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If $\hat{\mathbf{a}}, \hat{\mathbf{b}}$ and $\hat{\mathbf{c}}$ are unit vectors such that $\hat{\mathbf{a}}+\hat{\mathbf{b}}+\hat{\mathbf{c}}=\mathbf{0}$, then the $\hat{\mathbf{a}} \cdot \hat{\mathbf{b}}+\hat{\mathbf{b}} \cdot \hat{\mathbf{c}}+\hat{\mathbf{c}} \cdot \hat{\mathbf{a}}$ is equal to
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The correct answer is:
$\frac{-3}{2}$
Given that $\hat{\mathrm{a}}, \hat{\mathrm{b}}, \hat{\mathrm{c}}$ are unit vectors and $\hat{\mathrm{a}}+\hat{\mathrm{b}}+\hat{\mathrm{c}}=0$
$\therefore \quad|\hat{a}|=|\hat{b}|=|\hat{c}|=1$
Here, $\hat{a}+\hat{b}+\hat{c}=\hat{0}$
Then, $(\hat{a}+\hat{b}+\hat{c})^2=(\hat{0})^2$
$\begin{aligned}
& \Rightarrow \quad(\hat{a}+\hat{b}+\hat{c}) \cdot(\hat{a}+\hat{b}+\hat{c})=0 \\
& \Rightarrow \quad \hat{a} \cdot \hat{a}+\hat{a} \cdot \hat{b}+\hat{a} \cdot \hat{c}+\hat{b} \cdot \hat{a}+\hat{b} \cdot \hat{b} \\
& \hat{b} \cdot \hat{c}+\hat{c} \cdot \hat{a}+\hat{c} \cdot \hat{b}+\hat{c} \cdot \hat{c}=0 \\
& \Rightarrow \quad|\hat{a}|^2+\hat{a} \cdot \hat{b}+\hat{c} \cdot \hat{a}+\hat{a} \cdot \hat{b}+|\hat{b}|^2+\hat{b} \cdot \hat{c} \\
& +\hat{c} \cdot \hat{a}+\hat{b} \cdot \hat{c}+|\hat{c}|^2=0 \\
& \Rightarrow \quad(1)^2+2 \hat{a} \cdot \hat{b}+2 \hat{b} \cdot \hat{c}+2 \hat{c} \cdot \hat{a}+(1)^2+(1)^2=0 \\
& \Rightarrow \quad 2[\hat{a} \cdot \hat{b}+\hat{b} \cdot \hat{c}+\hat{c} \cdot \hat{a}]+3=0 \\
& 2[\hat{a} \cdot \hat{b}+\hat{b} \cdot \hat{c}+\hat{c} \cdot \hat{a}]=-3 \\
& \hat{a} \cdot \hat{b}+\hat{b} \cdot \hat{c}+\hat{c} \cdot \hat{a}=-\frac{3}{2}
\end{aligned}$
$\therefore \quad|\hat{a}|=|\hat{b}|=|\hat{c}|=1$
Here, $\hat{a}+\hat{b}+\hat{c}=\hat{0}$
Then, $(\hat{a}+\hat{b}+\hat{c})^2=(\hat{0})^2$
$\begin{aligned}
& \Rightarrow \quad(\hat{a}+\hat{b}+\hat{c}) \cdot(\hat{a}+\hat{b}+\hat{c})=0 \\
& \Rightarrow \quad \hat{a} \cdot \hat{a}+\hat{a} \cdot \hat{b}+\hat{a} \cdot \hat{c}+\hat{b} \cdot \hat{a}+\hat{b} \cdot \hat{b} \\
& \hat{b} \cdot \hat{c}+\hat{c} \cdot \hat{a}+\hat{c} \cdot \hat{b}+\hat{c} \cdot \hat{c}=0 \\
& \Rightarrow \quad|\hat{a}|^2+\hat{a} \cdot \hat{b}+\hat{c} \cdot \hat{a}+\hat{a} \cdot \hat{b}+|\hat{b}|^2+\hat{b} \cdot \hat{c} \\
& +\hat{c} \cdot \hat{a}+\hat{b} \cdot \hat{c}+|\hat{c}|^2=0 \\
& \Rightarrow \quad(1)^2+2 \hat{a} \cdot \hat{b}+2 \hat{b} \cdot \hat{c}+2 \hat{c} \cdot \hat{a}+(1)^2+(1)^2=0 \\
& \Rightarrow \quad 2[\hat{a} \cdot \hat{b}+\hat{b} \cdot \hat{c}+\hat{c} \cdot \hat{a}]+3=0 \\
& 2[\hat{a} \cdot \hat{b}+\hat{b} \cdot \hat{c}+\hat{c} \cdot \hat{a}]=-3 \\
& \hat{a} \cdot \hat{b}+\hat{b} \cdot \hat{c}+\hat{c} \cdot \hat{a}=-\frac{3}{2}
\end{aligned}$
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