QJEEM25Maa07a: JEE Main 2025 Mathematics - Class 12 JEEM25Maa07a Solution: The perpendicular distance, of the line from the point , is :

To find the perpendicular distance from the point $P(2,-10,1)$ to the line given by

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

follow these steps:

Parametrize the Line:

From the given symmetric equations, set the common parameter as $t$ :

$x = 1 + 2t$

$y = -2 - t$

$z = -3 + 2t$

This shows that:

A point on the line is $A(1,-2,-3)$ (when $t=0$ ).

The direction vector is $\vec{d} = \langle 2, -1, 2 \rangle.$

Determine the Vector from Point A to P:

Calculate

$\vec{AP} = P - A = \langle 2 - 1, \; -10 - (-2), \; 1 - (-3) \rangle = \langle 1, \; -8, \; 4 \rangle.$

Compute the Cross Product:

The formula for the perpendicular distance from a point to a line in 3D is:

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

First, find the cross product $\vec{AP} \times \vec{d}$ :

$\vec{AP} \times \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -8 & 4 \\ 2 & -1 & 2 \end{vmatrix} = \Big( (-8)(2) - (4)(-1), \; (4)(2) - (1)(2), \; (1)(-1) - (-8)(2) \Big).$

Evaluate each component:

First component: $(-8 \times 2) - (4 \times -1) = -16 + 4 = -12.$

Second component: $(4 \times 2) - (1 \times 2) = 8 - 2 = 6.$

Third component: $(1 \times -1) - (-8 \times 2) = -1 + 16 = 15.$

So,

$\vec{AP} \times \vec{d} = \langle -12, 6, 15 \rangle.$

Calculate the Magnitudes:

For the cross product:

$\|\vec{AP} \times \vec{d}\| = \sqrt{(-12)^2 + 6^2 + 15^2} = \sqrt{144 + 36 + 225} = \sqrt{405} = 9\sqrt{5}.$

For the direction vector:

$\|\vec{d}\| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3.$

Compute the Distance:

Substitute the magnitudes into the distance formula:

$d = \frac{9\sqrt{5}}{3} = 3\sqrt{5}.$

Thus, the perpendicular distance from the point $P(2,-10,1)$ to the line is $3\sqrt{5}$ .

Comparing with the options given, the correct answer is:

Option C: $3\sqrt{5}$ .