JEE Advanced · Mathematics · 30. Vector Algebra
Let \(\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\mathbf{c}=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}\) be three vectors. A vector \(\mathbf{v}\) in the plane of \(\mathbf{a}\) and \(\mathbf{b}\) whose projection of \(\mathbf{c}\) is \(\frac{1}{\sqrt{3}}\), is given by
- A
\(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\)
- B
\(-3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\hat{\mathbf{k}}\)
- C
\(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}\)
- D
\(\hat{\mathbf{i}}+3 \hat{\mathbf{k}}-3 \hat{\mathbf{k}}\)
Answer & Solution
Correct Answer
(C)
\(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}\)
Step-by-step Solution
Detailed explanation
Let \(\mathbf{v}=\mathbf{a}+\lambda \mathbf{b}\)
\[
\mathbf{v}=(1+\lambda) \hat{\mathbf{i}}+(1-\lambda) \hat{\mathbf{j}}+(1+\lambda) \hat{\mathbf{k}}
\]
Projection of \(\mathbf{v}\) on \(\mathbf{c}=\frac{1}{\sqrt{3}}\)
\[
\begin{array}{lc}
\Rightarrow & \frac{\mathbf{v} \cdot \mathbf{c}}{|\mathbf{c}|}=\frac{1}{\sqrt{3}} \\
\Rightarrow & \frac{(1+\lambda)-(1-\lambda)-(1+\lambda)}{\sqrt{3}}=\frac{1}{\sqrt{3}} \\
\Rightarrow & 1+\lambda-1+\lambda-1-\lambda=1 \\
\Rightarrow & \lambda-1=1 \Rightarrow \lambda=2 \\
\therefore & \mathbf{v}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}
\end{array}
\]
\[
\mathbf{v}=(1+\lambda) \hat{\mathbf{i}}+(1-\lambda) \hat{\mathbf{j}}+(1+\lambda) \hat{\mathbf{k}}
\]
Projection of \(\mathbf{v}\) on \(\mathbf{c}=\frac{1}{\sqrt{3}}\)
\[
\begin{array}{lc}
\Rightarrow & \frac{\mathbf{v} \cdot \mathbf{c}}{|\mathbf{c}|}=\frac{1}{\sqrt{3}} \\
\Rightarrow & \frac{(1+\lambda)-(1-\lambda)-(1+\lambda)}{\sqrt{3}}=\frac{1}{\sqrt{3}} \\
\Rightarrow & 1+\lambda-1+\lambda-1-\lambda=1 \\
\Rightarrow & \lambda-1=1 \Rightarrow \lambda=2 \\
\therefore & \mathbf{v}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}
\end{array}
\]
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