The lines \fracx-22=\fracy-2=\fracz-716 and \fracx+34=\fracy+23=\fracz+21 intersect at the point P . If the distance of P from the line \fracx+12=\fracy-13=\fracz-11 is l , then 14 l^2 is equal to \underline\hspace2cm .

3D Geometry PYQ — JEE Main 2024 · 27 Jan · Shift 2 · Q53 with Solution

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Correct answer: 108

For the first line, $\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}=t$ so $x=2+2t,\quad y=-2t,\quad z=7+16t.$

For the second line, $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}=s$ so $x=-3+4s,\quad y=-2+3s,\quad z=-2+s.$

Since the lines intersect, coordinates must be equal for some $t,s$ : $2+2t=-3+4s \quad ...(1)$ $-2t=-2+3s \quad ...(2)$ $7+16t=-2+s \quad ...(3)$

From (2), $-2t=-2+3s \implies 2t=2-3s \implies t=1-\frac{3s}{2}.$

Substitute into (1): $2+2\left(1-\frac{3s}{2}\right)=-3+4s$ $2+2-3s=-3+4s$ $4-3s=-3+4s$ $7=7s$ $s=1.$ Then $t=1-\frac{3}{2}=-\frac{1}{2}.$

Now find $P$ using the first line: $x=2+2\left(-\frac12\right)=1,$ $y=-2\left(-\frac12\right)=1,$ $z=7+16\left(-\frac12\right)=7-8=-1.$

Hence, $P=(1,1,-1).$

The line is $\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}=\lambda.$ So a point on it is $A=(-1,1,1)$ and its direction vector is $\vec d=(2,3,1).$

Distance from point $P$ to the line through $A$ with direction $\vec d$ is $l=\frac{\left|\overrightarrow{AP}\times \vec d\right|}{|\vec d|}.$

Now, $\overrightarrow{AP}=P-A=(1-(-1),1-1,-1-1)=(2,0,-2).$

Compute cross product:

\begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 0 & -2 \\ 2 & 3 & 1 \end{vmatrix}$$ $$=\hat i(0\cdot 1-(-2)\cdot 3)-\hat j(2\cdot 1-(-2)\cdot 2)+\hat k(2\cdot 3-0\cdot 2)$$ $$=6\hat i-6\hat j+6\hat k.$$ So, $$\left|\overrightarrow{AP}\times \vec d\right|=\sqrt{6^2+(-6)^2+6^2}=\sqrt{108}=6\sqrt3.$$ Also, $$|\vec d|=\sqrt{2^2+3^2+1^2}=\sqrt{14}.$$ Thus, $$l=\frac{6\sqrt3}{\sqrt{14}}.$$ Hence, $$l^2=\frac{108}{14}=\frac{54}{7}.$$ Therefore, $$14l^2=108.$$ 5. **Compare with stored answer** Stored correct answer = $108$. Our derived answer also is $108$, so it agrees.