JEE Mathematics 2023 Question with Solution

Step-by-step solution

Standard Method

Given: The line $L_1$ passes through $A(2,6,2)$ and is perpendicular to the plane $2x+y-2z=10$.

Find: The shortest distance between $L_1$ and $L_2: \frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$.

Since $L_1$ is perpendicular to the plane, its direction ratios are the normal vector of the plane. Therefore,

$$L_1: \frac{x-2}{2}=\frac{y-6}{1}=\frac{z-2}{-2}$$

and

$$L_2: \frac{x+1}{2}=\frac{y+4}{-3}=\frac{z}{2}$$

Take a point $A(2,6,2)$ on $L_1$ and a point $B(-1,-4,0)$ on $L_2$. Then

$$\overrightarrow{AB}=(-1-2)\hat{i}+(-4-6)\hat{j}+(0-2)\hat{k}=-3\hat{i}-10\hat{j}-2\hat{k}$$

The direction vectors of the two lines are

$$\vec{d_1}=2\hat{i}+\hat{j}-2\hat{k}, \qquad \vec{d_2}=2\hat{i}-3\hat{j}+2\hat{k}$$

Their cross product is

$$\vec{d_1}\times\vec{d_2}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 2 & -3 & 2 \end{vmatrix}=-4\hat{i}-8\hat{j}-8\hat{k}$$

So,

$$\left|\vec{d_1}\times\vec{d_2}\right|=\sqrt{(-4)^2+(-8)^2+(-8)^2}=\sqrt{144}=12$$

Now,

$$\overrightarrow{AB}\cdot(\vec{d_1}\times\vec{d_2})=(-3)(-4)+(-10)(-8)+(-2)(-8)=12+80+16=108$$

Hence the shortest distance between the skew lines is

$$d=\frac{\left|\overrightarrow{AB}\cdot(\vec{d_1}\times\vec{d_2})\right|}{\left|\vec{d_1}\times\vec{d_2}\right|}=\frac{108}{12}=9$$

Therefore, the shortest distance is $9$. The solution states option C, but the working clearly gives $9$, which matches option D.

Common mistakes

• Using the plane equation directly as the line equation is incorrect. Because the line is perpendicular to the plane, its direction ratios come from the plane's normal vector $\langle 2,1,-2 \rangle$; the line still must be written through the point $\left(2,6,2\right)$.

• Choosing the shortest distance formula for parallel lines is wrong here. These two lines are skew lines, so use $d=\frac{|\overrightarrow{AB}\cdot(\vec{d_1}\times\vec{d_2})|}{|\vec{d_1}\times\vec{d_2}|}$ instead.

• Errors in the cross product sign are common. A wrong sign in $\vec{d_1}\times\vec{d_2}$ changes the numerator or denominator. Expand the determinant carefully and then take magnitude at the end.