If vec{a}=2 hat{i}+hat{j}-hat{k}, vec{b}=hat{ | vedclass

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(A) Given $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+3 \hat{k}$.First, calculate the dot product $\vec{a} \cdot \vec{b} = (2)(1)

Vedclass · Accepted Answer

$17 \cos ^2 	heta$

Vedclass · Answer

(A) Given $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+3 \hat{k}$.First, calculate the dot product $\vec{a} \cdot \vec{b} = (2)(1) + (1)(-1) + (-1)(3) = 2 - 1 - 3 = -2$.Calculate the magnitudes squared: $|\vec{a}|^2 = 2^2 + 1^2 + (-1)^2 = 4 + 1 + 1 = 6$ and $|\vec{b}|^2 = 1^2 + (-1)^2 + 3^2 = 1 + 1 + 9 = 11$.Then, $\vec{x} = \left(\frac{-2}{11}ight) \vec{b} \implies |\vec{x}|^2 = \left(\frac{-2}{11}ight)^2 |\vec{b}|^2 = \frac{4}{121} 	imes 11 = \frac{4}{11}$.Similarly, $\vec{y} = \left(\frac{-2}{6}ight) \vec{a} \implies |\vec{y}|^2 = \left(\frac{-2}{6}ight)^2 |\vec{a}|^2 = \frac{4}{36} 	imes 6 = \frac{4}{6} = \frac{2}{3}$.However, the expression $x^2+y^2$ refers to the sum of the squares of the magnitudes $|\vec{x}|^2 + |\vec{y}|^2 = \frac{4}{11} + \frac{2}{3} = \frac{12+22}{33} = \frac{34}{33}$.Using $\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{-2}{\sqrt{6} \sqrt{11}} = \frac{-2}{\sqrt{66}}$, we have $\cos^2 	heta = \frac{4}{66} = \frac{2}{33}$.Thus, $x^2+y^2 = \frac{34}{33} = 17 	imes \frac{2}{33} = 17 \cos^2 	heta$.

If $\vec{a}=2 \hat{i}+\hat{j}-\hat{k}$, $\vec{b}=\hat{i}-\hat{j}+3 \hat{k}$, $\vec{x}=\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right) \vec{b}$, $\vec{y}=\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right) \vec{a}$ and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, then $x^2+y^2=$

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