The angle made by the vector 2hat{i}-hat{j}+h | vedclass

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(A) The normal vector to the plane is $\vec{n} = \hat{i} + \hat{j} + 2\hat{k}$.Let the given vector be $\vec{v} = 2\hat{i} - \hat{j} + \hat{k}$.The an

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) The normal vector to the plane is $\vec{n} = \hat{i} + \hat{j} + 2\hat{k}$.Let the given vector be $\vec{v} = 2\hat{i} - \hat{j} + \hat{k}$.The angle $\alpha$ between the vector $\vec{v}$ and the normal $\vec{n}$ is given by $\cos \alpha = \frac{|\vec{v} \cdot \vec{n}|}{|\vec{v}| |\vec{n}|}$.$\vec{v} \cdot \vec{n} = (2)(1) + (-1)(1) + (1)(2) = 2 - 1 + 2 = 3$.$|\vec{v}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$.$|\vec{n}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.$\cos \alpha = \frac{3}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$.Thus,$\alpha = 60^{\circ}$.The angle $	heta$ between the vector and the plane is given by $	heta = 90^{\circ} - \alpha = 90^{\circ} - 60^{\circ} = 30^{\circ}$.

The angle made by the vector $2\hat{i}-\hat{j}+\hat{k}$ with the plane represented by $\vec{r} \cdot(\hat{i}+\hat{j}+2\hat{k})=7$ is (in $^{\circ}$)

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