Question

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

Answer 1

Answer: 1

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