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MCQMediumJEE 2026Equation of Line in 3D

JEE Mathematics 2026 Question with Solution

Let (α,β,γ)(\alpha, \beta, \gamma) be the co-ordinates of the foot of the perpendicular drawn from the point (5,4,2)(5, 4, 2) on the line r⃗=(−i^+3j^+k^)+λ(2i^+3j^−k^)\vec{r}=(-\hat{i}+3\hat{j}+\hat{k})+\lambda(2\hat{i}+3\hat{j}-\hat{k}). Then the length of the projection of the vector αi^+βj^+γk^\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k} on the vector 6i^+2j^+3k^6\hat{i}+2\hat{j}+3\hat{k} is:

  • A

    33

  • B

    157\frac{15}{7}

  • C

    187\frac{18}{7}

  • D

    44

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The point is P(5,4,2)P(5,4,2) and the line is

r⃗=(−i^+3j^+k^)+λ(2i^+3j^−k^)\vec{r}=(-\hat{i}+3\hat{j}+\hat{k})+\lambda(2\hat{i}+3\hat{j}-\hat{k})

Find: The length of the projection of αi^+βj^+γk^\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k} on 6i^+2j^+3k^6\hat{i}+2\hat{j}+3\hat{k}, where (α,β,γ)(\alpha,\beta,\gamma) is the foot of the perpendicular from PP to the line.

Any point on the line is

F=(−1+2λ, 3+3λ, 1−λ)F=(-1+2\lambda,\ 3+3\lambda,\ 1-\lambda)

So,

PF⃗=F⃗−P⃗=((−1+2λ)−5)i^+((3+3λ)−4)j^+((1−λ)−2)k^\vec{PF}=\vec{F}-\vec{P}=(( -1+2\lambda)-5)\hat{i}+((3+3\lambda)-4)\hat{j}+((1-\lambda)-2)\hat{k} PF⃗=(2λ−6)i^+(3λ−1)j^+(−λ−1)k^\vec{PF}=(2\lambda-6)\hat{i}+(3\lambda-1)\hat{j}+(-\lambda-1)\hat{k}

Since PF⃗\vec{PF} is perpendicular to the direction vector d⃗=2i^+3j^−k^\vec{d}=2\hat{i}+3\hat{j}-\hat{k},

PF⃗⋅d⃗=0\vec{PF}\cdot \vec{d}=0 (2λ−6)(2)+(3λ−1)(3)+(−λ−1)(−1)=0(2\lambda-6)(2)+(3\lambda-1)(3)+(-\lambda-1)(-1)=0 4λ−12+9λ−3+λ+1=04\lambda-12+9\lambda-3+\lambda+1=0 14λ−14=014\lambda-14=0 λ=1\lambda=1

Therefore,

α=−1+2(1)=1,β=3+3(1)=6,γ=1−1=0\alpha=-1+2(1)=1,\quad \beta=3+3(1)=6,\quad \gamma=1-1=0

So the foot of the perpendicular is F(1,6,0)F(1,6,0).

Now,

v⃗=αi^+βj^+γk^=i^+6j^\vec{v}=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}=\hat{i}+6\hat{j}

and

u⃗=6i^+2j^+3k^\vec{u}=6\hat{i}+2\hat{j}+3\hat{k}

The length of projection of v⃗\vec{v} on u⃗\vec{u} is

∣v⃗⋅u⃗∣∣u⃗∣\frac{|\vec{v}\cdot \vec{u}|}{|\vec{u}|}

Now,

v⃗⋅u⃗=(1)(6)+(6)(2)+(0)(3)=18\vec{v}\cdot \vec{u}=(1)(6)+(6)(2)+(0)(3)=18

and

∣u⃗∣=62+22+32=36+4+9=49=7|\vec{u}|=\sqrt{6^2+2^2+3^2}=\sqrt{36+4+9}=\sqrt{49}=7

Hence, the projection length is

187\frac{18}{7}

Therefore, the correct option is C.

Common mistakes

  • Taking the vector from the point on the line to PP incorrectly. If PF⃗\vec{PF} or FP⃗\vec{FP} is written with wrong coordinates, the perpendicularity equation becomes wrong. Compute the coordinate differences carefully before applying the dot product condition.

  • Using the projection formula without dividing by the magnitude of the vector on which projection is taken. The scalar projection is ∣v⃗⋅u⃗∣∣u⃗∣\frac{|\vec{v}\cdot\vec{u}|}{|\vec{u}|}, not just v⃗⋅u⃗\vec{v}\cdot\vec{u}.

  • Substituting the value of λ\lambda incorrectly into the line coordinates. After finding λ=1\lambda=1, evaluate all three coordinates of the foot carefully to get (1,6,0)(1,6,0).

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