91Ó°ÊÓ

It has been given that A and B is a unit vector with an angle $\mathrm{θ}$ between them.

A quantity that has magnitude as well as direction is called a vector. It is typically denoted by an arrow in which the head determines the direction of the vector and the length determines it magnitude.

In order to evaluate the expression, First evaluate individual terms. Use the fact that$\hat{C}$ is normal to both$\hat{A}$ and$\hat{B}$ so and of unit magnitude.

$\left(\hat{A}×\hat{B}\right)=\left|\hat{A}\right|\left|\hat{B}\right|\sin \left(\mathrm{θ}\right)\hat{C}$

Both and are unit vectors, then$\left|\hat{A}\right|=\left|\hat{B}\right|=1$ .

$\left(\hat{A}×\hat{B}\right)=\sin \left(\mathrm{θ}\right)\hat{C}$

Now let role="math" localid="1659236656738" $\hat{D}$be a normal vector to both $\hat{C}$and role="math" localid="1659236668230" $\hat{B}$with unit magnitude as well, this means role="math" localid="1659236687766" $\hat{D}$is in the same plane as $\hat{A}$and $\hat{B}$and it makes and angle $\frac{\mathrm{π}}{2}-\mathrm{θ}$with,

$\left(\hat{B}×\hat{C}\right)=\left|\hat{C}\right|\left|\hat{B}\right|\sin \left(\frac{\mathrm{π}}{2}\right)\hat{D}=\hat{D}$

Now let$\hat{E}$ be a normal vector to both$\hat{C}$ and$\hat{A}$ with unit magnitude as well, this means$\hat{E}$ is in the same plane as$\hat{A}$ and$\hat{B}$ and it makes and angle$\frac{\mathrm{π}}{2}-\mathrm{θ}$ with,$\hat{B}$ which means and angle of $\mathrm{π}-\mathrm{θ}$with $\hat{D}$.

$\left(\hat{C}×\hat{A}\right)=\left|\hat{C}\right|\left|\hat{A}\right|\sin \left(\frac{\mathrm{π}}{2}\right)\hat{E}=\hat{E}$

Now write the expression.

$\left[\sin \left(\mathrm{θ}\right)\hat{C}×\hat{D}\right]×\hat{E}=-\hat{E}×\left[\sin \left(\mathrm{θ}\right)\hat{C}×\hat{D}\right]$

Use the triplet product property known as$BAC-CAB$ rule.

$-\left[\sin \left(\mathrm{θ}\right)\hat{C}\left(\hat{E}.\hat{D}\right)-\hat{D}\left(\sin \left(\mathrm{θ}\right)\hat{E}.\hat{C}\right)\right]$

Since$\hat{E}$ and$\hat{C}$ are normal to each other, then$\left(\hat{E}.\hat{C}\right)=0$

It must be known that$\left(\hat{E}.\hat{D}\right)=\left|\hat{E}\right|\left|\hat{D}\right|\cos \left(\mathrm{π}-\mathrm{θ}\right)=-\cos \left(\mathrm{θ}\right)$ .

Thus, our expression is:$-\left[-\sin \left(\mathrm{θ}\right)\cos \left(\mathrm{θ}\right)\hat{C}-0\right]=\sin \left(\mathrm{θ}\right)\cos \left(\mathrm{θ}\right)\hat{C}$

Hence the evaluation is found to be$[(\hat{\mathrm{A}}×\hat{\mathrm{B}})×(\hat{\mathrm{B}}×\hat{\mathrm{C}})×(\hat{\mathrm{C}}×\hat{\mathrm{A}})]=\sin \left(\mathrm{θ}\right)\cos \left(\mathrm{θ}\right)\hat{\mathrm{C}}$ .