The perpendicular distance of the line x - 12 = y + 2-1 = z + 32 from the point P(2, -10, 1) is:

Given: The line is x - 12 = y + 2-1 = z + 32 and the point is P(2,-10,1). Find: The perpendicular distance of the point from the line. Write the line in parametric form: x = 2t + 1, y = -t - 2, z = 2t - 3 So, a point on the line is A(1,-2,-3) and the direction vector is d = 2,-1,2. Now, AP = 2-1, -10+2, 1+3 = 1,-8,4 Using the formula for distance of a point from a line in three dimensions, d = AP d d Compute the cross product: AP d = vmatrix i & j & k 1 & -8 & 4 2 & -1 & 2 vmatrix = i((-8)(2) - 4(-1)) - j((1)(2) - 4(2)) + k((1)(-1) - (-8)(2)) = i(-12) - j(-6) + k(15) = (-12,6,15) Therefore, AP d = (-12)^2 + 6^2 + 15^2 = 144 + 36 + 225 = 405 = 9 5 and d = 2^2 + (-1)^2 + 2^2 = 9 = 3 Hence, d = 9 53 = 3 5 Therefore, the perpendicular distance is 3 5, so the correct option is C. The solution's also shows option A, but the extracted working gives 3 5, which matches option C.

The perpendicular distance of the line x - 12 = y + 2-1 = z + 32… — Solution | JEE Mathematics

MCQMediumJEE 2025Equation of Line in 3D

The perpendicular distance of the line $\frac{x - 1}{2} = \frac{y + 2}{-1} = \frac{z + 3}{2}$ from the point $P(2, -10, 1)$ is:

A
$4\sqrt{3}$
B
$5\sqrt{2}$
C
$3\sqrt{5}$
D
$6$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The line is $\frac{x - 1}{2} = \frac{y + 2}{-1} = \frac{z + 3}{2}$ and the point is $P(2,-10,1)$ .

Find: The perpendicular distance of the point from the line.

Write the line in parametric form:

x = 2t + 1, y = -t - 2, z = 2t - 3

So, a point on the line is $A(1,-2,-3)$ and the direction vector is $\vec{d} = \langle 2,-1,2 \rangle$ .

Now,

\overrightarrow{AP} = \langle 2-1,\,-10+2,\,1+3 \rangle = \langle 1,-8,4 \rangle

Using the formula for distance of a point from a line in three dimensions,

d = \frac{\|\overrightarrow{AP} \times \vec{d}\|}{\|\vec{d}\|}

Cross Product Computation

Compute the cross product:

\overrightarrow{AP} \times \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -8 & 4 \\ 2 & -1 & 2 \end{vmatrix}

= \mathbf{i}((-8)(2) - 4(-1)) - \mathbf{j}((1)(2) - 4(2)) + \mathbf{k}((1)(-1) - (-8)(2))

= \mathbf{i}(-12) - \mathbf{j}(-6) + \mathbf{k}(15) = (-12,6,15)

Therefore,

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

and

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

Hence,

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

Therefore, the perpendicular distance is $3\sqrt{5}$ , so the correct option is C.

The solution's also shows option A, but the extracted working gives $3\sqrt{5}$ , which matches option C.

Common mistakes

Using the wrong point on the line. A point on the line is $A(1,-2,-3)$ from the symmetric form; choosing coordinates not satisfying the line gives an incorrect vector and hence a wrong distance.
Applying the point-to-plane or projection formula incorrectly. For a point-to-line distance in three dimensions, use $d = \frac{\|\overrightarrow{AP} \times \vec{d}\|}{\|\vec{d}\|}$ , not a dot-product formula meant for a plane or for resolving along the line.
Sign error in the cross product expansion. The middle term has a minus sign in the determinant expansion, and missing it changes the vector components and the final magnitude.

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