JEE MainMathematics

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Foot of a Perpendicular from a Point to a 3D Line

Q: Why do we add the parametric expressions before substituting k?

Adding the expressions for α, β, γ first gives 10k - 6, which simplifies the arithmetic since 19 × 10k = 5 × 38k, cleanly canceling the denominator.

β γ be the foot of perpendicular from the point (1,2,3) on the line (x+3)/5=(y-1)/2=(z+4)/3, then 19(α+β+γ) is equal to:

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A102B101C99D100

Step-by-step solutionView

Correct answer

The coordinates of the foot of perpendicular $(α, β, γ)$ give $α + β + γ = \frac{101}{19}$ , so $19 (α + β + γ) = 101$ .

Option analysis

Why each option works or fails

A · 102

Arithmetic addition error when computing $10 (18) - 114$ , leading to an extra unit. Ensure exact arithmetic when computing $19 (α + β + γ) = 10 (18) - 114 = 180 - 114 = 101$ (with correct sum formulation: $10 λ - 6 ⟹ 190 λ / 19 - 114 = 180 - 114 = 66$ , but with correct $(α + β + γ) = (5 λ - 3) + (2 λ + 1) + (3 λ - 4) = 10 λ - 6$ , carefully verify each term).

B · 101

This is the correct value. Correctly parameterize the line, use the dot product condition with the direction vector to find $λ = \frac{18}{19}$ , and evaluate $19 (α + β + γ) = 19 (10 λ - 6) = 101$ .

C · 99

Sign error when finding the vector connecting the given point to the parametric point on the line. Check the signs: the vector is $(5 λ - 3 - 1, 2 λ + 1 - 2, 3 λ - 4 - 3) = (5 λ - 4, 2 λ - 1, 3 λ - 7)$ .

D · 100

Arithmetic off-by-one subtraction slip when evaluating the numerator in the dot product or the linear combination. Re-check the dot product: $5 (5 λ - 4) + 2 (2 λ - 1) + 3 (3 λ - 7) = 38 λ - 43 = 0 ⟹ λ = \frac{43}{38}$ , then $10 λ - 6 = 10 (\frac{43}{38}) - 6 = \frac{215 - 114}{19} = \frac{101}{19}$ .

Reviewed route

Solution

StepWorking

01Given

Given the point $A (1, 2, 3)$ and the line $L : \frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3} = k$ . Let the foot of the perpendicular from $A$ to $L$ be $P (α, β, γ)$ .

02Goal

Find the value of $19 (α + β + γ)$ .

03Approach

Express the coordinates of an arbitrary point $P$ on the line in terms of parameter $k$ . Find the direction ratios of the vector $\vec{A P}$ and use the orthogonality condition $\vec{A P} \cdot \vec{d} = 0$ , where $\vec{d} = ⟨ 5, 2, 3 ⟩$ is the direction vector of the line, to solve for $k$ .

04Execute

The general point on the line is $P (5 k - 3, 2 k + 1, 3 k - 4)$ . The direction ratios of $A P$ are $(5 k - 3 - 1, 2 k + 1 - 2, 3 k - 4 - 3) = (5 k - 4, 2 k - 1, 3 k - 7)$ . Since $A P ⊥ L$ : $5 (5 k - 4) + 2 (2 k - 1) + 3 (3 k - 7) = 0$ $(25 k - 20) + (4 k - 2) + (9 k - 21) = 0$ $38 k - 43 = 0 ⟹ k = \frac{43}{38}$

05Execute

The coordinates of the foot are $α = 5 k - 3$ , $β = 2 k + 1$ , $γ = 3 k - 4$ . Their sum is: $α + β + γ = (5 k - 3) + (2 k + 1) + (3 k - 4) = 10 k - 6$ Now compute $19 (α + β + γ)$ : $19 (10 k - 6) = 19 (10 \times \frac{43}{38} - 6) = 19 (\frac{215}{19} - 6) = 215 - 114 = 101$

✓Verify

Substitute $k = \frac{43}{38}$ back to get $(α, β, γ) = (\frac{101}{38}, \frac{124}{38}, \frac{- 23}{38})$ . Sum $α + β + γ = \frac{101 + 124 - 23}{38} = \frac{202}{38} = \frac{101}{19}$ . Multiplying by $19$ gives exactly $101$ .

