If the line frac{x-3}{2}=frac{y+5}{-1}=frac{z | vedclass

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(D) The line passes through the point $P(3, -5, -2)$. Since the line lies in the plane $\alpha x + 3y - z + \beta = 0$,the point $P$ must satisfy the

Vedclass · Accepted Answer

$\frac{5}{2}, \frac{11}{2}$

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(D) The line passes through the point $P(3, -5, -2)$. Since the line lies in the plane $\alpha x + 3y - z + \beta = 0$,the point $P$ must satisfy the plane equation: $\alpha(3) + 3(-5) - (-2) + \beta = 0$ $3\alpha - 15 + 2 + \beta = 0$ $3\alpha + \beta = 13$ (Equation $1$). Also,the direction vector of the line $\vec{v} = 2\hat{i} - \hat{j} + 2\hat{k}$ must be perpendicular to the normal vector of the plane $\vec{n} = \alpha\hat{i} + 3\hat{j} - \hat{k}$. Thus,$\vec{v} \cdot \vec{n} = 0$: $(2)(\alpha) + (-1)(3) + (2)(-1) = 0$ $2\alpha - 3 - 2 = 0$ $2\alpha = 5$ $\alpha = \frac{5}{2}$. Substituting $\alpha = \frac{5}{2}$ into Equation $1$: $3(\frac{5}{2}) + \beta = 13$ $\frac{15}{2} + \beta = 13$ $\beta = 13 - \frac{15}{2} = \frac{26-15}{2} = \frac{11}{2}$. Therefore,the values are $\alpha = \frac{5}{2}$ and $\beta = \frac{11}{2}$.

If the line $\frac{x-3}{2}=\frac{y+5}{-1}=\frac{z+2}{2}$ lies in the plane $\alpha x+3y-z+\beta=0$,then the values of $\alpha$ and $\beta$ respectively are ....

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