The vector xhat{i} + yhat{j} + zhat{k} makes | vedclass

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(C) Let $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ and the two vectors defining the plane be $\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} =

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$x(y + z) = yz$

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(C) Let $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ and the two vectors defining the plane be $\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$.The normal vector $\vec{n}$ to the plane is given by $\vec{a} 	imes \vec{b}$:$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & -1 \ 1 & -1 & 2 \end{vmatrix} = \hat{i}(6 - 1) - \hat{j}(4 + 1) + \hat{k}(-2 - 3) = 5\hat{i} - 5\hat{j} - 5\hat{k}$.We can take the normal vector as $\vec{n} = \hat{i} - \hat{j} - \hat{k}$.Let $	heta$ be the angle between the vector $\vec{v}$ and the plane. Then the angle between $\vec{v}$ and the normal $\vec{n}$ is $90^{\circ} - 	heta$.Given $\cot 	heta = \sqrt{2}$,we have $	an 	heta = \frac{1}{\sqrt{2}}$,so $\sin 	heta = \frac{1}{\sqrt{3}}$ and $\cos 	heta = \sqrt{\frac{2}{3}}$.Using the dot product formula: $\cos(90^{\circ} - 	heta) = \sin 	heta = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|}$.$\frac{1}{\sqrt{3}} = \frac{|x - y - z|}{\sqrt{x^2 + y^2 + z^2} \sqrt{1^2 + (-1)^2 + (-1)^2}}$.$\frac{1}{\sqrt{3}} = \frac{|x - y - z|}{\sqrt{x^2 + y^2 + z^2} \sqrt{3}}$.Squaring both sides: $\frac{1}{3} = \frac{(x - y - z)^2}{3(x^2 + y^2 + z^2)}$.$x^2 + y^2 + z^2 = x^2 + y^2 + z^2 - 2xy + 2yz - 2zx$.$0 = -2xy + 2yz - 2zx$.$xy + zx = yz$,which is $x(y + z) = yz$.

The vector $x\hat{i} + y\hat{j} + z\hat{k}$ makes an acute angle $\cot^{-1} \sqrt{2}$ with the plane containing the vectors $(2, 3, -1)$ and $(1, -1, 2)$. Then,

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