Match the following. List-I List-II (A) Length of perpendicular from ( 1 , 2 , 3 ) to the line 3 x − 6 = 2 y − 7 = − 2 z − 7 is (p) 11 (B) Length of perpendicular from ( 2 , 3 , 7 ) to the pla

Question

Match the following.

List-I	List-II
(A) Length of perpendicular from $(1, 2, 3)$ to the line $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}$ is	(p) $11$
(B) Length of perpendicular from $(2, 3, 7)$ to the plane $3 x - y - z = 7$ is	(q) $1$
(C) Distance of $(1, - 2, 3)$ from the plane $x - y + z = 5$ measured along a line parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6}$ is	(r) $7$
(D) The distance of the point of intersection of the line $\frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2}$ and the plane $x + y + z = 17$ from $A (3, 4, 5)$ is	(s) $3$

TestHub · Accepted Answer

(A) Vector from (6, 7, 7) to (1, 2, 3) is (-5, -5, -4); ​ ( − 5 , − 5 , − 4 ) × 17 ​ ( 3 , 2 , − 2 ) ​ ​ = 17 ​ ∣ ( 18 , − 22 , 5 ) ∣ ​ = 17 ​ 833 ​ ​ = 7 . (B) 9 + 1 + 1 ​ ∣3 ( 2 ) − 3 − 7 − 7∣ ​ = 11 ​ 11 ​ = 11 ​ . (C) Perpendicular distance = 3 ​ ∣1 + 2 + 3 − 5∣ ​ = 3 ​ 1 ​ ; the direction ( 2 , 3 , − 6 ) makes angle with the plane normal ( 1 , − 1 , 1 ) : cos γ = 7 3 ​ ∣2 − 3 − 6∣ ​ = 3 ​ 1 ​ ; required distance = 1/ 3 ​ 1/ 3 ​ ​ = 1 . (D) Intersection: ( 3 + t , 4 + 2 t , 5 + 2 t ) with 12 + 5 t = 17 , t = 1 , point ( 4 , 6 , 7 ) ; distance from ( 3 , 4 , 5 ) = 1 + 4 + 4 ​ = 3 .

Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

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