Match the following. List-I List-II (a) If the plane a x − b y + cz = 0 contains the line a x − a = b y − 2 d = c z − c then d b is equal to. (p) 1 (b) The distance of the point ( 1 , − 2 , 3

Question

Match the following.

List-I	List-II
(a) If the plane $a x - b y + cz = 0$ contains the line $\frac{x - a}{a} = \frac{y - 2 d}{b} = \frac{z - c}{c}$ then $\frac{b}{d}$ is equal to.	(p) $1$
(b) The distance of the point $(1, - 2, 3)$ from the plane $x - y + z - 5 = 0$ measured parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z - 1}{- 6}$ is equal to.	(q) $2$
(c) If the straight lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{4 - z}{k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ intersect then $k$ is equal to.	(r) $0$
(d) If a line makes an angle $θ$ with x and y axis then $cot θ$ can be equal to.	(s) $- 3$

TestHub · Accepted Answer

(a) The line passes through ( a , 2 d , c ) with direction ( a , b , c ) . Since the plane contains the line: a ⋅ a − b ⋅ b + c ⋅ c = 0 ⇒ b 2 = a 2 + c 2 and a 2 − 2 b d + c 2 = 0 ⇒ b 2 = 2 b d ⇒ b = 2 d , so d b ​ = 2 → q. (b) Point on the parallel line: ( 2 r + 1 , 3 r − 2 , 3 − 6 r ) ; it lies on the plane when 2 r + 1 − 3 r + 2 + 3 − 6 r − 5 = 0 ⇒ r = 7 1 ​ ; distance = ∣ r ∣ 4 + 9 + 36 ​ = 7 7 ​ = 1 → p . (c) Equating parametric points ( r 1 ​ + 2 , r 1 ​ + 3 , 4 − k r 1 ​ ) and ( 1 + k r 2 ​ , 4 + 2 r 2 ​ , 5 + r 2 ​ ) and eliminating gives k = 0 or k = − 3 → r, s. cos 2 θ + cos 2 θ + cos 2 γ = 1 ⇒ cos 2 γ = − cos 2 θ ≥ 0 ⇒ cos 2 θ ≤ 0 ⇒ θ ∈ [ 4 π ​ , 4 3 π ​ ] , so cot θ ∈ [ − 1 , 1 ] ; possible values 1 and 0 → p , r .

Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

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