Hints that build this answer step by step

Let the line be parameterized by $\frac{x+3}{5}=\frac{y-1}{2}= rac{z+4}{3}=\lambda$ . What are the coordinates of a general point on this line?
$(5 λ - 3, 2 λ + 1, 3 λ - 4)$
Let $P = (1, 2, 3)$ and $Q = (5 λ - 3, 2 λ + 1, 3 λ - 4)$ . Since $P Q$ is perpendicular to the line with direction vector $\vec{d} = ⟨ 5, 2, 3 ⟩$ , what equation must $λ$ satisfy?
$5 (5 λ - 4) + 2 (2 λ - 1) + 3 (3 λ - 7) = 0$
Solving $5 (5 λ - 4) + 2 (2 λ - 1) + 3 (3 λ - 7) = 0$ gives which value of $λ$ ?
$λ = \frac{43}{38}$
With $λ = \frac{43}{38}$ , what is the value of $19 (α + β + γ)$ ?
$101$

✓ Source and academic review

Question type: Single correct
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 do we add the parametric expressions before substituting

k

Adding the expressions for $α, β, γ$ first gives $10 k - 6$ , which simplifies the arithmetic since $19 \times 10 k = 5 \times 38 k$ , cleanly canceling the denominator.

Answer

The coordinates of the foot of perpendicular $(α, β, γ)$ give $α + β + γ = \frac{101}{19}$ , so $19 (α + β + γ) = 101$ .

Why each option works or fails

A: 102 - Arithmetic addition error when computing $10 (18) - 114$ , leading to an extra unit. Ensure exact arithmetic when computing $19 (α + β + γ) = 10 (18) - 114 = 180 - 114 = 101$ (with correct sum formulation: $10 λ - 6 ⟹ 190 λ / 19 - 114 = 180 - 114 = 66$ , but with correct $(α + β + γ) = (5 λ - 3) + (2 λ + 1) + (3 λ - 4) = 10 λ - 6$ , carefully verify each term).
B · correct: 101 - This is the correct value. Correctly parameterize the line, use the dot product condition with the direction vector to find $λ = \frac{18}{19}$ , and evaluate $19 (α + β + γ) = 19 (10 λ - 6) = 101$ .
C: 99 - Sign error when finding the vector connecting the given point to the parametric point on the line. Check the signs: the vector is $(5 λ - 3 - 1, 2 λ + 1 - 2, 3 λ - 4 - 3) = (5 λ - 4, 2 λ - 1, 3 λ - 7)$ .
D: 100 - Arithmetic off-by-one subtraction slip when evaluating the numerator in the dot product or the linear combination. Re-check the dot product: $5 (5 λ - 4) + 2 (2 λ - 1) + 3 (3 λ - 7) = 38 λ - 43 = 0 ⟹ λ = \frac{43}{38}$ , then $10 λ - 6 = 10 (\frac{43}{38}) - 6 = \frac{215 - 114}{19} = \frac{101}{19}$ .

Step-by-step solution

Given: Given the point $A (1, 2, 3)$ and the line $L : \frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3} = k$ . Let the foot of the perpendicular from $A$ to $L$ be $P (α, β, γ)$ .
Goal: Find the value of $19 (α + β + γ)$ .
Approach: Express the coordinates of an arbitrary point $P$ on the line in terms of parameter $k$ . Find the direction ratios of the vector $\vec{A P}$ and use the orthogonality condition $\vec{A P} \cdot \vec{d} = 0$ , where $\vec{d} = ⟨ 5, 2, 3 ⟩$ is the direction vector of the line, to solve for $k$ .
Execute: The general point on the line is $P (5 k - 3, 2 k + 1, 3 k - 4)$ . The direction ratios of $A P$ are $(5 k - 3 - 1, 2 k + 1 - 2, 3 k - 4 - 3) = (5 k - 4, 2 k - 1, 3 k - 7)$ . Since $A P ⊥ L$ : $5 (5 k - 4) + 2 (2 k - 1) + 3 (3 k - 7) = 0$ $(25 k - 20) + (4 k - 2) + (9 k - 21) = 0$ $38 k - 43 = 0 ⟹ k = \frac{43}{38}$
Execute: The coordinates of the foot are $α = 5 k - 3$ , $β = 2 k + 1$ , $γ = 3 k - 4$ . Their sum is: $α + β + γ = (5 k - 3) + (2 k + 1) + (3 k - 4) = 10 k - 6$ Now compute $19 (α + β + γ)$ : $19 (10 k - 6) = 19 (10 \times \frac{43}{38} - 6) = 19 (\frac{215}{19} - 6) = 215 - 114 = 101$
Verify: Substitute $k = \frac{43}{38}$ back to get $(α, β, γ) = (\frac{101}{38}, \frac{124}{38}, \frac{- 23}{38})$ . Sum $α + β + γ = \frac{101 + 124 - 23}{38} = \frac{202}{38} = \frac{101}{19}$ . Multiplying by $19$ gives exactly $101$ .

Foot of a Perpendicular from a Point to a 3D Line

What feels right?

The coordinates of the foot of perpendicular $(α, β, γ)$ give $α + β + γ = \frac{101}{19}$ , so $19 (α + β + γ) = 101$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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