Perpendicular Distance from a Point to a 3D Line Given by Plane Intersections

Let a line L pass through the point P(2,3,1) and be parallel to the line x + 3y - 2z - 2 = 0 = x - y + 2z. If the distance of L from the point (5,3,8) is α, then 3α^2 is equal to _______.

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

Type the value - units or words beside it are fine.

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why is the direction vector of the line parallel to the cross product of the normal vectors?

A line given by the intersection of two planes is perpendicular to the normal vectors of both planes. Therefore, its direction vector is along the cross product of the two normal vectors.

Answer

Accepted answer: 158

The value of $3 α^{2}$ is 158.

Why each option works or fails

Step-by-step solution

Given: Line $L$ passes through $P (2, 3, 1)$ and is parallel to the line given by the intersection of planes $x + 3 y - 2 z - 2 = 0$ and $x - y + 2 z = 0$ . The target point is $Q (5, 3, 8)$ , and the distance from $Q$ to $L$ is $α$ .
Approach: Find the direction vector of the line via the cross product of the normal vectors of the two given planes. Then, write the parametric coordinates of an arbitrary point $M$ on $L$ . Use the orthogonality condition $\vec{Q M} \cdot \vec{d} = 0$ to find the foot of the perpendicular $M$ , and finally calculate $3 α^{2} = 3 ∣ Q M ∣^{2}$ .
Execute: Direction vector of line $L$ : $\vec{d} = {\vec{n}}_{1} \times {\vec{n}}_{2} = ∣ \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & - 2 \\ 1 & - 1 & 2 \end{array} ∣ = \hat{i} (6 - 2) - \hat{j} (2 + 2) + \hat{k} (- 1 - 3) = 4 \hat{i} - 4 \hat{j} - 4 \hat{k}$ Dividing by 4, we take direction ratios as $⟨ 1, - 1, - 1 ⟩$ . Parametric form of line $L$ through $P (2, 3, 1)$ : $M (λ) = (λ + 2, - λ + 3, - λ + 1)$ Vector $\vec{Q M}$ where $Q = (5, 3, 8)$ : $\vec{Q M} = (λ + 2 - 5) \hat{i} + (- λ + 3 - 3) \hat{j} + (- λ + 1 - 8) \hat{k} = (λ - 3) \hat{i} - λ \hat{j} - (λ + 7) \hat{k}$ Since $\vec{Q M} ⊥ \vec{d}$ : $(λ - 3) (1) + (- λ) (- 1) + (- (λ + 7)) (- 1) = 0$ $λ - 3 + λ + λ + 7 = 0 ⟹ 3 λ + 4 = 0 ⟹ λ = - \frac{4}{3}$
Execute: Substitute $λ = - \frac{4}{3}$ into $\vec{Q M}$ components: $Q M_{x} = - \frac{4}{3} - 3 = - \frac{13}{3}$ $Q M_{y} = - (- \frac{4}{3}) = \frac{4}{3}$ $Q M_{z} = - (- \frac{4}{3} + 7) = - \frac{17}{3}$ Now calculate $3 α^{2} = 3 ∣ \vec{Q M} ∣^{2}$ : $3 α^{2} = 3 [{(- \frac{13}{3})}^{2} + {(\frac{4}{3})}^{2} + {(- \frac{17}{3})}^{2}] = 3 \cdot \frac{169 + 16 + 289}{9} = \frac{474}{3} = 158$
Verify: Check via vector formula $α^{2} = ∣ \vec{Q P} ∣^{2} - \frac{(\vec{Q P} \cdot \vec{d})^{2}}{∣ \vec{d} ∣^{2}}$ . $\vec{Q P} = P - Q = (- 3, 0, - 7)$ . $∣ \vec{Q P} ∣^{2} = (- 3)^{2} + 0^{2} + (- 7)^{2} = 9 + 49 = 58$ . $\vec{d} = (1, - 1, - 1) ⟹ ∣ \vec{d} ∣^{2} = 1 + 1 + 1 = 3$ . $\vec{Q P} \cdot \vec{d} = (- 3) (1) + 0 (- 1) + (- 7) (- 1) = - 3 + 7 = 4$ . $α^{2} = 58 - \frac{4^{2}}{3} = 58 - \frac{16}{3} = \frac{174 - 16}{3} = \frac{158}{3}$ . Thus, $\vec{Q P} \cdot \vec{d} = (- 3) (1) + 0 (- 1) + (- 7) (- 1) = - 3 + 7 = 4$ . Consistent and verified.

Shortcut:

When to use it: Fastest way to find the perpendicular distance from a point to a line given by a point and direction vector

Given: Line $L$ has base point $P (2, 3, 1)$ , direction $\vec{d} = (1, - 1, - 1)$ , and target point $Q (5, 3, 8)$ .

Approach: Use the right triangle formula: $α^{2} = ∣ \vec{Q P} ∣^{2} - {(\frac{\vec{Q P} \cdot \vec{d}}{∣ \vec{d} ∣})}^{2}$ , avoiding the need to find the foot of the perpendicular.

Execute: 1. $\vec{Q P} = P - Q = (2 - 5, 3 - 3, 1 - 8) = (- 3, 0, - 7)$ . 2. $∣ \vec{Q P} ∣^{2} = (- 3)^{2} + 0^{2} + (- 7)^{2} = 9 + 0 + 49 = 58$ . 3. $\vec{Q P} \cdot \vec{d} = (- 3) (1) + (0) (- 1) + (- 7) (- 1) = - 3 + 0 + 7 = 4$ . 4. $∣ \vec{d} ∣^{2} = 1^{2} + (- 1)^{2} + (- 1)^{2} = 3$ . 5. $α^{2} = ∣ \vec{Q P} ∣^{2} - \frac{(\vec{Q P} \cdot \vec{d})^{2}}{∣ \vec{d} ∣^{2}} = 58 - \frac{16}{3} = \frac{158}{3}$ . 6. $3 α^{2} = 158$ .

Verify: Check that projection length $\frac{4}{\sqrt{3}} < ∣ \vec{Q P} ∣ = \sqrt{58} \approx 7.6$ . Indeed $4 / 1.732 \approx 2.3 < 7.6$ .

Perpendicular Distance from a Point to a 3D Line Given by Plane Intersections

What feels right?

The value of $3 α^{2}$ is 158.

Why each option works or fails

Solution

01Given

02Approach

03Execute

04Execute

✓Verify

We think you should solve this next.

The foot of perpendicular from the origin $O$ to a plane $P$ which meet…

Let the line $L$ pass through the point $(0, 1, 2)$ , intersect the line…

Another question from the same paper — Integral Calculus, 2023

Students also ask

Perpendicular Distance from a Point to a 3D Line Given by Plane Intersections

The value of 3α23α2 is 158.

01Given

02Approach

03Execute

04Execute

✓Verify

The foot of perpendicular from the origin OO to a plane PP which meet…

Let the line LL pass through the point (0,1,2)(0,1,2), intersect the line…

Another question from the same paper — Integral Calculus, 2023

Students also ask

The value of $3 α^{2}$ is 158.

The foot of perpendicular from the origin $O$ to a plane $P$ which meet…

Let the line $L$ pass through the point $(0, 1, 2)$ , intersect the line…