A vector overrightarrow{V} in the first octan | vedclass

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(C) Let the unit vector be $\hat{v} = \cos 60^{\circ} \hat{i} + \cos 45^{\circ} \hat{j} + \cos \gamma \hat{k}$.Since $\cos^2 \alpha + \cos^2 \beta + \

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$a + \sqrt{2} b + c = 1$

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(C) Let the unit vector be $\hat{v} = \cos 60^{\circ} \hat{i} + \cos 45^{\circ} \hat{j} + \cos \gamma \hat{k}$.Since $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$,we have $\cos^2 60^{\circ} + \cos^2 45^{\circ} + \cos^2 \gamma = 1$.$\Rightarrow \frac{1}{4} + \frac{1}{2} + \cos^2 \gamma = 1$.$\Rightarrow \cos^2 \gamma = 1 - \frac{3}{4} = \frac{1}{4}$.Since $\gamma$ is an acute angle,$\cos \gamma = \frac{1}{2}$.Thus,the normal vector to the plane is $\vec{n} = \frac{1}{2} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{2} \hat{k}$.The equation of the plane passing through $(\sqrt{2}, -1, 1)$ is $\frac{1}{2}(x - \sqrt{2}) + \frac{1}{\sqrt{2}}(y + 1) + \frac{1}{2}(z - 1) = 0$.Multiplying by $2$,we get $(x - \sqrt{2}) + \sqrt{2}(y + 1) + (z - 1) = 0$.$\Rightarrow x - \sqrt{2} + \sqrt{2} y + \sqrt{2} + z - 1 = 0$.$\Rightarrow x + \sqrt{2} y + z = 1$.Since the point $(a, b, c)$ lies on the plane,we have $a + \sqrt{2} b + c = 1$.

$A$ vector $\overrightarrow{V}$ in the first octant is inclined to the $x$-axis at $60^{\circ}$,to the $y$-axis at $45^{\circ}$ and to the $z$-axis at an acute angle. If a plane passing through the points $(\sqrt{2}, -1, 1)$ and $(a, b, c)$ is normal to $\overrightarrow{V}$,then:

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