Let P be a plane passing through the points ( | vedclass

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Question

(A) The normal vector $\vec{n}_P$ to the plane $P$ is given by the cross product of two vectors in the plane. Let $A=(1,0,1), B=(1,-2,1), C=(0,1,-2)$.

Vedclass · Accepted Answer

$81$

Vedclass · Answer

(A) The normal vector $\vec{n}_P$ to the plane $P$ is given by the cross product of two vectors in the plane. Let $A=(1,0,1), B=(1,-2,1), C=(0,1,-2)$.$\vec{AB} = (1-1, -2-0, 1-1) = (0, -2, 0)$$\vec{AC} = (0-1, 1-0, -2-1) = (-1, 1, -3)$$\vec{n}_P = \vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & -2 & 0 \ -1 & 1 & -3 \end{vmatrix} = 6\hat{i} + 0\hat{j} - 2\hat{k} = 2(3\hat{i} - \hat{k})$.Since $\vec{a}$ is parallel to the plane $P$,$\vec{a}$ is perpendicular to $\vec{n}_P$. Also,$\vec{a}$ is perpendicular to $\vec{v} = \hat{i} + 2\hat{j} + 3\hat{k}$.Thus,$\vec{a} = k(\vec{n}_P 	imes \vec{v}) = k \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 0 & -1 \ 1 & 2 & 3 \end{vmatrix} = k(2\hat{i} - 10\hat{j} + 6\hat{k})$.Given $\vec{a} \cdot (\hat{i} + \hat{j} + 2\hat{k}) = 2$,we have $k(2(1) - 10(1) + 6(2)) = 2 \Rightarrow k(2 - 10 + 12) = 2 \Rightarrow 4k = 2 \Rightarrow k = 1/2$.So,$\vec{a} = \frac{1}{2}(2\hat{i} - 10\hat{j} + 6\hat{k}) = \hat{i} - 5\hat{j} + 3\hat{k}$.Here $\alpha = 1, \beta = -5, \gamma = 3$.Then $(\alpha - \beta + \gamma)^2 = (1 - (-5) + 3)^2 = (1 + 5 + 3)^2 = 9^2 = 81$.

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