Let the line x-23-y-1-5-z-22 lies in the plane x-3y- z-0- Then - equals

Dr's of line =(3, 5,2) Dr's of normal to the plane =(1,3, α) Line is perpendicular to normal nbsp; ⇒ 3(1) 5(3)+2( α)=0 ⇒ 3 15 2α=0 ⇒α= 6 Also (2,1, 2) lies o ...

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Byju's Answer

Standard XII

Mathematics

Equation of a Plane Passing through a Point and Parallel to the Two Given Vectors

Let the line ...

Question

Let the line $\frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2}$ lies in the plane $x + 3 y - α z + β = 0.$ Then $(α, β)$ equals

(6, - 17)

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(- 6, 7)

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(5, - 15)

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(- 5, 15)

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Solution

The correct option is B $(- 6, 7)$
Dr's of line $= (3, - 5, 2)$
Dr's of normal to the plane $= (1, 3, - α)$
Line is perpendicular to normal
$\Rightarrow 3 (1) - 5 (3) + 2 (- α) = 0 \Rightarrow 3 - 15 - 2 α = 0 \Rightarrow α = - 6$
Also $(2, 1, - 2)$ lies on the plane
$2 + 3 + 6 (- 2) + β = 0 \Rightarrow β = 7$
$∴ (α, β) = (- 6, 7)$

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Let the line x−23=y−1−5=z+22 lies in the plane x+3y−αz+β=0. Then (α,β) equals

The correct option is B (−6,7)Dr's of line =(3,−5,2) Dr's of normal to the plane =(1,3,−α) Line is perpendicular to normal ⇒3(1)−5(3)+2(−α)=0⇒3−15−2α=0⇒α=−6 Also (2,1,−2) lies on the plane 2+3+6(−2)+β=0⇒β=7 ∴(α,β)=(−6,7)

Let the line $\frac{x - 2}{3} = \frac{y - 1}{- 5} = \frac{z + 2}{2}$ lies in the plane $x + 3 y - α z + β = 0.$ Then $(α, β)$ equals