Let a = 2i - j + k,b = i + 2j - k,and c = i + | vedclass

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(A) Any vector $r$ in the plane of $b$ and $c$ can be written as $r = \lambda b + \mu c$. For simplicity,let $r = b + tc$. Substituting the vectors,$r

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$2i + 3j - 3k$ and $-2i - j + 5k$

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(A) Any vector $r$ in the plane of $b$ and $c$ can be written as $r = \lambda b + \mu c$. For simplicity,let $r = b + tc$. Substituting the vectors,$r = (i + 2j - k) + t(i + j - 2k) = (1 + t)i + (2 + t)j - (1 + 2t)k$. The projection of $r$ on $a$ is given by $\frac{|r \cdot a|}{|a|} = \sqrt{2/3}$. First,calculate $r \cdot a = (1 + t)(2) + (2 + t)(-1) + (-1 - 2t)(1) = 2 + 2t - 2 - t - 1 - 2t = -t - 1$. Also,$|a| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$. Thus,$\frac{|-t - 1|}{\sqrt{6}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} = \frac{\sqrt{2} \cdot \sqrt{2}}{\sqrt{3} \cdot \sqrt{2}} = \frac{2}{\sqrt{6}}$. So,$|-t - 1| = 2$,which implies $-t - 1 = 2$ or $-t - 1 = -2$. If $-t - 1 = 2$,then $t = -3$. Substituting $t = -3$ into $r$,we get $r = (1 - 3)i + (2 - 3)j - (1 - 6)k = -2i - j + 5k$. If $-t - 1 = -2$,then $t = 1$. Substituting $t = 1$ into $r$,we get $r = (1 + 1)i + (2 + 1)j - (1 + 2)k = 2i + 3j - 3k$. Therefore,the vectors are $2i + 3j - 3k$ and $-2i - j + 5k$.

Let $a = 2i - j + k$,$b = i + 2j - k$,and $c = i + j - 2k$ be three vectors. $A$ vector in the plane of $b$ and $c$ whose projection on $a$ is of magnitude $\sqrt{2/3}$ is

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