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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

k = 8 + sin α

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

k = 8 - cos β

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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 $α$

$\Rightarrow k = 8 + sin α$

Suggest Corrections

Similar questions

Q. If

sin α - sin β = 1 / 3

and

cos β - cos α = 1 / 2,

then

cot \frac{α + β}{2} = \frac{k}{3}

. Find the value of

k

Q. If

a = cos α + i sin α, b = cos β + i sin β,

c = cos γ + i sin γ

and

\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = 1,

then

Q. If

x = sin α, y = sin β, z = sin (α + β)

, then

cos (α + β)

is equal to

Q. If cos(α + β) = 0, then sin(α − β) = ?
(a) sin α
(b) cos β
(c) sin 2α
(d) cos 2β

cosα+cosβ=a, sinα+sinβ=b and α-β = 2θ. If $\frac{c o s 3 θ}{c o s θ}$ = K, when a= $\frac{1}{3}$ ,b= $\frac{1}{2}$ Find the value of

-36K.

Join BYJU'S Learning Program

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