Line passes through (2,−3,1) with direction (2,4,1).
Vector from (3,2,0) to (2,−3,1): (−1,−5,1).
Normal =(−1,−5,1)×(2,4,1)=(−9,3,6), or (3,−1,−2).
Plane: 3(x−3)−(y−2)−2z=0, i.e. 3x−y−2z=7.
CUET UG 2022 — Mathematics Geometry
The equation of plane passing through the point of (3, 2, 0) and containing the line 2x−2=4y+3=1z−1 is
Held on 6 Aug 2022 · Verified 13 Jul 2026.
3x−y−2z=7
3x−y+2z=7
x+y−2z=5
x−y+2z=1
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Consider the line $\vec{r} = -2\hat{i} + 3\hat{j} + \hat{k} + \lambda(5\hat{i} - 3\hat{j} - \hat{k})$. Match List-I with List-II | List-I | List-II | |---|---| | (A) A point on the given line | (I) $\left(\frac{5}{\sqrt{35}}, \frac{-3}{\sqrt{35}}, \frac{-1}{\sqrt{35}}\right)$ | | (B) Direction ratios of the given line | (II) (2, 3, 1) | | (C) Direction cosines of the given line | (III) (5, -3, -1) | | (D) Direction ratios of a line perpendicular to given line | (IV) (-2, 3, 1) | Choose the correct answer from the options given below:
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