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

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(B) The formula for the orthogonal projection of vector $\overrightarrow{b}$ on vector $\overrightarrow{a}$ is given by $\frac{(\overrightarrow{b} \cd

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$2$

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

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

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