KCET · Maths · Vector Algebra
If \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{c}\) are unit vectors such that \(\mathbf{a}+\mathbf{b}+\mathbf{c}=0\), then \(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}\) is equal to
- A \(\frac{3}{2}\)
- B \(-\frac{3}{2}\)
- C \(\frac{2}{3}\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(B) \(-\frac{3}{2}\)
Step-by-step Solution
Detailed explanation
Given that, \(a, b\) and \(c\) are unit vectors.
i.e., \(\quad|a|=|b|=|c|=1...(i)\)
Now, we have
\(\begin{gathered}
(a+b+c)=0 \\
\Rightarrow \quad|a+b+c|^{2}=|0|^{2} \\
\Rightarrow \quad|a|^{2}+|b|^{2}+|c|^{2}+2(a \cdot b+b \cdot c+c \cdot a)=0 \\
\Rightarrow \quad 1+1+1+2(a \cdot b+b \cdot c+c \cdot a)=0 [from Eq. (i)] \\
\Rightarrow \quad(a \cdot b+b \cdot c+c \cdot a)=-\frac{3}{2}
\end{gathered}\)
i.e., \(\quad|a|=|b|=|c|=1...(i)\)
Now, we have
\(\begin{gathered}
(a+b+c)=0 \\
\Rightarrow \quad|a+b+c|^{2}=|0|^{2} \\
\Rightarrow \quad|a|^{2}+|b|^{2}+|c|^{2}+2(a \cdot b+b \cdot c+c \cdot a)=0 \\
\Rightarrow \quad 1+1+1+2(a \cdot b+b \cdot c+c \cdot a)=0 [from Eq. (i)] \\
\Rightarrow \quad(a \cdot b+b \cdot c+c \cdot a)=-\frac{3}{2}
\end{gathered}\)
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