Given that, $|\mathbf{x}|=|\mathbf{y}|=|\mathbf{z}|=\sqrt{2}$ and $\theta=60^{\circ}$

Thus, $\mathbf{x} \cdot \mathbf{y}=|\mathbf{x}||\mathbf{y}| \cos \theta$

$$ \begin{aligned} & =(\sqrt{2})(\sqrt{2}) \cos 60^{\circ} \\ & =2 \times \frac{1}{2}=1 \end{aligned} $$

Similarly $\mathbf{y} \cdot \mathbf{z}=\mathbf{z} \cdot \mathbf{x}=1$

and $\mathbf{x} \cdot \mathbf{x}=|\mathbf{x}||\mathbf{x}|$

$$ \begin{aligned} \cos 0^{\circ} & =(\sqrt{2})(\sqrt{2}) \times 1=2 \\ \mathbf{y} \cdot \mathbf{y} & =2, \mathbf{z} \cdot \mathbf{z}=2 \end{aligned} $$

Now, $\mathbf{a}=\mathbf{x} \times(\mathbf{y} \times \mathbf{z})=(\mathbf{x} \cdot \mathbf{z}) \mathbf{y}-(\mathbf{x} \cdot \mathbf{y}) \mathbf{z}$

$$ =1 \cdot \mathbf{y}-1 \cdot \mathbf{z} $$

$$ \begin{equation*} a=y-z \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $$

Also, $\quad \mathbf{b}=\mathbf{y} \times(\mathbf{z} \times \mathbf{x})=(\mathbf{y} \cdot \mathbf{z}) \mathbf{z}-(\mathbf{y} \cdot \mathbf{z}) \mathbf{x}$

$$ =1 \cdot \mathbf{z}-1 \cdot \mathbf{x} $$

$$ \begin{equation*} \mathbf{b}=\mathbf{z}-\mathbf{x} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{equation*} $$

Adding Eqs. (i) and (ii), we get

$$ \begin{align*} & \mathbf{a}+\mathbf{b}=(\mathbf{y}-\mathbf{z})+(\mathbf{z}-\mathbf{x}) \\ & \mathbf{y}-\mathbf{x}=\mathbf{a}+\mathbf{b}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{align*} $$

Again $\quad \mathbf{c}=\mathbf{x} \times \mathbf{y}$

Taking cross product with $\mathbf{x}$ on both sides

$$ \begin{align*} & \mathbf{x} \times \mathbf{c}=\mathbf{x} \times(\mathbf{x} \times \mathbf{y})=(\mathbf{x} \cdot \mathbf{y}) \mathbf{x}-(\mathbf{x} \cdot \mathbf{x}) \mathbf{y} \\ & \mathbf{x} \times \mathbf{c}=(1) \mathbf{x}-(2) \mathbf{y} \\ & \Rightarrow \mathbf{x} \times \mathbf{c}=\mathbf{x}-2 \mathbf{y} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv) \end{align*} $$

Finally, $\mathbf{c}=\mathbf{x} \times \mathbf{y}$

Taking cross product with $\mathbf{y}$ on both sides

$$ \begin{align*} & \mathbf{y} \times \mathbf{c}=\mathbf{y} \times(\mathbf{x} \times \mathbf{y}) \\ & \mathbf{y} \times \mathbf{c}=(\mathbf{y} \cdot \mathbf{y}) \mathbf{x}-(\mathbf{y} \cdot \mathbf{x}) \mathbf{y}=(2) \mathbf{x}-(1) \mathbf{y} \\ & \mathbf{y} \times \mathbf{c}=2 \mathbf{x}-\mathbf{y} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v) \end{align*} $$

Subtracting Eq. (v) from Eq. (iv), we get

$$ \begin{aligned} \mathbf{x} \times \mathbf{c}-\mathbf{y} \times \mathbf{c} & =(+\mathbf{x}-2 \mathbf{y})-(2 \mathbf{x}-\mathbf{y}) \\ (\mathbf{x}-\mathbf{y}) \times \mathbf{c} & =-\mathbf{x}-\mathbf{y} \end{aligned} $$

or $\quad \mathbf{x}+\mathbf{y}=(\mathbf{y}-\mathbf{x}) \times \mathbf{c}$

From Eq. (iii), $\mathbf{y}-\mathbf{x}=\mathbf{a}+\mathbf{b}$

$$ \begin{equation*} \therefore \quad \mathbf{x}+\mathbf{y}=(\mathbf{a}+\mathbf{b}) \times \mathbf{c} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi)\end{equation*} $$

Subtracting Eq. (iii) from Eq. (vi), we get

$$ \begin{aligned} & (\mathbf{x}+\mathbf{y})-(\mathbf{y}-\mathbf{x})=(\mathbf{a}+\mathbf{b}) \times \mathbf{c}-(\mathbf{a}+\mathbf{b}) \\ & 2 \mathbf{x}=(\mathbf{a}+\mathbf{b}) \times \mathbf{c}-(\mathbf{a}+\mathbf{b}) \\ & \text { or } \quad \mathbf{x}=\frac{1}{2}[(\mathbf{a}+\mathbf{b}) \times \mathbf{c}-(\mathbf{a}+\mathbf{b})] \end{aligned} $$

Vector Algebra

Continue this paper

Similar Vector Algebra PYQs

More practice from this question