Match the following. List-I List-II (A) The vector equation of the plane perpendicular to the line 1 x − 1 = − 1 y − 2 = 2 z + 1 and passing through the point ( 3 , 6 , 2 ) is. (p) r ⋅ ( i ^ − 3

Question

Match the following.

List-I	List-II
(A) The vector equation of the plane perpendicular to the line $\frac{x - 1}{1} = \frac{y - 2}{- 1} = \frac{z + 1}{2}$ and passing through the point $(3, 6, 2)$ is.	(p) $r \cdot (\hat{i} - 3 \hat{j} - 2 \hat{k}) = 3$
(B) The vector equation of the plane through the point $(5, - 2, 4)$ and parallel to the plane $4 x - 12 y - 8 z = 7$ .	(q) $r \cdot (\hat{i} - \hat{j} - \hat{k}) = 2$
(C) The vector equation of the plane containing the line $r = 2 \hat{i} + λ (\hat{j} - \hat{k})$ and perpendicular to the plane $r \cdot (\hat{i} + \hat{k}) = 3$ is.	(r) $r \cdot (\hat{i} - \hat{j} - \hat{k}) = 0$
(D) The vector equation of the plane containing the lines $\frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z}{- 1}$ ; $\frac{x - 1}{- 1} = \frac{y - 1}{1} = \frac{z}{- 2}$ is.	(s) $r \cdot (\hat{i} - \hat{j} + 2 \hat{k}) = 1$
	(t) $r \cdot (\hat{i} - \hat{j} - \hat{k}) = 1$

TestHub · Accepted Answer

(A) Normal to the plane is along the line: n = i ^ − j ^ ​ + 2 k ^ ; plane through ( 3 , 6 , 2 ) : r ⋅ ( i ^ − j ^ ​ + 2 k ^ ) = 3 − 6 + 4 = 1 → s. (B) Parallel plane has normal ( 4 , − 12 , − 8 ) ∼ ( 1 , − 3 , − 2 ) ; through ( 5 , − 2 , 4 ) : r ⋅ ( i ^ − 3 j ^ ​ − 2 k ^ ) = 5 + 6 − 8 = 3 → p. (C) Plane contains direction (0, 1, -1) and is perpendicular to normal (1, 0, 1), so its normal is ( 0 , 1 , − 1 ) × ( 1 , 0 , 1 ) = ( 1 , − 1 , − 1 ) ; it passes through ( 2 , 0 , 0 ) : r ⋅ ( i ^ − j ^ ​ − k ^ ) = 2 → q. (D) Normal = ( 1 , 2 , − 1 ) × ( − 1 , 1 , − 2 ) = ( − 3 , 3 , 3 ) ∼ ( 1 , − 1 , − 1 ) ; through ( 1 , 1 , 0 ) : r ⋅ ( i ^ − j ^ ​ − k ^ ) = 0 → r .

Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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