Question

A body is rotating with angular velocity $\overrightarrow{\omega}=\left(3\hat{i}-4\hat{j}+\hat{k}\right)$. The linear velocity of a point having position vector $\overrightarrow{r}=\left(5\hat{i}-6\hat{j}+6\hat{k}\right)$ is

• A. $6\hat{i}+2\hat{j}-3\hat{k}$
• B. $18\hat{i}+3\hat{j}-2\hat{k}$
• C. $-18\hat{i}-13\hat{j}+2\hat{k}$
• D. $6\hat{i}-2\hat{j}+8\hat{k}$

Answer 1

Answer: (C)

Here, $\vec{\omega }=3\hat{i}-4\hat{j}+\hat{k}$
$\overrightarrow{r}=5\hat{i}-6\hat{j}+6\hat{k}$
As $\overrightarrow{\upsilon}=\vec{\omega} \times\overrightarrow{r}$
$=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ 3&-4&1\\ 5&-6&6\end{vmatrix}$
$=\hat{i} (-24-\left(-6\right))+\hat{j} \left(5-18\right)+\hat{k} (-18-\left(-20\right))$
$=-18 \hat{i}-13\hat{j}+2\hat{k}$