Q76 / 160 · TS EAMCET 2013 · Mathematics
If $\mathbf{a}$ and $\mathbf{b}$ are two non-zero perpendicular vectors, then a vector $\mathbf{y}$ satisfying equations $\mathbf{a} \cdot \mathbf{y}=c$ (where, $c$ is scalar) and $\mathbf{a} \times \mathbf{y}=\mathbf{b}$ is
- A$|\mathbf{a}|^2[\mathrm{c} \mathbf{a}-(\mathbf{a} \times \mathbf{b})]$
- B$|\mathbf{a}|^2 \cdot[\mathbf{c} \mathbf{a}+(\mathbf{a} \times \mathbf{b})]$
- Correct answerC$\frac{1}{|\mathbf{a}|^2}[c \mathbf{a}-(\mathbf{a} \times \mathbf{b})]$
- D$\frac{1}{|\mathbf{a}|^2}[c \mathbf{a}+(\mathbf{a} \times \mathbf{b})]$
Solution
Since, $\mathbf{a}, \mathbf{b}$ and $\mathbf{a} \times \mathbf{b}$ are non-coplanar, hence
$\mathbf{y}=\mathbf{a} x+t b \mathbf{x}+(\mathbf{a} \times \mathbf{b}) z$
For some scalars $x, t$ and $z$,
Now,
$\begin{aligned}<br />& \mathbf{b}=\mathbf{a} \times \mathbf{y} \\<br />& \Rightarrow \mathbf{b}=(\mathbf{a} \times \mathbf{a}) x+(\mathbf{a} \times \mathbf{b}) t+\mathbf{a} \times(\mathbf{a} \times \mathbf{b}) z \\<br />& =0 x+(\mathbf{a} \times \mathbf{b}) t+[(\mathbf{a} \cdot \mathbf{b}) \mathbf{a}-(\mathbf{a} \cdot \mathbf{a}) \mathbf{b}] z \\<br />& {[\because \mathbf{a} \times \mathbf{a}=0, \mathbf{a} \cdot \mathbf{b}=0]} \\<br />& \Rightarrow \quad b=(\mathbf{a} \times \mathbf{b}) t-(\mathbf{a} \cdot \mathbf{a}) \mathbf{b} \cdot \mathbf{z} \\<br />& \Rightarrow \quad t=0 \text { and } z=\frac{-1}{|\mathbf{a}|^2}<br />\end{aligned}$
Also,
$\begin{aligned}<br />C & =\mathbf{a} \cdot \mathbf{y}=\mathbf{a} \cdot \mathbf{a} x+\mathbf{a} \cdot \mathbf{b} t+\mathbf{a} \cdot(\mathbf{a} \times \mathbf{b}) z \\<br />& =|\mathbf{a}|^2 x+0 t+0 z \\<br />& =|\mathbf{a}|^2 x \quad[\because \mathbf{a} \cdot \mathbf{b}=0,(\mathbf{a} \mathbf{a} \mathbf{b})=0]<br />\end{aligned}$
$\begin{aligned} & \Rightarrow \quad x=\frac{c}{|\mathbf{a}|^2} \\ & \therefore \quad \mathbf{y}=\mathbf{a} x+\mathbf{b} t+(\mathbf{a} \times \mathbf{b}) z \\ & \mathbf{y}=\frac{\mathbf{a c}}{|\mathbf{a}|^2}+\mathbf{b} 0+(\mathbf{a} \times \mathbf{b})\left\{\frac{-1}{|\mathbf{a}|^2}\right\} \\ & \Rightarrow \quad \mathbf{y}=\frac{1}{|\mathbf{a}|^2}[\mathbf{a c}-(\mathbf{a} \times \mathbf{b})] \\ & \end{aligned}$
$\mathbf{y}=\mathbf{a} x+t b \mathbf{x}+(\mathbf{a} \times \mathbf{b}) z$
For some scalars $x, t$ and $z$,
Now,
$\begin{aligned}<br />& \mathbf{b}=\mathbf{a} \times \mathbf{y} \\<br />& \Rightarrow \mathbf{b}=(\mathbf{a} \times \mathbf{a}) x+(\mathbf{a} \times \mathbf{b}) t+\mathbf{a} \times(\mathbf{a} \times \mathbf{b}) z \\<br />& =0 x+(\mathbf{a} \times \mathbf{b}) t+[(\mathbf{a} \cdot \mathbf{b}) \mathbf{a}-(\mathbf{a} \cdot \mathbf{a}) \mathbf{b}] z \\<br />& {[\because \mathbf{a} \times \mathbf{a}=0, \mathbf{a} \cdot \mathbf{b}=0]} \\<br />& \Rightarrow \quad b=(\mathbf{a} \times \mathbf{b}) t-(\mathbf{a} \cdot \mathbf{a}) \mathbf{b} \cdot \mathbf{z} \\<br />& \Rightarrow \quad t=0 \text { and } z=\frac{-1}{|\mathbf{a}|^2}<br />\end{aligned}$
Also,
$\begin{aligned}<br />C & =\mathbf{a} \cdot \mathbf{y}=\mathbf{a} \cdot \mathbf{a} x+\mathbf{a} \cdot \mathbf{b} t+\mathbf{a} \cdot(\mathbf{a} \times \mathbf{b}) z \\<br />& =|\mathbf{a}|^2 x+0 t+0 z \\<br />& =|\mathbf{a}|^2 x \quad[\because \mathbf{a} \cdot \mathbf{b}=0,(\mathbf{a} \mathbf{a} \mathbf{b})=0]<br />\end{aligned}$
$\begin{aligned} & \Rightarrow \quad x=\frac{c}{|\mathbf{a}|^2} \\ & \therefore \quad \mathbf{y}=\mathbf{a} x+\mathbf{b} t+(\mathbf{a} \times \mathbf{b}) z \\ & \mathbf{y}=\frac{\mathbf{a c}}{|\mathbf{a}|^2}+\mathbf{b} 0+(\mathbf{a} \times \mathbf{b})\left\{\frac{-1}{|\mathbf{a}|^2}\right\} \\ & \Rightarrow \quad \mathbf{y}=\frac{1}{|\mathbf{a}|^2}[\mathbf{a c}-(\mathbf{a} \times \mathbf{b})] \\ & \end{aligned}$
Hence, the correct answer is option: 'C'.