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Physics Question on System of Particles & Rotational Motion

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

A

6i^+2j^−3k^6\hat{i}+2\hat{j}-3\hat{k}

B

18i^+3j^−2k^18\hat{i}+3\hat{j}-2\hat{k}

C

−18i^−13j^+2k^-18\hat{i}-13\hat{j}+2\hat{k}

D

6i^−2j^+8k^6\hat{i}-2\hat{j}+8\hat{k}

Answer

−18i^−13j^+2k^-18\hat{i}-13\hat{j}+2\hat{k}

Explanation

Solution

Here, ω⃗=3i^−4j^+k^\vec{\omega }=3\hat{i}-4\hat{j}+\hat{k} r→=5i^−6j^+6k^\overrightarrow{r}=5\hat{i}-6\hat{j}+6\hat{k} As υ→=ω⃗×r→\overrightarrow{\upsilon}=\vec{\omega} \times\overrightarrow{r} =∣i^j^k^ 3−41 5−66∣=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\\ 3&-4&1\\\ 5&-6&6\end{vmatrix} =i^(−24−(−6))+j^(5−18)+k^(−18−(−20))=\hat{i} (-24-\left(-6\right))+\hat{j} \left(5-18\right)+\hat{k} (-18-\left(-20\right)) =−18i^−13j^+2k^=-18 \hat{i}-13\hat{j}+2\hat{k}