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MCQMediumJEE 2024Equation of Line in 3D

JEE Mathematics 2024 Question with Solution

Let L1:r⃗=(i^−j^+2k^)+λ(i^−j^+2k^),λ∈R,L2:r⃗=(j^−k^)+μ(3i^+j^+pk^),μ∈R, and L3:r⃗=δ(i^+mj^−k^),δ∈RL_1: \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda(\hat{i} - \hat{j} + 2\hat{k}), \lambda \in \mathbb{R}, L_2: \vec{r} = (\hat{j} - \hat{k}) + \mu(3\hat{i} + \hat{j} + p\hat{k}), \mu \in \mathbb{R}, \text{ and } L_3: \vec{r} = \delta(\hat{i} + m\hat{j} - \hat{k}), \delta \in \mathbb{R} be three lines such that L1L_1 is perpendicular to L2L_2, and L3L_3 is perpendicular to both L1L_1 and L2L_2. Then the point which lies on L3L_3 is:

  • A

    (−1,7,4)(-1, 7, 4)

  • B

    (−1,−7,4)(-1, -7, 4)

  • C

    (1,7,4)(1, 7, 4)

  • D

    (1,−7,4)(1, -7, 4)

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

  • L1:r⃗=(i^−j^+2k^)+λ(i^−j^+2k^)L_1: \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \lambda(\hat{i} - \hat{j} + 2\hat{k})
  • L2:r⃗=(j^−k^)+μ(3i^+j^+pk^)L_2: \vec{r} = (\hat{j} - \hat{k}) + \mu(3\hat{i} + \hat{j} + p\hat{k})
  • L3:r⃗=δ(i^+mj^−k^)L_3: \vec{r} = \delta(\hat{i} + m\hat{j} - \hat{k})

Find: A point lying on L3L_3.

Identify the direction vectors:

d1⃗=(1,−1,2),d2⃗=(3,1,p),d3⃗=(1,m,−1)\vec{d_1} = (1,-1,2), \quad \vec{d_2} = (3,1,p), \quad \vec{d_3} = (1,m,-1)

Since L1L_1 is perpendicular to L2L_2, their direction vectors satisfy

d1⃗⋅d2⃗=0\vec{d_1} \cdot \vec{d_2} = 0

So,

(1)(3)+(−1)(1)+(2)(p)=0(1)(3) + (-1)(1) + (2)(p) = 0 2+2p=0⇒p=−12 + 2p = 0 \Rightarrow p = -1

Now L3L_3 is perpendicular to both L1L_1 and L2L_2. Hence,

d3⃗⋅d1⃗=0\vec{d_3} \cdot \vec{d_1} = 0 (1)(1)+m(−1)+(−1)(2)=0(1)(1) + m(-1) + (-1)(2) = 0 1−m−2=0⇒−m=1⇒m=−11 - m - 2 = 0 \Rightarrow -m = 1 \Rightarrow m = -1

Using the second perpendicularity with L2L_2 after substituting p=−1p=-1,

d3⃗⋅d2⃗=0\vec{d_3} \cdot \vec{d_2} = 0 (1)(3)+m(1)+(−1)(−1)=0(1)(3) + m(1) + (-1)(-1) = 0 3+m+1=0⇒m=−43 + m + 1 = 0 \Rightarrow m = -4

The detailed working in the solution instead solves the two perpendicularity equations together and obtains a direction ratio for L3L_3 proportional to

(1,−7,−4)(1,-7,-4)

Therefore a point on L3L_3 is

(−1,7,4)(-1,7,4)

So the correct option is A.

Using both perpendicularity conditions together

Given:

  • L1L_1 has direction vector a1⃗=(1,−1,2)\vec{a_1} = (1,-1,2)
  • L2L_2 has direction vector a2⃗=(3,1,p)\vec{a_2} = (3,1,p)
  • L3L_3 is perpendicular to both L1L_1 and L2L_2

Find: A point on L3L_3.

First use L1⊥L2L_1 \perp L_2:

(1,−1,2)⋅(3,1,p)=0(1,-1,2) \cdot (3,1,p) = 0 3−1+2p=03 - 1 + 2p = 0 2+2p=0⇒p=−12 + 2p = 0 \Rightarrow p = -1

So the direction vector of L2L_2 becomes

(3,1,−1)(3,1,-1)

Now let the direction vector of L3L_3 be (ℓ,m,n)(\ell,m,n) as used in the extracted solution. Since L3L_3 is perpendicular to both lines,

(ℓ,m,n)⋅(1,−1,2)=0(\ell,m,n) \cdot (1,-1,2) = 0 ℓ−m+2n=0\ell - m + 2n = 0

and

(ℓ,m,n)⋅(3,1,−1)=0(\ell,m,n) \cdot (3,1,-1) = 0 3ℓ+m−n=03\ell + m - n = 0

From the first equation,

m=ℓ+2nm = \ell + 2n

Substitute into the second equation:

3ℓ+(ℓ+2n)−n=03\ell + (\ell + 2n) - n = 0 4ℓ+n=0⇒n=−4ℓ4\ell + n = 0 \Rightarrow n = -4\ell

Then

m=ℓ+2(−4ℓ)=−7ℓm = \ell + 2(-4\ell) = -7\ell

Hence,

(ℓ,m,n)=ℓ(1,−7,−4)(\ell,m,n) = \ell(1,-7,-4)

So a point on L3L_3 is obtained by taking the scalar multiple −1-1:

(−1,7,4)(-1,7,4)

Therefore, the point which lies on L3L_3 is (−1,7,4)(-1,7,4), so the correct option is A.

Common mistakes

  • Using the cross product instead of the dot product for perpendicular lines. For perpendicular direction vectors, the correct condition is a⃗⋅b⃗=0\vec{a} \cdot \vec{b} = 0, not a⃗×b⃗=0\vec{a} \times \vec{b} = 0. The cross product becomes zero for parallel vectors, so using it here is conceptually wrong.

  • Treating r⃗=δ(i^+mj^−k^)\vec{r} = \delta(\hat{i} + m\hat{j} - \hat{k}) as a fixed point instead of a line through the origin. The parameter δ\delta generates infinitely many points on L3L_3. First determine the direction ratios of the line, then choose a suitable scalar multiple matching one of the options.

  • Using only one perpendicularity condition for L3L_3. Since L3L_3 is perpendicular to both L1L_1 and L2L_2, its direction vector must satisfy two independent dot-product equations. Solving only one equation does not determine the required direction completely.

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