JEE Mathematics 2025 Question with Solution

Given: The vertices $Q$ and $R$ lie on the line

$$\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$$

with $QR = 5$, and $P = (0,2,3)$.

Find: The correct relation satisfied by $m$ and $n$ if the area of triangle $PQR$ is

$$\frac{m}{n}$$

A point on the given line is $A(-3,1,-4)$ and its direction vector is

$$\vec{d} = 5\hat{i} + 2\hat{j} + 3\hat{k}$$

Vector from $A$ to $P$ is

$$\vec{AP} = (0-(-3))\hat{i} + (2-1)\hat{j} + (3-(-4))\hat{k} = 3\hat{i} + \hat{j} + 7\hat{k}$$

The perpendicular distance from point $P$ to the line containing $QR$ is

$$h = \frac{|\vec{AP} \times \vec{d}|}{|\vec{d}|}$$

Now,

$$\vec{AP} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 7 \\ 5 & 2 & 3 \end{vmatrix}$$ $$= \hat{i}(1\cdot 3 - 7\cdot 2) - \hat{j}(3\cdot 3 - 7\cdot 5) + \hat{k}(3\cdot 2 - 1\cdot 5)$$ $$= -11\hat{i} + 26\hat{j} + \hat{k}$$

Therefore,

$$|\vec{AP} \times \vec{d}| = \sqrt{(-11)^2 + 26^2 + 1^2} = \sqrt{798}$$

and

$$|\vec{d}| = \sqrt{5^2 + 2^2 + 3^2} = \sqrt{38}$$

So the height is

$$h = \frac{\sqrt{798}}{\sqrt{38}} = \sqrt{21}$$

Using the area formula,

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ $$\text{Area} = \frac{1}{2} \times 5 \times \sqrt{21} = \frac{5\sqrt{21}}{2}$$

Hence,

$$\frac{m}{n} = \frac{5\sqrt{21}}{2}$$

which gives

$$2m = 5\sqrt{21}n$$ $$2m - 5\sqrt{21}n = 0$$

Therefore, the correct option is B.

Given: The base $QR$ lies on the line

$$\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$$

and $QR = 5$, with vertex $P=(0,2,3)$.

Find: The relation among $m$ and $n$ when the area is $\frac{m}{n}$.

Use the idea that the area of triangle $PQR$ equals half the product of the base $QR$ and the perpendicular distance from $P$ to the line containing $QR$.

Take a point on the line as

$$A=(-3,1,-4)$$

and direction vector

$$\vec{d}=(5,2,3)$$

Then

$$\vec{AP}=(3,1,7)$$

The perpendicular distance is

$$h = \frac{|\vec{AP} \times \vec{d}|}{|\vec{d}|}$$

Compute the cross product:

$$\vec{AP} \times \vec{d} = (-11,26,1)$$

So,

$$|\vec{AP} \times \vec{d}| = \sqrt{121+676+1} = \sqrt{798}$$ $$|\vec{d}| = \sqrt{25+4+9} = \sqrt{38}$$

Hence,

$$h = \sqrt{\frac{798}{38}} = \sqrt{21}$$

Now area of triangle $PQR$ is

$$\frac{1}{2}\times 5 \times \sqrt{21} = \frac{5\sqrt{21}}{2}$$

Thus,

$$\frac{m}{n} = \frac{5\sqrt{21}}{2}$$

which implies

$$2m - 5\sqrt{21}n = 0$$

Therefore, the correct option is B.