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If $\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=+\lambda \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and the orthogonal projection of $\overrightarrow{\mathbf{b}}$ on $\overrightarrow{\mathbf{a}}$ is $\frac{4}{3}(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})$, then $\lambda$ is equal to
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The correct answer is:
$2$
Given orthogonal projection of $\overrightarrow{\mathbf{b}}$ on $\overrightarrow{\mathbf{a}}$ is
$\begin{array}{r}\frac{(\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}}) \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|^2}=\frac{4}{3}(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ \Rightarrow \frac{\{(\lambda \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})\}(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})}{(1+1+1)}\end{array}$
$=\frac{4}{3}(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})$
$\begin{aligned} & \Rightarrow(\lambda+3-1)(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})=4(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ & \Rightarrow \quad(\lambda+2)(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})=4(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})\end{aligned}$
On equating the coefficient of $\hat{\mathbf{i}}$, we get
$\begin{aligned} \lambda+2 & =4 \\ \lambda & =2\end{aligned}$
$\begin{array}{r}\frac{(\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}}) \overrightarrow{\mathbf{a}}}{|\overrightarrow{\mathbf{a}}|^2}=\frac{4}{3}(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ \Rightarrow \frac{\{(\lambda \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})\}(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})}{(1+1+1)}\end{array}$
$=\frac{4}{3}(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})$
$\begin{aligned} & \Rightarrow(\lambda+3-1)(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})=4(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ & \Rightarrow \quad(\lambda+2)(\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}})=4(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})\end{aligned}$
On equating the coefficient of $\hat{\mathbf{i}}$, we get
$\begin{aligned} \lambda+2 & =4 \\ \lambda & =2\end{aligned}$
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