Q64 / 160 · TS EAMCET 2017 · Mathematics
If $\mathbf{a}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}$, then $(\mathbf{a} \times \hat{\mathbf{i}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}})+(\mathbf{a} \times \hat{\mathbf{j}}) \cdot(\hat{\mathbf{j}}+\hat{\mathbf{k}})+(\mathbf{a} \times \hat{\mathbf{k}}) \cdot(\hat{\mathbf{k}}+\hat{\mathbf{i}})=$
- A$x-y+z$
- Correct answerB$x+y+z$
- C$x+y-z$
- D$-x+y+z$
Solution
We have,
$\begin{aligned}<br />& \mathbf{a}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\<br />& (\mathbf{a} \times \hat{\mathbf{i}})(\hat{\mathbf{i}}+\hat{\mathbf{j}})+(\hat{\mathbf{a}} \times \hat{\mathbf{j}})(\hat{\mathbf{j}}+\hat{\mathbf{k}})+(\mathbf{a} \times \hat{\mathbf{k}})(\hat{\mathbf{k}}+\hat{\mathbf{i}}) \\<br />& =[\mathbf{a} \hat{\mathbf{i}} \hat{\mathbf{i}}]+[\mathbf{a} \hat{\mathbf{i}} \hat{\mathbf{j}}]+[\mathbf{a} \hat{\mathbf{j}} \hat{\mathbf{j}}]+[\mathbf{a} \hat{\mathbf{j}} \hat{\mathbf{k}}]+[\mathbf{a} \hat{\mathbf{k}} \hat{\mathbf{k}}]+[\mathbf{a} \hat{\mathbf{k}} \hat{\mathbf{i}}] \\<br />& =[\mathbf{a} \hat{\mathbf{i}} \hat{\mathbf{j}}]+[\mathbf{a} \hat{\mathbf{j}} \hat{\mathbf{k}}]+[\mathbf{a} \hat{\mathbf{k}} \hat{\mathbf{i}}] \\<br />& \qquad[\because[\mathbf{a} \hat{\mathbf{i}} \hat{\mathbf{i}}]=[\mathbf{a} \hat{\mathbf{j}} \hat{\mathbf{j}}]=[\mathbf{a} \hat{\mathbf{k}} \hat{\mathbf{k}}]=0] \\<br />& =\mathbf{a} \cdot(\hat{\mathbf{i}} \times \hat{\mathbf{j}})+\mathbf{a} \cdot(\hat{\mathbf{j}} \times \hat{\mathbf{k}})+\mathbf{a} \cdot(\hat{\mathbf{k}} \times \hat{\mathbf{i}}) \\<br />& =\mathbf{a} \cdot \hat{\mathbf{k}}+\mathbf{a} \cdot \hat{\mathbf{i}}+\mathbf{a} \cdot \hat{\mathbf{j}}=\mathbf{a} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\<br />& =(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=x+y+z<br />\end{aligned}$
$\begin{aligned}<br />& \mathbf{a}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\<br />& (\mathbf{a} \times \hat{\mathbf{i}})(\hat{\mathbf{i}}+\hat{\mathbf{j}})+(\hat{\mathbf{a}} \times \hat{\mathbf{j}})(\hat{\mathbf{j}}+\hat{\mathbf{k}})+(\mathbf{a} \times \hat{\mathbf{k}})(\hat{\mathbf{k}}+\hat{\mathbf{i}}) \\<br />& =[\mathbf{a} \hat{\mathbf{i}} \hat{\mathbf{i}}]+[\mathbf{a} \hat{\mathbf{i}} \hat{\mathbf{j}}]+[\mathbf{a} \hat{\mathbf{j}} \hat{\mathbf{j}}]+[\mathbf{a} \hat{\mathbf{j}} \hat{\mathbf{k}}]+[\mathbf{a} \hat{\mathbf{k}} \hat{\mathbf{k}}]+[\mathbf{a} \hat{\mathbf{k}} \hat{\mathbf{i}}] \\<br />& =[\mathbf{a} \hat{\mathbf{i}} \hat{\mathbf{j}}]+[\mathbf{a} \hat{\mathbf{j}} \hat{\mathbf{k}}]+[\mathbf{a} \hat{\mathbf{k}} \hat{\mathbf{i}}] \\<br />& \qquad[\because[\mathbf{a} \hat{\mathbf{i}} \hat{\mathbf{i}}]=[\mathbf{a} \hat{\mathbf{j}} \hat{\mathbf{j}}]=[\mathbf{a} \hat{\mathbf{k}} \hat{\mathbf{k}}]=0] \\<br />& =\mathbf{a} \cdot(\hat{\mathbf{i}} \times \hat{\mathbf{j}})+\mathbf{a} \cdot(\hat{\mathbf{j}} \times \hat{\mathbf{k}})+\mathbf{a} \cdot(\hat{\mathbf{k}} \times \hat{\mathbf{i}}) \\<br />& =\mathbf{a} \cdot \hat{\mathbf{k}}+\mathbf{a} \cdot \hat{\mathbf{i}}+\mathbf{a} \cdot \hat{\mathbf{j}}=\mathbf{a} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\<br />& =(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})=x+y+z<br />\end{aligned}$
Hence, the correct answer is option: 'B'.