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

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(D) Since the vector $r$ lies in the plane of $a$ and $b$,it can be expressed as a linear combination: $r = a + tb$ (assuming the vector is not parall

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$2i + j + 2k$

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(D) Since the vector $r$ lies in the plane of $a$ and $b$,it can be expressed as a linear combination: $r = a + tb$ (assuming the vector is not parallel to $b$).$r = (i + 2j + k) + t(i - j + k) = (1 + t)i + (2 - t)j + (1 + t)k$.Given that the projection of $r$ on $c$ is $\frac{1}{\sqrt{3}}$,we use the formula: $\frac{r \cdot c}{|c|} = \frac{1}{\sqrt{3}}$.First,calculate $|c| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$.Now,calculate the dot product $r \cdot c = ((1 + t)i + (2 - t)j + (1 + t)k) \cdot (i + j - k) = (1 + t) + (2 - t) - (1 + t) = 2 - t$.Substituting these into the projection formula: $\frac{2 - t}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.This implies $2 - t = 1$,so $t = 1$.Substituting $t = 1$ back into the expression for $r$: $r = (1 + 1)i + (2 - 1)j + (1 + 1)k = 2i + j + 2k$.

Let $a = i + 2j + k$,$b = i - j + k$,and $c = i + j - k$. $A$ vector in the plane of $a$ and $b$ has a projection of $\frac{1}{\sqrt{3}}$ on $c$. Then,one such vector is:

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