If vec{b}=2 hat{i}-hat{j}-hat{k}, vec{a}=3 ha | vedclass

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(B) Given, $\vec{b}=2 \hat{i}-\hat{j}-\hat{k}$ and $\vec{a}=3 \hat{i}+4 \hat{j}-5 \hat{k}$.First, calculate the dot products:$\vec{b} \cdot \vec{b} =

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the orthogonal projection of $\vec{a}$ on $\vec{b}$ and equal to $\frac{7}{\sqrt{6}}$

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(B) Given, $\vec{b}=2 \hat{i}-\hat{j}-\hat{k}$ and $\vec{a}=3 \hat{i}+4 \hat{j}-5 \hat{k}$.First, calculate the dot products:$\vec{b} \cdot \vec{b} = (2)^2 + (-1)^2 + (-1)^2 = 4 + 1 + 1 = 6$.$\vec{b} \cdot \vec{a} = (2)(3) + (-1)(4) + (-1)(-5) = 6 - 4 + 5 = 7$.Using the vector triple product formula $\vec{b} 	imes(\vec{a} 	imes \vec{b})=(\vec{b} \cdot \vec{b}) \vec{a}-(\vec{b} \cdot \vec{a}) \vec{b}$, we get:$\vec{b} 	imes(\vec{a} 	imes \vec{b}) = 6 \vec{a} - 7 \vec{b}$.We can rewrite this as $\frac{\vec{a} - (7/6) \vec{b}}{1/6}$.Comparing this with the given expression $\frac{\vec{a}-k \vec{b}}{l}$, we find $k = 7/6$ and $l = 1/6$.Now, calculate $|\vec{b}| = \sqrt{2^2 + (-1)^2 + (-1)^2} = \sqrt{6}$.Finally, $\frac{k}{l|\vec{b}|} = \frac{7/6}{(1/6) 	imes \sqrt{6}} = \frac{7}{\sqrt{6}}$.This value represents the scalar projection of $\vec{a}$ on $\vec{b}$, which is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{7}{\sqrt{6}}$.

If $\vec{b}=2 \hat{i}-\hat{j}-\hat{k}$, $\vec{a}=3 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\vec{b} \times(\vec{a} \times \vec{b})=\frac{\vec{a}-k \vec{b}}{l}$, then $\frac{k}{l|\vec{b}|}$ is

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