JEE Mathematics 2025 Question with Solution

Given: Point $A(1,2,2)$ and line $L_1: \frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-2}{2}$. Also, line $L_2$ is

$$\mathbf{r}=(-\hat{i}+\hat{j}-2\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$$

Find: $2(PQ)^2$, where $P$ is the foot of the perpendicular from $A$ to $L_1$ and $Q$ is the intersection point of $L_1$ and $L_2$.

A general point on $L_1$ is

$$(1+t,-1-t,2+2t)$$

So let

$$P=(1+t,-1-t,2+2t)$$

The direction vector of $L_1$ is

$$\vec{d}_1=(1,-1,2)$$

Then

$$\overrightarrow{AP}=(t,-3-t,2t)$$

Since $P$ is the foot of the perpendicular, we use

$$\overrightarrow{AP}\cdot \vec{d}_1=0$$

Thus,

$$(t,-3-t,2t)\cdot(1,-1,2)=0$$ $$t+(-3-t)(-1)+2t\cdot 2=0$$ $$t+3+t+4t=0$$ $$6t+3=0$$ $$t=-\frac{1}{2}$$

Hence,

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

Now write $L_2$ in parametric form:

$$(x,y,z)=(-1+\lambda,1-\lambda,-2+\lambda)$$

For the intersection point $Q$ of $L_1$ and $L_2$,

$$1+t=-1+\lambda,$$ $$-1-t=1-\lambda,$$ $$2+2t=-2+\lambda$$

From the first equation,

$$\lambda=t+2$$

Substitute into the third equation:

$$2+2t=-2+(t+2)$$ $$2+2t=t$$ $$t=-2$$

Then

$$\lambda=0$$

So,

$$Q=(1-2,-1+2,2-4)=(-1,1,-2)$$

Now,

$$PQ^2=\left(\frac{1}{2}+1\right)^2+\left(-\frac{1}{2}-1\right)^2+(1+2)^2$$ $$PQ^2=\left(\frac{3}{2}\right)^2+\left(-\frac{3}{2}\right)^2+3^2$$ $$PQ^2=\frac{9}{4}+\frac{9}{4}+9=\frac{27}{2}$$

Therefore,

$$2(PQ)^2=2\cdot \frac{27}{2}=27$$

So, the correct option is A.

Given: The point is $(1,2,2)$, the line $L$ is $\frac{x-1}{1}=\frac{y+1}{-1}=\frac{z-2}{2}$, and the second line is $\mathbf{r}=(-\hat{i}+\hat{j}-2\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$.

Find: The value of $2(PQ)^2$.

From the solution, the coordinates of the foot of the perpendicular are obtained first. A general point on the line is

$$Q(1+\mu,-1-\mu,2+2\mu)$$

Using perpendicularity with direction vector $(1,-1,2)$,

$$(\mu,-3-\mu,2\mu)\cdot(1,-1,2)=0$$ $$\mu+(-3-\mu)(-1)+2\mu\times 2=0$$ $$\mu+3+\mu+4\mu=0$$ $$6\mu=-3$$ $$\mu=-\frac{1}{2}$$

So the foot point is

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

Next, find the intersection point of the two lines. For $L_2$,

$$(x,y,z)=(-1+\lambda,1-\lambda,-2+\lambda)$$

Equating coordinates with $L_1$,

$$1+t=-1+\lambda,$$ $$-1-t=1-\lambda,$$ $$2+2t=-2+\lambda$$

From these,

$$t=-2,\quad \lambda=0$$

Hence,

$$Q=(-1,1,-2)$$

Now compute the distance:

$$PQ^2=\left(\frac{1}{2}+1\right)^2+\left(-\frac{1}{2}-1\right)^2+(1+2)^2$$ $$=\frac{9}{4}+\frac{9}{4}+9$$ $$=\frac{27}{2}$$

Therefore,

$$2(PQ)^2=27$$

Hence, the correct option is A.