A non-zero vector vec{a} is parallel to the l | vedclass

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(A) The first plane is defined by vectors $\vec{n}_1 = \vec{i}$ and $\vec{n}_2 = \vec{i} + \vec{j}$. The normal to this plane is $\vec{N}_1 = \vec{n}_

Vedclass · Accepted Answer

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

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(A) The first plane is defined by vectors $\vec{n}_1 = \vec{i}$ and $\vec{n}_2 = \vec{i} + \vec{j}$. The normal to this plane is $\vec{N}_1 = \vec{n}_1 	imes \vec{n}_2 = \vec{i} 	imes (\vec{i} + \vec{j}) = \vec{k}$.The second plane is defined by vectors $\vec{n}_3 = \vec{i} - \vec{j}$ and $\vec{n}_4 = \vec{i} + \vec{k}$. The normal to this plane is $\vec{N}_2 = \vec{n}_3 	imes \vec{n}_4 = (\vec{i} - \vec{j}) 	imes (\vec{i} + \vec{k}) = \vec{i} 	imes \vec{i} + \vec{i} 	imes \vec{k} - \vec{j} 	imes \vec{i} - \vec{j} 	imes \vec{k} = 0 - \vec{j} + \vec{k} - \vec{i} = -\vec{i} - \vec{j} + \vec{k}$.The vector $\vec{a}$ is parallel to the line of intersection,so $\vec{a}$ is parallel to $\vec{N}_1 	imes \vec{N}_2 = \vec{k} 	imes (-\vec{i} - \vec{j} + \vec{k}) = -(\vec{k} 	imes \vec{i}) - (\vec{k} 	imes \vec{j}) + (\vec{k} 	imes \vec{k}) = -\vec{j} + \vec{i} + 0 = \vec{i} - \vec{j}$.Let $\vec{b} = \vec{i} - 2\vec{j} + 2\vec{k}$. The angle $	heta$ between $\vec{a}$ and $\vec{b}$ is given by $\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|}$.$\vec{a} \cdot \vec{b} = (1)(1) + (-1)(-2) + (0)(2) = 1 + 2 = 3$.$|\vec{a}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.$|\vec{b}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = 3$.$\cos 	heta = \frac{3}{\sqrt{2} \cdot 3} = \frac{1}{\sqrt{2}}$.Thus,$	heta = \frac{\pi}{4}$. Since the vector $\vec{a}$ could be in the opposite direction,the angle could also be $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$.

$A$ non-zero vector $\vec{a}$ is parallel to the line of intersection of the planes defined by the vectors $\vec{i}, \vec{i} + \vec{j}$ and $\vec{i} - \vec{j}, \vec{i} + \vec{k}$. The angle between $\vec{a}$ and the vector $\vec{i} - 2\vec{j} + 2\vec{k}$ is .....

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