If the line frac{2x - 8}{sin beta} = frac{y - | vedclass

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(B) The line passes through the point $P(4, \sin \alpha, 1)$. Since the line lies in the plane $2x - (\sin \beta)y + (\cos \beta)z = k$,the point $P$

Vedclass · Accepted Answer

$k = 8 + \sin \alpha$

Vedclass · Answer

(B) The line passes through the point $P(4, \sin \alpha, 1)$. Since the line lies in the plane $2x - (\sin \beta)y + (\cos \beta)z = k$,the point $P$ must satisfy the plane equation:$2(4) - (\sin \beta)(\sin \alpha) + (\cos \beta)(1) = k$$8 - \sin \alpha \sin \beta + \cos \beta = k \quad \dots(1)$Also,the direction vector of the line $\vec{v} = (\frac{1}{2} \sin \beta, 1, \cos \alpha)$ must be perpendicular to the normal of the plane $\vec{n} = (2, -\sin \beta, \cos \beta)$:$\vec{v} \cdot \vec{n} = 0$$(\frac{1}{2} \sin \beta)(2) + (1)(-\sin \beta) + (\cos \alpha)(\cos \beta) = 0$$\sin \beta - \sin \beta + \cos \alpha \cos \beta = 0$$\cos \alpha \cos \beta = 0$Since this must hold for all $\alpha \in R$,we must have $\cos \beta = 0$.Substituting $\cos \beta = 0$ into equation $(1)$:$8 - \sin \alpha \sin \beta + 0 = k$Since $\cos \beta = 0$,$\sin \beta = \pm 1$. Given $\sin \beta 
eq 1$,we have $\sin \beta = -1$.Thus,$k = 8 - \sin \alpha (-1) = 8 + \sin \alpha$.

If the line $\frac{2x - 8}{\sin \beta} = \frac{y - \sin \alpha}{1} = \frac{z - 1}{\cos \alpha}$,where $\beta \in R$ and $\sin \beta \neq 1$,lies in the plane $2x - (\sin \beta)y + (\cos \beta)z = k$ for all $\alpha \in R$,then:

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