Let P be the plane passing through the line f | vedclass

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(D) The line passes through $A(1, 2, -5)$ and has direction vector $\vec{v} = \langle 1, -3, 7 \rangle$. The plane also passes through $B(2, 4, -3)$.V

Vedclass · Accepted Answer

$10$

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(D) The line passes through $A(1, 2, -5)$ and has direction vector $\vec{v} = \langle 1, -3, 7 angle$. The plane also passes through $B(2, 4, -3)$.Vector $\vec{AB} = \langle 2-1, 4-2, -3-(-5) angle = \langle 1, 2, 2 angle$.The normal vector to the plane is $\vec{n} = \vec{v} 	imes \vec{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -3 & 7 \ 1 & 2 & 2 \end{vmatrix} = \hat{i}(-6-14) - \hat{j}(2-7) + \hat{k}(2+3) = -20\hat{i} + 5\hat{j} + 5\hat{k}$.We can take the normal vector as $\vec{n} = \langle 4, -1, -1 angle$.The equation of the plane is $4(x-1) - 1(y-2) - 1(z+5) = 0$,which simplifies to $4x - y - z = 3$.Let the point be $Q(-1, 3, 4)$. The image $(\alpha, \beta, \gamma)$ is given by $\frac{\alpha - (-1)}{4} = \frac{\beta - 3}{-1} = \frac{\gamma - 4}{-1} = -2 \frac{4(-1) - 3 - 4 - 3}{4^2 + (-1)^2 + (-1)^2} = -2 \frac{-14}{18} = \frac{14}{9}$.Wait,re-evaluating the plane equation: $4(1) - 2 - (-5) = 4-2+5 = 7$. So $4x-y-z=7$.Then $\frac{\alpha+1}{4} = \frac{\beta-3}{-1} = \frac{\gamma-4}{-1} = -2 \frac{4(-1)-3-4-7}{16+1+1} = -2 \frac{-18}{18} = 2$.$\alpha+1 = 8 \implies \alpha = 7$.$\beta-3 = -2 \implies \beta = 1$.$\gamma-4 = -2 \implies \gamma = 2$.$\alpha+\beta+\gamma = 7+1+2 = 10$.

Let $P$ be the plane passing through the line $\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}$ and the point $(2,4,-3)$. If the image of the point $(-1,3,4)$ in the plane $P$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta+\gamma$ is equal to

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