Q72 / 160 · TS EAMCET 2018 (05 May Shift 2) · Mathematics
If $\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathrm{d}$ are 4 vectors such that $\mathbf{a} \cdot \mathbf{b}=0$, $|\mathbf{a} \times \mathbf{c}|=|\mathbf{a}\|\mathbf{c}|,| \mathbf{a} \times \mathbf{d}|=| \mathbf{a}\| \mathbf{d}|$, then $[\mathbf{b c d}]=$
- A$\mathbf{a}|| \mathbf{b} \| \mathbf{c} \mid$
- B$|\mathbf{b}\|\mathbf{c}\| \mathbf{d}|$
- C$\frac{1}{6}$
- Correct answerD0
Solution
We have, four vectors $a, b, c$ and $d$ such that $\mathbf{a} \cdot \mathbf{b}=\mathbf{0}$,
$|\mathbf{a} \times \mathbf{c}|=|\mathbf{a}| \mathbf{c}|,| \mathbf{a} \times \mathbf{d}|=| \mathbf{a}|| \mathbf{d} \mid$
From the given condition, we get
$\mathbf{a} \perp \mathbf{c}$ and $\mathbf{a} \perp \mathbf{d}$
$\left[\because|| x \times y|=| x|| y \mid \sin \theta \text { and } \sin \theta=1 \Rightarrow \theta=\frac{\pi}{2}\right]$
$\Rightarrow \quad \underline{\mathbf{a}}||(\underline{\mathbf{c}} \times \underline{\mathbf{d}})$
$\Rightarrow \mathbf{c} \times \mathbf{d}=\lambda \mathbf{a}$, where $\lambda$ is a constant.
Now, consider $[\mathbf{b} \mathbf{c} \mathbf{d}]=\mathbf{b} \cdot(\mathbf{c} \times \mathbf{d})$
$\begin{aligned}<br />& =\mathbf{b} \cdot(\lambda \mathbf{a})=\lambda(\mathbf{b} \cdot \mathbf{a})=\lambda(\mathbf{a} \cdot \mathbf{b}) \\<br />& =\lambda \times 0=0 \\<br />& \quad[\because \mathbf{a} \cdot \mathbf{b}=0]<br />\end{aligned}$
$|\mathbf{a} \times \mathbf{c}|=|\mathbf{a}| \mathbf{c}|,| \mathbf{a} \times \mathbf{d}|=| \mathbf{a}|| \mathbf{d} \mid$
From the given condition, we get
$\mathbf{a} \perp \mathbf{c}$ and $\mathbf{a} \perp \mathbf{d}$
$\left[\because|| x \times y|=| x|| y \mid \sin \theta \text { and } \sin \theta=1 \Rightarrow \theta=\frac{\pi}{2}\right]$
$\Rightarrow \quad \underline{\mathbf{a}}||(\underline{\mathbf{c}} \times \underline{\mathbf{d}})$
$\Rightarrow \mathbf{c} \times \mathbf{d}=\lambda \mathbf{a}$, where $\lambda$ is a constant.
Now, consider $[\mathbf{b} \mathbf{c} \mathbf{d}]=\mathbf{b} \cdot(\mathbf{c} \times \mathbf{d})$
$\begin{aligned}<br />& =\mathbf{b} \cdot(\lambda \mathbf{a})=\lambda(\mathbf{b} \cdot \mathbf{a})=\lambda(\mathbf{a} \cdot \mathbf{b}) \\<br />& =\lambda \times 0=0 \\<br />& \quad[\because \mathbf{a} \cdot \mathbf{b}=0]<br />\end{aligned}$
Hence, the correct answer is option: 'D'.