Q73 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics
If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are three non-coplanar vectors and $\mathbf{d}$ is any unit vector, then $|(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \times \mathbf{c})+(\mathbf{b} \cdot \mathbf{d})(\mathbf{c} \times \mathbf{a})+(\mathbf{c} \cdot \mathbf{d})(\mathbf{a} \times \mathbf{b})|=$
- A$2 \mid[\mathrm{abc}]$
- B$\frac{1}{2} \mid[a b c]$
- Correct answerC$[\mathrm{abc}] \mid$
- D$\frac{1}{6}[\mathbf{a b c}]$
Solution
a, b, c are non-coplanar.
$\therefore \mathbf{b} \times \mathbf{c}, \mathbf{c} \times \mathbf{a}$ and $\mathbf{a} \times \mathbf{b}$ are also non-coplanar.
So, any vector can be expressed as a linear combination of these vectors.
Let, $\mathbf{d}=\lambda(\mathbf{b} \times \mathbf{c})+\mu(\mathbf{c} \times \mathbf{a})+m(\mathbf{a} \times \mathbf{b})$
$\mathbf{a} \cdot \mathbf{d}=\lambda[\mathbf{a} \mathbf{b} \mathbf{c}]$
$\begin{gathered}<br />\mathbf{b} \cdot \mathbf{d}=\mu(\mathbf{b} \mathbf{c} \mathbf{a}) \\<br />\mathbf{c} \cdot \mathbf{d}=m(\mathbf{c} \mathbf{a}) \\<br />\therefore \mathrm{d}=\frac{(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \times \mathbf{c})}{[\mathbf{a} \mathbf{b} \mathbf{c}]}+\frac{(\mathbf{b} \cdot \mathbf{d})(\mathbf{c} \times \mathbf{a})}{[\mathbf{a} \mathbf{b} \mathbf{c}]}+\frac{(\mathbf{c} \cdot \mathbf{d})(\mathbf{a} \times \mathbf{b})}{[\mathbf{a b} \mathbf{c}]} \\<br />\Rightarrow(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \times \mathbf{c})+(\mathbf{b} \cdot \mathbf{d})(\mathbf{c} \times \mathbf{a})+(\mathbf{c}-\mathbf{d})(\mathbf{a} \times \mathbf{b})= \\<br />\mathbf{d}[\mathbf{a} \mathbf{b} \mathbf{c}] \\<br />\Rightarrow(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \times \mathbf{c}+(\mathbf{b} \cdot \mathbf{d})(\mathbf{c} \times \mathbf{a})+(\mathbf{c} \cdot \mathbf{d})(\mathbf{a} \times \mathbf{b}) \mid \\<br />=[\mathbf{a} \mathbf{b} \mathbf{c}]<br />\end{gathered}$
$\therefore \mathbf{b} \times \mathbf{c}, \mathbf{c} \times \mathbf{a}$ and $\mathbf{a} \times \mathbf{b}$ are also non-coplanar.
So, any vector can be expressed as a linear combination of these vectors.
Let, $\mathbf{d}=\lambda(\mathbf{b} \times \mathbf{c})+\mu(\mathbf{c} \times \mathbf{a})+m(\mathbf{a} \times \mathbf{b})$
$\mathbf{a} \cdot \mathbf{d}=\lambda[\mathbf{a} \mathbf{b} \mathbf{c}]$
$\begin{gathered}<br />\mathbf{b} \cdot \mathbf{d}=\mu(\mathbf{b} \mathbf{c} \mathbf{a}) \\<br />\mathbf{c} \cdot \mathbf{d}=m(\mathbf{c} \mathbf{a}) \\<br />\therefore \mathrm{d}=\frac{(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \times \mathbf{c})}{[\mathbf{a} \mathbf{b} \mathbf{c}]}+\frac{(\mathbf{b} \cdot \mathbf{d})(\mathbf{c} \times \mathbf{a})}{[\mathbf{a} \mathbf{b} \mathbf{c}]}+\frac{(\mathbf{c} \cdot \mathbf{d})(\mathbf{a} \times \mathbf{b})}{[\mathbf{a b} \mathbf{c}]} \\<br />\Rightarrow(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \times \mathbf{c})+(\mathbf{b} \cdot \mathbf{d})(\mathbf{c} \times \mathbf{a})+(\mathbf{c}-\mathbf{d})(\mathbf{a} \times \mathbf{b})= \\<br />\mathbf{d}[\mathbf{a} \mathbf{b} \mathbf{c}] \\<br />\Rightarrow(\mathbf{a} \cdot \mathbf{d})(\mathbf{b} \times \mathbf{c}+(\mathbf{b} \cdot \mathbf{d})(\mathbf{c} \times \mathbf{a})+(\mathbf{c} \cdot \mathbf{d})(\mathbf{a} \times \mathbf{b}) \mid \\<br />=[\mathbf{a} \mathbf{b} \mathbf{c}]<br />\end{gathered}$
Hence, the correct answer is option: 'C'.