If vec{a} = hat{i} - hat{j} - hat{k} and vec{ | vedclass

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(B) The formula for the orthogonal projection of vector $\vec{b}$ on vector $\vec{a}$ is given by $\frac{(\vec{b} \cdot \vec{a})\vec{a}}{|\vec{a}|^2}$

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$2$

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(B) The formula for the orthogonal projection of vector $\vec{b}$ on vector $\vec{a}$ is given by $\frac{(\vec{b} \cdot \vec{a})\vec{a}}{|\vec{a}|^2}$.Given that the projection is $\frac{4}{3}(\hat{i} - \hat{j} - \hat{k})$,we have:$\frac{(\vec{b} \cdot \vec{a})\vec{a}}{|\vec{a}|^2} = \frac{4}{3}(\hat{i} - \hat{j} - \hat{k})$First,calculate $|\vec{a}|^2 = (1)^2 + (-1)^2 + (-1)^2 = 1 + 1 + 1 = 3$.Next,calculate the dot product $\vec{b} \cdot \vec{a} = (\lambda \hat{i} - 3\hat{j} + \hat{k}) \cdot (\hat{i} - \hat{j} - \hat{k}) = \lambda(1) + (-3)(-1) + (1)(-1) = \lambda + 3 - 1 = \lambda + 2$.Substituting these into the projection formula:$\frac{(\lambda + 2)(\hat{i} - \hat{j} - \hat{k})}{3} = \frac{4}{3}(\hat{i} - \hat{j} - \hat{k})$Comparing the coefficients,we get $\frac{\lambda + 2}{3} = \frac{4}{3}$.$\lambda + 2 = 4 \Rightarrow \lambda = 2$.

If $\vec{a} = \hat{i} - \hat{j} - \hat{k}$ and $\vec{b} = \lambda \hat{i} - 3\hat{j} + \hat{k}$ and the orthogonal projection of $\vec{b}$ on $\vec{a}$ is $\frac{4}{3}(\hat{i} - \hat{j} - \hat{k})$,then $\lambda$ is equal to

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