Q73 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Mathematics

$\mathbf{x}, \mathbf{y}, \mathbf{z}$ are three vectors each of magnitude $\sqrt{2}$ and each making an angle $60^{\circ}$ with one another. If $\mathbf{a}=\mathbf{x} \times(\mathbf{y} \times \mathbf{z}), \mathbf{b}=\mathbf{y} \times(\mathbf{z} \times \mathbf{x})$, $\mathbf{c}=\mathbf{x} \times \mathbf{y}$, then $\mathbf{x}=$

  • Correct answer
    A
    $\frac{1}{2}[(a+b) \times c-(a+b)]$
  • B
    $\frac{1}{2}[c+a-b]$
  • C
    $\frac{1}{2}[(a+b) \times c+(a+b)]$
  • D
    $\frac{1}{2}[(a \times b) \times c-a+b]$

Solution

Given that, $|\mathbf{x}|=|\mathbf{y}|=|\mathbf{z}|=\sqrt{2}$ and $\theta=60^{\circ}$ Thus, $\mathbf{x} \cdot \mathbf{y}=|\mathbf{x}||\mathbf{y}| \cos \theta=(\sqrt{2})(\sqrt{2}) \cos 60^{\circ}$ $=2 \times \frac{1}{2}=1$ Similarly $\mathbf{y} \cdot \mathbf{z}=\mathbf{z} \cdot \mathbf{x}=1$ and $\mathbf{x} \cdot \mathbf{x}=|\mathbf{x}||\mathbf{x}| \cos 0^{\circ}=(\sqrt{2})(\sqrt{2}) \times 1=2$ $\mathbf{y} \cdot \mathbf{y}=2, \mathbf{z} \cdot \mathbf{z}=2$ Now, $\mathbf{a}=\mathbf{x} \times(\mathbf{y} \times \mathbf{z})=(\mathbf{x} \cdot \mathbf{z}) \mathbf{y}-(\mathbf{x} \cdot \mathbf{y}) \mathbf{z}$ $=1 \cdot y-1 \cdot z$ $\mathbf{a}=\mathbf{y}-\mathbf{z}$ $\ldots(\mathrm{i})$ Also, $\quad \mathbf{b}=\mathbf{y} \times(\mathbf{z} \times \mathbf{x})=(\mathbf{y} \cdot \mathbf{z}) \mathbf{z}-(\mathbf{y} \cdot \mathbf{z}) \mathbf{x}$ $=1 \cdot \mathbf{z}-1 \cdot \mathbf{x}$ $\mathbf{b}=\mathbf{z}-\mathbf{x}$ $\ldots(\mathrm{ii})$ Adding Eqs. (i) and (ii), we get $\mathbf{a}+\mathbf{b}=(\mathbf{y}-\mathbf{z})+(\mathbf{z}-\mathbf{x})$ $y-x=a+b$ $\ldots(\mathrm{iii})$ Again $\quad \mathbf{c}=\mathbf{x} \times \mathbf{y}$ Taking cross product with $\mathbf{x}$ on both sides $\mathbf{x} \times \mathbf{c}=\mathbf{x} \times(\mathbf{x} \times \mathbf{y})=(\mathbf{x} \cdot \mathbf{y}) \mathbf{x}-(\mathbf{x} \cdot \mathbf{x}) \mathbf{y}$ $\mathbf{x} \times \mathbf{c}=(1) \mathbf{x}-(2) \mathbf{y} \Rightarrow \mathbf{x} \times \mathbf{c}=\mathbf{x}-2 \mathbf{y}$ $\ldots(\mathrm{iv})$ Finally, $\mathbf{c}=\mathbf{x} \times \mathbf{y}$ Taking cross product with $\mathbf{y}$ on both sides $\mathbf{y} \times \mathbf{c}=\mathbf{y} \times(\mathbf{x} \times \mathbf{y})$ $\mathbf{y} \times \mathbf{c}=(\mathbf{y} \cdot \mathbf{y}) \mathbf{x}-(\mathbf{y} \cdot \mathbf{x}) \mathbf{y}=(2) \mathbf{x}-(1) \mathbf{y}$ $\mathbf{y} \times \mathbf{c}=2 \mathbf{x}-\mathbf{y}$ $\ldots(v)$ Subtracting Eq. (v) from Eq. (iv), we get $\mathbf{x} \times \mathbf{c}-\mathbf{y} \times \mathbf{c}=(+\mathbf{x}-2 \mathbf{y})-(2 \mathbf{x}-\mathbf{y})$ $(\mathbf{x}-\mathbf{y}) \times \mathbf{c}=-\mathbf{x}-\mathbf{y}$ or $\mathbf{x}+\mathbf{y}=(\mathbf{y}-\mathbf{x}) \times \mathbf{c}$ From Eq. (iii), $\mathbf{y}-\mathbf{x}=\mathbf{a}+\mathbf{b}$ $\therefore \quad \mathbf{x}+\mathbf{y}=(\mathbf{a}+\mathbf{b}) \times \mathbf{c}$ $\ldots(v i)$ Subtracting Eq. (iii) from Eq. (vi), we get $(\mathbf{x}+\mathbf{y})-(\mathbf{y}-\mathbf{x})=(\mathbf{a}+\mathbf{b}) \times \mathbf{c}-(\mathbf{a}+\mathbf{b})$ $2 \mathbf{x}=(\mathbf{a}+\mathbf{b}) \times \mathbf{c}-(\mathbf{a}+\mathbf{b})$ $\mathbf{x}=\frac{1}{2}[(\mathbf{a}+\mathbf{b}) \times \mathbf{c}-(\mathbf{a}+\mathbf{b})]$

Hence, the correct answer is option: 'A'.

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