If the point (2, alpha, beta) lies on the pla | vedclass

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(B) Let the points be $A(7, 0, 6)$ and $B(3, 4, 2)$.The vector $\vec{AB} = (3-7)\hat{i} + (4-0)\hat{j} + (2-6)\hat{k} = -4\hat{i} + 4\hat{j} - 4\hat{k

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) Let the points be $A(7, 0, 6)$ and $B(3, 4, 2)$.The vector $\vec{AB} = (3-7)\hat{i} + (4-0)\hat{j} + (2-6)\hat{k} = -4\hat{i} + 4\hat{j} - 4\hat{k}$.We can simplify this direction vector as $\vec{v} = \hat{i} - \hat{j} + \hat{k}$.The plane is also perpendicular to the plane $2x - 5y = 15$,which has a normal vector $\vec{n_1} = 2\hat{i} - 5\hat{j} + 0\hat{k}$.The normal vector $\vec{n}$ to the required plane is the cross product of $\vec{v}$ and $\vec{n_1}$:$\vec{n} = \vec{v} 	imes \vec{n_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \ 2 & -5 & 0 \end{vmatrix} = \hat{i}(0 - (-5)) - \hat{j}(0 - 2) + \hat{k}(-5 - (-2)) = 5\hat{i} + 2\hat{j} - 3\hat{k}$.The equation of the plane passing through $(7, 0, 6)$ is $5(x-7) + 2(y-0) - 3(z-6) = 0$,which simplifies to $5x + 2y - 3z = 17$.Since the point $(2, \alpha, \beta)$ lies on this plane,we substitute the coordinates:$5(2) + 2(\alpha) - 3(\beta) = 17$$10 + 2\alpha - 3\beta = 17$$2\alpha - 3\beta = 7$.

If the point $(2, \alpha, \beta)$ lies on the plane which passes through the points $(3, 4, 2)$ and $(7, 0, 6)$ and is perpendicular to the plane $2x - 5y = 15$,then $2\alpha - 3\beta$ is equal to:

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