Chapter 6.2, Problem 68E

Solution

i.

To explain: Can vector and vector be parallel?

   only when either $\sin t=0$ or .

Given:

A trigonometric expression .

Concept used:

Two vectors and are said to be parallel only when where $k\ne 0$ and it is scalar.

Calculation:

   and be parallel only when $\sin t=0$ or .

If $\sin t=0$ then,

  $\begin{array}{c}\cos t=\sqrt{1-{\sin }^{2}t}\\ =\sqrt{1-0}\\ =1\end{array}$

Thus, which means and are parallel.

And when then again and are parallel.

Conclusion:

Hence, and be parallel only when either $\sin t=0$ or .

ii.

To explain: Can vector and vector be parallel?

   only when either $\cos t=0$ or .

Given:

A trigonometric expression .

Concept used:

Two vectors and are said to be parallel only when where $k\ne 0$ and it is scalar.

Calculation:

   and be parallel only when $\cos t=0$ or .

If $\cos t=0$ then,

  $\begin{array}{c}\sin t=\sqrt{1-{\cos }^{2}t}\\ =\sqrt{1-0}\\ =1\end{array}$

Thus, which means and are parallel.

And when then again and are parallel.

Conclusion:

   and be parallel only when $\cos t=0$ or .

iii.

To explain: Can vector and vector be parallel?

   only when $\sin t=\cos t$ where $k=\frac{1}{\sqrt{2}}$

Given:

A trigonometric expression .

Concept used:

Two vectors and are said to be parallel only when where $k\ne 0$ and it is scalar.

Calculation:

   and be parallel only when $\sin t=\cos t$ .

For example, consider $t={45}^{\circ }$ then $\sin t=\cos t=\frac{1}{\sqrt{2}}$ . So,

  

Thus, and are parallel.

Conclusion:

Hence, and be parallel only when $\sin t=\cos t$ .