Logo
Jeeify
MCQMediumJEE 2024Equation of Line in 3D

JEE Mathematics 2024 Question with Solution

The distance of the point Q(0,2,−2)Q(0, 2, -2) from the line passing through the point P(5,−4,3)P(5, -4, 3) and perpendicular to the lines r⃗=(−3i⃗+2k⃗)+λ(2i⃗+3j⃗+5k⃗),λ∈R\vec{r} = (-3\vec{i} + 2\vec{k}) + \lambda(2\vec{i} + 3\vec{j} + 5\vec{k}), \lambda \in \mathbb{R} and r⃗=(i⃗−2j⃗+k⃗)+μ(−i⃗+3j⃗+2k⃗),μ∈R\vec{r} = (\vec{i} - 2\vec{j} + \vec{k}) + \mu(-\vec{i} + 3\vec{j} + 2\vec{k}), \mu \in \mathbb{R} is:

  • A

    86\sqrt{86}

  • B

    20\sqrt{20}

  • C

    54\sqrt{54}

  • D

    74\sqrt{74}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The required line passes through P(5,−4,3)P(5,-4,3) and is perpendicular to the two lines with direction vectors a⃗1=2i^+3j^+5k^\vec{a}_1 = 2\hat{i}+3\hat{j}+5\hat{k} and a⃗2=−i^+3j^+2k^\vec{a}_2 = -\hat{i}+3\hat{j}+2\hat{k}.

Find: The distance of Q(0,2,−2)Q(0,2,-2) from this line.

A direction vector of the required line is perpendicular to both given direction vectors, so take their cross product:

d⃗=a⃗1×a⃗2=∣i^j^k^235−132∣\vec{d} = \vec{a}_1 \times \vec{a}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ -1 & 3 & 2 \end{vmatrix} =i^(3⋅2−5⋅3)−j^(2⋅2−5⋅(−1))+k^(2⋅3−3⋅(−1))= \hat{i}(3\cdot2-5\cdot3)-\hat{j}(2\cdot2-5\cdot(-1))+\hat{k}(2\cdot3-3\cdot(-1)) =−9i^−9j^+9k^= -9\hat{i}-9\hat{j}+9\hat{k}

So a convenient direction vector is i^+j^−k^\hat{i}+\hat{j}-\hat{k}.

Hence the required line is

r⃗=(5i^−4j^+3k^)+t(i^+j^−k^)\vec{r} = (5\hat{i}-4\hat{j}+3\hat{k}) + t(\hat{i}+\hat{j}-\hat{k})

Let the foot of the perpendicular from QQ to the line be the point

A=(5+t,−4+t,3−t)A = (5+t, -4+t, 3-t)

Then

QA⃗=(5+t,−6+t,5−t)\vec{QA} = (5+t, -6+t, 5-t)

Since QA⃗\vec{QA} is perpendicular to the line direction (1,1,−1)(1,1,-1), we use

QA⃗⋅(i^+j^−k^)=0\vec{QA}\cdot(\hat{i}+\hat{j}-\hat{k})=0 (5+t)+(−6+t)−(5−t)=0(5+t)+(-6+t)-(5-t)=0 −6+3t=0  ⟹  t=2-6+3t=0 \implies t=2

Therefore,

QA⃗=(7,−4,3)\vec{QA} = (7,-4,3)

and the distance is

∣QA⃗∣=72+(−4)2+32=49+16+9=74|\vec{QA}| = \sqrt{7^2+(-4)^2+3^2} = \sqrt{49+16+9} = \sqrt{74}

Therefore, the distance is 74\sqrt{74} and the correct option is D.

The first approach shown in the source contains an incorrect distance formula application and inconsistent simplification, but its stated conclusion matches the correct result. The second approach gives the defensible working leading to 74\sqrt{74}.

Using nearest point on the line

Given: The line must pass through P(5,−4,3)P(5,-4,3) and be perpendicular to both given lines.

Find: The distance of Q(0,2,−2)Q(0,2,-2) from that line.

First get the line direction from the cross product of the two given direction vectors:

(2,3,5)×(−1,3,2)=(−9,−9,9)(2,3,5)\times(-1,3,2)=(-9,-9,9)

So use the simpler parallel direction (1,1,−1)(1,1,-1).

A general point on the required line is

A(5+λ,−4+λ,3−λ)A(5+\lambda,-4+\lambda,3-\lambda)

Then

QA⃗=(5+λ,−6+λ,5−λ)\vec{QA}=(5+\lambda,-6+\lambda,5-\lambda)

For the nearest point, QA⃗\vec{QA} must be perpendicular to the line direction:

(5+λ)+(−6+λ)−(5−λ)=0(5+\lambda)+(-6+\lambda)-(5-\lambda)=0 3λ−6=0  ⟹  λ=23\lambda-6=0 \implies \lambda=2

So the nearest point is A(7,−2,1)A(7,-2,1) and hence

QA=(7−0)2+(−2−2)2+(1−(−2))2=49+16+9=74QA = \sqrt{(7-0)^2+(-2-2)^2+(1-(-2))^2} = \sqrt{49+16+9} = \sqrt{74}

Therefore, the correct option is D.

Common mistakes

  • Using the formula ∣(q⃗−p⃗)⋅d⃗∣∣d⃗∣\dfrac{|(\vec{q}-\vec{p})\cdot\vec{d}|}{|\vec{d}|} for distance from a point to a line. That expression gives the component parallel to the line direction, not the perpendicular distance. Instead, either use the nearest-point condition or the cross-product form for point-to-line distance.

  • Making an error in the cross product of the two direction vectors. A wrong perpendicular direction changes the entire required line. Compute the determinant carefully and then simplify only after obtaining the correct vector.

  • Forgetting that any non-zero scalar multiple of the direction vector represents the same line direction. After getting (−9,−9,9)(-9,-9,9), it is valid and easier to use (1,1,−1)(1,1,-1) in the perpendicularity condition.

Practice more Equation of Line in 3D questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Equation of Line in 3D

Related questions