JEE Mathematics 2023 Question with Solution

Given: The lines are

$$\frac{x+1}{1}=\frac{y}{\frac{1}{2}}=\frac{z}{-\frac{1}{12}}$$

and

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

So, take

$$\vec{a}=(-1,0,0),\quad \vec{b}=(0,-2,1),\quad \vec{p}=\left(1,\frac{1}{2},-\frac{1}{12}\right),\quad \vec{q}=\left(1,1,\frac{1}{6}\right)$$

Find: The shortest distance between the two lines.

Use the formula

$$\text{S.D.}=\frac{|(\vec{b}-\vec{a})\cdot(\vec{p}\times\vec{q})|}{|\vec{p}\times\vec{q}|}$$

Now,

$$\vec{p}\times\vec{q}=\begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\ 1&\frac{1}{2}&-\frac{1}{12}\\ 1&1&\frac{1}{6} \end{vmatrix}=\frac{1}{6}\hat{i}-\frac{1}{4}\hat{j}+\frac{1}{2}\hat{k}$$

which is proportional to

$$2\hat{i}-3\hat{j}+6\hat{k}$$

Also,

$$\vec{b}-\vec{a}=(0,-2,1)-(-1,0,0)=(1,-2,1)$$

Hence,

$$(\vec{b}-\vec{a})\cdot(2\hat{i}-3\hat{j}+6\hat{k})=1\cdot 2+(-2)(-3)+1\cdot 6=14$$

and

$$|2\hat{i}-3\hat{j}+6\hat{k}|=\sqrt{2^2+(-3)^2+6^2}=\sqrt{49}=7$$

Therefore,

$$\text{S.D.}=\frac{14}{7}=2$$

So, the shortest distance is $2$ and the correct option is C. The option text on the page is inconsistent with the numeric value shown beside C.

Given:

$$x+1=2y=-12z$$

represents a line through $(-1,0,0)$ with direction ratios proportional to

$$\left(1,\frac{1}{2},-\frac{1}{12}\right)$$

And

$$x=y+2=6z-6$$

represents a line through $(0,-2,1)$ with direction ratios proportional to

$$\left(1,1,\frac{1}{6}\right)$$

Find: Shortest distance.

A common error in one provided approach is taking the second point as $(0,-2,-1)$ or dropping the $\hat{k}$ component of the cross product. From

$$6z-6=x$$

we get $z=1$ when $x=0$, so the correct point is $(0,-2,1)$.

Now compute

$$\vec{p}\times\vec{q}=\begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\ 1&\frac{1}{2}&-\frac{1}{12}\\ 1&1&\frac{1}{6} \end{vmatrix}$$ $$=\hat{i}\left(\frac{1}{12}+\frac{1}{12}\right)-\hat{j}\left(\frac{1}{6}+\frac{1}{12}\right)+\hat{k}\left(1-\frac{1}{2}\right)$$ $$=\frac{1}{6}\hat{i}-\frac{1}{4}\hat{j}+\frac{1}{2}\hat{k}$$

Multiply by $12$ to simplify:

$$\vec{p}\times\vec{q}\propto 2\hat{i}-3\hat{j}+6\hat{k}$$

Now,

$$\vec{b}-\vec{a}=(1,-2,1)$$

So,

$$(1,-2,1)\cdot(2,-3,6)=2+6+6=14$$

And,

$$\sqrt{2^2+(-3)^2+6^2}=7$$

Thus,

$$\text{S.D.}=\frac{14}{7}=2$$

Therefore, the shortest distance equals $2$.