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MathematicsSingle correct+4 / -1Question 16 of 90

JEE Main 2022 · 30 Jun · Shift 1, Mathematics question 16

The distance of the point (3,2,−1)(3,2,-1) from the plane 3x−y+4z+1=03x-y+4z+1=0 along the line 2−x2=y−32=z+11\dfrac{2-x}2=\dfrac{y-3}2=\dfrac{z+1}1 is equal to :

Choose one option
Answer and worked solution

Official answer: C

Along the line (3−2t,2+2t,−1+t)(3-2t,2+2t,-1+t): 4−4t=04-4t=0, t=1t=1; direction length 3, so distance 3.

(C)

Topics:Angles and distances
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