Let L_1: =( i- j+2 k)+λ ( i- j+2 k),λ ∈ R, L_2: =( j- k)+μ (3 i+ j+p k),μ ∈ R and L_3: =δ (l i+m j+n k),δ ∈ R be three lines such that L_1 is…
JEE Main 2024 — Mathematics Vectors & 3D Geometry
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Let L1:r=(i^−j^+2k^)+λ(i^−j^+2k^),λ∈R, L2:r=(j^−k^)+μ(3i^+j^+pk^),μ∈R and L3:r=δ(li^+mj^+nk^),δ∈R be three lines such that L1 is perpendicular to L2 and L3 is perpendicular to both L1 and L2. Then the point which lies on L3 is
Official previous-year question
Held on 30 Jan 2024 · Verified 6 Jul 2026.
Options
A
(−1,7,4)
B
(−1,−7,4)
C
(1,7,−4)
D
(1,−7,4)
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Solution
Given: L1⊥L2
⇒(i^−j^+2k^).(3i^+j^+pk^)=0
⇒3−1+2p=0
⇒p=−1
Also, L3⊥L1,L2
So, L3∥(L1×L2)
⇒L1×L2=∣i^13j^−11k^2−1∣
⇒L1×L2=−i^+7j^+4k^
On comparing with L3:r=δ(li^+mj^+nk^), we get that (−δ,7δ,4δ) will lie on L3.
Now, for δ=1 the point will be (−1,7,4).
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