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 $	ext{proj}_{\vec{a}} \vec{b} = \frac{(\vec{b} \cdo

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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 $	ext{proj}_{\vec{a}} \vec{b} = \frac{(\vec{b} \cdot \vec{a})}{|\vec{a}|^2} \vec{a}$.Given $\vec{a} = \hat{i} - \hat{j} - \hat{k}$,we have $|\vec{a}|^2 = (1)^2 + (-1)^2 + (-1)^2 = 1 + 1 + 1 = 3$.Given $\vec{b} = \lambda\hat{i} - 3\hat{j} + \hat{k}$,the dot product is $\vec{b} \cdot \vec{a} = (\lambda)(1) + (-3)(-1) + (1)(-1) = \lambda + 3 - 1 = \lambda + 2$.Substituting these into the projection formula:$\frac{(\lambda + 2)}{3} (\hat{i} - \hat{j} - \hat{k}) = \frac{4}{3} (\hat{i} - \hat{j} - \hat{k})$.Comparing the coefficients of the vectors,we get $\frac{\lambda + 2}{3} = \frac{4}{3}$.$\lambda + 2 = 4$.$\lambda = 2$.

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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