p=2 hat{i}-3 hat{j}+hat{k}, q=hat{i}+hat{j}-h | vedclass

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(C) Given $p=2 \hat{i}-3 \hat{j}+\hat{k}$ and $q=\hat{i}+\hat{j}-\hat{k}$.First, calculate the dot product $p \cdot q = (2)(1) + (-3)(1) + (1)(-1) = 2

Vedclass · Accepted Answer

$\frac{2 \hat{i}+3 \hat{j}+5 \hat{k}}{2}$

Vedclass · Answer

(C) Given $p=2 \hat{i}-3 \hat{j}+\hat{k}$ and $q=\hat{i}+\hat{j}-\hat{k}$.First, calculate the dot product $p \cdot q = (2)(1) + (-3)(1) + (1)(-1) = 2 - 3 - 1 = -2$.Calculate the magnitudes squared: $|p|^2 = 2^2 + (-3)^2 + 1^2 = 4 + 9 + 1 = 14$ and $|q|^2 = 1^2 + 1^2 + (-1)^2 = 3$.Vector $a$ (projection of $p$ on $q$) is given by $a = \frac{p \cdot q}{|q|^2} q = \frac{-2}{3}(\hat{i}+\hat{j}-\hat{k})$.Vector $b$ (projection of $q$ on $p$) is given by $b = \frac{q \cdot p}{|p|^2} p = \frac{-2}{14}(2 \hat{i}-3 \hat{j}+\hat{k}) = \frac{-1}{7}(2 \hat{i}-3 \hat{j}+\hat{k})$.Now, $a 	imes b = \left(\frac{-2}{3}ight) \left(\frac{-1}{7}ight) [(\hat{i}+\hat{j}-\hat{k}) 	imes (2 \hat{i}-3 \hat{j}+\hat{k})] = \frac{2}{21} [\hat{i}(1-3) - \hat{j}(1+2) + \hat{k}(-3-2)] = \frac{2}{21} (-2 \hat{i}-3 \hat{j}-5 \hat{k})$.Also, $a \cdot b = \left(\frac{-2}{3}ight) \left(\frac{-1}{7}ight) [(\hat{i}+\hat{j}-\hat{k}) \cdot (2 \hat{i}-3 \hat{j}+\hat{k})] = \frac{2}{21} (2 - 3 - 1) = \frac{2}{21} (-2) = \frac{-4}{21}$.Finally, $\frac{a 	imes b}{a \cdot b} = \frac{\frac{2}{21} (-2 \hat{i}-3 \hat{j}-5 \hat{k})}{\frac{-4}{21}} = \frac{-2 \hat{i}-3 \hat{j}-5 \hat{k}}{-2} = \frac{2 \hat{i}+3 \hat{j}+5 \hat{k}}{2}$.

$p=2 \hat{i}-3 \hat{j}+\hat{k}, q=\hat{i}+\hat{j}-\hat{k}$. If the vectors $a$ and $b$ are the orthogonal projections of $p$ on $q$ and $q$ on $p$ respectively, then $\frac{a \times b}{a \cdot b}=$

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