Question

If $\hat{i}\times \left[\left(\overset{ \rightarrow }{a} - \hat{j}\right) \times \hat{i}\right]+\hat{j}\times \left[\left(\overset{ \rightarrow }{a} - \hat{k}\right) \times \hat{j}\right]+\hat{k}\times \left[\left(\overset{ \rightarrow }{a} - \hat{i}\right) \times \hat{k}\right]=0$ and $\overset{ \rightarrow }{a}=x\hat{i}+y\hat{j}+z\hat{k}$ , then the value of $8\left(x^{3} - x y + z x\right)$ is equal to

Answer 1

$\hat{i} \times[(\vec{a}-\hat{j}) \times \hat{i}]=(\hat{i} . \hat{i})(\vec{a}-\hat{j})-(\hat{i} \cdot(\vec{a}-\hat{j})) \hat{i}$
$=\overset{ \rightarrow }{a}-\hat{j}-\left(\hat{i} . \overset{ \rightarrow }{a}\right)\hat{i}$
Therefore, $\overset{ \rightarrow }{a}-\hat{j}-\left(\hat{i} . \overset{ \rightarrow }{a}\right)\hat{i}+\overset{ \rightarrow }{a}-\hat{k}+\left(\hat{j} . \overset{ \rightarrow }{a}\right)\hat{j}+\overset{ \rightarrow }{a}-\hat{i}-\left(\hat{k} . \overset{ \rightarrow }{a}\right)\hat{k}=0$
$\Rightarrow 3\overset{ \rightarrow }{a}-\left(\hat{i} + \hat{j} + \hat{k}\right)-\overset{ \rightarrow }{a}=0$
$\overset{ \rightarrow }{a}=\frac{1}{2}\left(\hat{i} + \hat{j} + \hat{k}\right)=x\hat{i}+y\hat{j}+z\hat{k}$
$x=y=z=\frac{1}{2}$
$8\left(x^{3} - x y + z x\right)=8\left(x^{3} - x^{2} + x^{2}\right)=8\times \frac{1}{8}=1$