If line 2 x -8-sin- y -sin-1- z -1-cos- - R - sin-1 lies in the plane 2 x -sin- y -cos- z - k - R- thenA-k -8-sin-B- k -8-cos-C- None of theseD- k -8-sin-

The correct option is D k=8+sinαPoint (4, sinα, 1) lies in the plane nbsp;2x (sinβ)y+(cosβ)z=k ∀ α∈ R ⇒ 8 sinα sinβ + cosβ = k …… (1) Line lie in plan ...

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Byju's Answer

Standard XII

Mathematics

Perpendicular Distance of a Point from a Line

If line 2 x -...

Question

If line $\frac{2 x - 8}{sin β} = \frac{y - sin α}{1} = \frac{z - 1}{cos α}, β \in R, sin β \neq 1$ lies in the plane $2 x - (sin β) y + (cos β) z = k \forall α \in R$ , then

k = 8 - sin α

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k = 8 + sin α

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k = 8 - cos β

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None of these

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Solution

The correct option is B $k = 8 + sin α$
Point $(4, sin α, 1)$ lies in the plane $2 x - (sin β) y + (cos β) z = k \forall α \in R$

$\Rightarrow 8 - sin α sin β + cos β = k \dots \dots (1)$

Line lie in plane

$\Rightarrow sin β - sin β + cos α cos β = 0 \dots \dots (2) \Rightarrow cos β = 0 & sin β = - 1$

$(1)$ and $(2)$ are satisfied for all real values of $α$

$\Rightarrow k = 8 + sin α$

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If line 2x−8sinβ=y−sinα1=z−1cosα, β∈R, sinβ≠1 lies in the plane 2x−(sinβ)y+(cosβ)z=k ∀ α∈R, then

The correct option is B k=8+sinαPoint (4, sinα, 1) lies in the plane 2x−(sinβ)y+(cosβ)z=k ∀ α∈R ⇒8−sinα sinβ+cosβ=k……(1) Line lie in plane ⇒sinβ−sinβ+cosα cosβ=0……(2)⇒cosβ=0 &sinβ=−1 (1) and (2) are satisfied for all real values of α ⇒k=8+sinα

If line $\frac{2 x - 8}{sin β} = \frac{y - sin α}{1} = \frac{z - 1}{cos α}, β \in R, sin β \neq 1$ lies in the plane $2 x - (sin β) y + (cos β) z = k \forall α \in R$ , then