KCET · Maths · Vector Algebra
If \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{c}\) are non-coplanar, then the value of \(\mathbf{a} \cdot\left\{\frac{\mathbf{b} \times \mathbf{c}}{3 \mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})}\right\}-\mathbf{b} \cdot\left\{\frac{\mathbf{c} \times \mathbf{a}}{2 \mathbf{c} \cdot(\mathbf{a} \times \mathbf{b})}\right\}\) is
- A \(-\frac{1}{2}\)
- B \(-\frac{1}{3}\)
- C \(-\frac{1}{6}\)
- D \(\frac{1}{6}\)
Answer & Solution
Correct Answer
(C) \(-\frac{1}{6}\)
Step-by-step Solution
Detailed explanation
Given, [a b c] \(\neq 0\), i.e., non-coplanar.
\[
\begin{aligned}
&=\mathbf{a} \cdot\left\{\frac{\mathbf{a} \times \mathbf{c}}{3 \mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})}\right\}-\mathbf{b} \cdot\left\{\frac{\mathbf{c} \times \mathbf{a}}{2 \mathbf{c} \cdot(\mathbf{a} \times \mathbf{c})}\right\} \\
&=\frac{\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})}{3 \mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})}-\frac{\mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})}{2 \mathbf{c} \cdot(\mathbf{a} \times \mathbf{b})} \\
&=\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{3[\mathbf{b} \mathbf{c} \mathbf{a}]}-\frac{[\mathbf{b} \mathbf{c} \mathbf{a}]}{2[\mathbf{c} \mathbf{a} \mathbf{b}]}(\because \mathbf{a} \cdot(\mathbf{b} \times \mathbf{b})=[\mathbf{a} \mathbf{b} \mathbf{c}] \\
&=\frac{[\mathbf{b} \mathbf{c} \mathbf{a}]=[\mathbf{a} \mathbf{b} \mathbf{c}],[\mathbf{c} \mathbf{a} \mathbf{b}]=[\mathbf{b} \mathbf{c} \mathbf{a}]}{3[\mathbf{a} \mathbf{b ~ c}]}-\frac{[\mathbf{b} \mathbf{c} \mathbf{a}]}{2[\mathbf{b} \mathbf{c} \mathbf{a}]} \\
&=1 / 3-1 / 2=-1 / 6
\end{aligned}
\]
\[
\begin{aligned}
&=\mathbf{a} \cdot\left\{\frac{\mathbf{a} \times \mathbf{c}}{3 \mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})}\right\}-\mathbf{b} \cdot\left\{\frac{\mathbf{c} \times \mathbf{a}}{2 \mathbf{c} \cdot(\mathbf{a} \times \mathbf{c})}\right\} \\
&=\frac{\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})}{3 \mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})}-\frac{\mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})}{2 \mathbf{c} \cdot(\mathbf{a} \times \mathbf{b})} \\
&=\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{3[\mathbf{b} \mathbf{c} \mathbf{a}]}-\frac{[\mathbf{b} \mathbf{c} \mathbf{a}]}{2[\mathbf{c} \mathbf{a} \mathbf{b}]}(\because \mathbf{a} \cdot(\mathbf{b} \times \mathbf{b})=[\mathbf{a} \mathbf{b} \mathbf{c}] \\
&=\frac{[\mathbf{b} \mathbf{c} \mathbf{a}]=[\mathbf{a} \mathbf{b} \mathbf{c}],[\mathbf{c} \mathbf{a} \mathbf{b}]=[\mathbf{b} \mathbf{c} \mathbf{a}]}{3[\mathbf{a} \mathbf{b ~ c}]}-\frac{[\mathbf{b} \mathbf{c} \mathbf{a}]}{2[\mathbf{b} \mathbf{c} \mathbf{a}]} \\
&=1 / 3-1 / 2=-1 / 6
\end{aligned}
\]
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