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If the vector $\mathbf{a}=3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ is the sum of two vectors $\mathbf{a}_1$ and $\mathbf{a}_2$, vector $\mathbf{a}_1$ is parallel to $\mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and vector $\mathbf{a}_2$ is perpendicular to $\mathbf{b}$, then $\mathbf{a}_1=$
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2202 Upvotes
Verified Answer
The correct answer is:
$\frac{3}{2}(\hat{\mathbf{i}}+\hat{\mathbf{j}})$
Given,
$\begin{aligned}
& \mathbf{a}=3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \text { and } \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}} \\
& \mathbf{a}=\mathbf{a}_1+\mathbf{a}_2
\end{aligned}$
$\begin{aligned}
& \mathbf{a}_1 \text { is paralleI to } \mathbf{b} \text {. } \\
& \therefore \mathbf{a}_1=\lambda \mathbf{b}=\lambda(\hat{\mathbf{i}}+\hat{\mathrm{j}}) \\
& \mathbf{a}_2 \text { is perpendicular to } \mathbf{b} \text {. } \\
& \therefore \mathbf{a}_2 \cdot \mathbf{b}=0 \\
& \left(\mathbf{a}-\mathbf{a}_1\right) \cdot \mathbf{b}=\mathbf{0} \\
& (3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}-\lambda(\hat{\mathbf{i}}+\hat{\mathrm{j}}) \cdot(\hat{\mathbf{i}}+\hat{\mathrm{j}})=0 \\
& (-\lambda \hat{\mathbf{I}}+(3-\lambda) \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot(\hat{\mathrm{i}}+\hat{\mathrm{J}})=0 \\
& \Rightarrow \quad-\lambda+3-\lambda=0 \Rightarrow \lambda=\frac{3}{2} \\
& \therefore \mathbf{a}_1=\frac{3}{2}(\hat{\mathbf{i}}+\hat{\mathrm{j}})
\end{aligned}$
$\begin{aligned}
& \mathbf{a}=3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \text { and } \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}} \\
& \mathbf{a}=\mathbf{a}_1+\mathbf{a}_2
\end{aligned}$
$\begin{aligned}
& \mathbf{a}_1 \text { is paralleI to } \mathbf{b} \text {. } \\
& \therefore \mathbf{a}_1=\lambda \mathbf{b}=\lambda(\hat{\mathbf{i}}+\hat{\mathrm{j}}) \\
& \mathbf{a}_2 \text { is perpendicular to } \mathbf{b} \text {. } \\
& \therefore \mathbf{a}_2 \cdot \mathbf{b}=0 \\
& \left(\mathbf{a}-\mathbf{a}_1\right) \cdot \mathbf{b}=\mathbf{0} \\
& (3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}-\lambda(\hat{\mathbf{i}}+\hat{\mathrm{j}}) \cdot(\hat{\mathbf{i}}+\hat{\mathrm{j}})=0 \\
& (-\lambda \hat{\mathbf{I}}+(3-\lambda) \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot(\hat{\mathrm{i}}+\hat{\mathrm{J}})=0 \\
& \Rightarrow \quad-\lambda+3-\lambda=0 \Rightarrow \lambda=\frac{3}{2} \\
& \therefore \mathbf{a}_1=\frac{3}{2}(\hat{\mathbf{i}}+\hat{\mathrm{j}})
\end{aligned}$
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