A non-zero vector a is parallel to the line o | vedclass

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(A) The plane containing vectors $i$ and $i + j$ has a normal vector $n_1 = i 	imes (i + j) = i 	imes i + i 	imes j = 0 + k = k.$ The equation of t

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

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(A) The plane containing vectors $i$ and $i + j$ has a normal vector $n_1 = i 	imes (i + j) = i 	imes i + i 	imes j = 0 + k = k.$ The equation of this plane is $r \cdot k = 0,$ which implies $z = 0.$ The plane containing vectors $i - j$ and $i + k$ has a normal vector $n_2 = (i - j) 	imes (i + k) = i 	imes i + i 	imes k - j 	imes i - j 	imes k = 0 - j + k - i = -i - j + k.$ The equation of this plane is $r \cdot (-i - j + k) = 0,$ which implies $-x - y + z = 0$ or $x + y - z = 0.$ The vector $a$ is parallel to the line of intersection of these two planes,so $a$ must be parallel to the cross product of the normals $n_1$ and $n_2$: $v = n_1 	imes n_2 = k 	imes (-i - j + k) = -(k 	imes i) - (k 	imes j) + (k 	imes k) = -j + i + 0 = i - j.$ Thus,$a$ is parallel to $i - j.$ Let $	heta$ be the angle between $a = i - j$ and $b = i - 2j + 2k.$ $\cos 	heta = \frac{a \cdot b}{|a| |b|} = \frac{(1)(1) + (-1)(-2) + (0)(2)}{\sqrt{1^2 + (-1)^2} \sqrt{1^2 + (-2)^2 + 2^2}} = \frac{1 + 2}{\sqrt{2} \sqrt{1 + 4 + 4}} = \frac{3}{\sqrt{2} \cdot 3} = \frac{1}{\sqrt{2}}.$ Since $a$ can be in the direction of $i - j$ or $-(i - j)$,the angle $	heta$ satisfies $\cos 	heta = \pm \frac{1}{\sqrt{2}}.$ Therefore,$	heta = \frac{\pi}{4}$ or $	heta = \frac{3\pi}{4}$.

$A$ non-zero vector $a$ is parallel to the line of intersection of the plane determined by the vectors $i, i + j$ and the plane determined by the vectors $i - j, i + k.$ The angle between $a$ and the vector $i - 2j + 2k$ is

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