Skip to content
dronacharya.ai
← JEE Main 2022 · 26 Jun · Shift 2
MathematicsSingle correct+4 / -1Question 15 of 90

JEE Main 2022 · 26 Jun · Shift 2, Mathematics question 15

If the lines r⃗=(i^−j^+k^)+λ(3j^−k^)\vec r=\left(\hat i-\hat j+\hat k\right)+\lambda\left(3\hat j-\hat k\right) and r⃗=(αi^−j^)+μ(2i^−3k^)\vec r=\left(\alpha\hat i-\hat j\right)+\mu\left(2\hat i-3\hat k\right) are co-planar, then the distance of the plane containing these two lines from the point (α,0,0)(\alpha,0,0) is:

Choose one option
Answer and worked solution

Official answer: B

Normal. (0,3,−1)×(2,0,−3)=(−9,−2,−6)(0,3,-1)\times(2,0,-3)=(-9,-2,-6).

Coplanar. (α−1,0,−1)⋅(−9,−2,−6)=−9(α−1)+6=0(\alpha-1,0,-1)\cdot(-9,-2,-6)=-9(\alpha-1)+6=0, so α=53\alpha=\frac53.

Plane through (1,−1,1)(1,-1,1): 9x+2y+6z=139x+2y+6z=13.

Distance from (53,0,0)\left(\frac53,0,0\right): ∣15−13∣121=211\dfrac{|15-13|}{\sqrt{121}}=\dfrac2{11}.

(B)

Topics:Skew linesEquation of a planeAngles and distances
← Previous questionNext question →