A vector vec{a} is parallel to the line of in | vedclass

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(A) The normal to the first plane is $\vec{n}_{1} = \hat{i} 	imes (\hat{i} + \hat{j}) = \hat{k}$.The normal to the second plane is $\vec{n}_{2} = (\h

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$\frac{3\pi}{4}$

Vedclass · Answer

(A) The normal to the first plane is $\vec{n}_{1} = \hat{i} 	imes (\hat{i} + \hat{j}) = \hat{k}$.The normal to the second plane is $\vec{n}_{2} = (\hat{i} - \hat{j}) 	imes (\hat{i} + \hat{k}) = \hat{i} + \hat{j} + \hat{k}$.The line of intersection is parallel to $\vec{v} = \vec{n}_{1} 	imes \vec{n}_{2} = \hat{k} 	imes (\hat{i} + \hat{j} + \hat{k}) = -\hat{i} + \hat{j}$.Thus,$\vec{a} = -\hat{i} + \hat{j}$.Given $\vec{b} = \hat{i} - 2\hat{j} + 2\hat{k}$.The cosine of the angle $	heta$ between $\vec{a}$ and $\vec{b}$ is given by $\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.$\vec{a} \cdot \vec{b} = (-1)(1) + (1)(-2) + (0)(2) = -1 - 2 = -3$.$|\vec{a}| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$.$|\vec{b}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = 3$.$\cos 	heta = \frac{-3}{\sqrt{2} 	imes 3} = -\frac{1}{\sqrt{2}}$.Since $\cos 	heta = -\frac{1}{\sqrt{2}}$,the obtuse angle is $	heta = \frac{3\pi}{4}$.

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