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Let $\mathbf{A}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$. If $\mathbf{B}$ is a vector in $X Y$ plane such that $(\mathbf{A}+\mathbf{B}) \cdot \mathbf{B}=15$ and $\mathbf{A} \cdot \mathbf{B}=6$, then $|\mathbf{B}|$ is
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The correct answer is:
$3$
$\mathbf{A}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}$
Let $\mathbf{B}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}$
$\because(\mathbf{A}+\mathbf{B}) \cdot \mathbf{B}=15$ and $\mathbf{A} \cdot \mathbf{B}=6$
$\Rightarrow \mathbf{A} \cdot \mathbf{B}+\mathbf{B} \cdot \mathbf{B}=15$
$\begin{array}{cc}\Rightarrow & 6+|\mathbf{B}|^2=15 \\ \Rightarrow & |\mathbf{B}|^2=15-6=9 \\ \Rightarrow & |\mathbf{B}|=3\end{array}$
Let $\mathbf{B}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}$
$\because(\mathbf{A}+\mathbf{B}) \cdot \mathbf{B}=15$ and $\mathbf{A} \cdot \mathbf{B}=6$
$\Rightarrow \mathbf{A} \cdot \mathbf{B}+\mathbf{B} \cdot \mathbf{B}=15$
$\begin{array}{cc}\Rightarrow & 6+|\mathbf{B}|^2=15 \\ \Rightarrow & |\mathbf{B}|^2=15-6=9 \\ \Rightarrow & |\mathbf{B}|=3\end{array}$
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