\hat{i} - 2\hat{k}$.The distance $l$ is given by $l = \frac{|\vec{AP} 	imes \vec{v}|}{|\vec{v}|}$.$\vec{AP} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 0 & -2 \ 2 & 3 & 1 \end{vmatrix} = \hat{i}(0 - (-6)) - \hat{j}(2 - (-4)) + \hat{k}(6 - 0) = 6\hat{i} - 6\hat{j} + 6\hat{k}$.$|\vec{AP} 	imes \vec{v}| = \sqrt{6^2 + (-6)^2 + 6^2} = \sqrt{36+36+36} = \sqrt{108} = 6\sqrt{3}$.$|\vec{v}| = \sqrt{2^2+3^2+1^2} = \sqrt{4+9+1} = \sqrt{14}$.$l = \frac{6\sqrt{3}}{\sqrt{14}} \Rightarrow l^2 = \frac{36 	imes 3}{14} = \frac{108}{14}$.Therefore,$14l^2 = 108$.

The lines $\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}$ and $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}$ intersect at the point $P$. If the distance of $P$ from the line $\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$ is $l$,then $14 l^2$ is equal to:

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