The radius vector r→=2i^+5k^ and the angular velocity of a particle is ω→=3i^−4k^. Then the linear velocity of the particle is

r¯=2i^+5k^ω¯=3i^−4k^v¯=ω¯×r¯ v¯=ijk30−4205=i(0)−j(15+8)+k(0)=−23j^

The radius vector r→=2i^+5k^ and the angular velocity of a particle is ω→=3i^−4k^. Then the linear velocity of the particle is

r¯=2i^+5k^ω¯=3i^−4k^v¯=ω¯×r¯ v¯=ijk30−4205=i(0)−j(15+8)+k(0)=−23j^

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The radius vector $\vec{r} = 2 \hat{i} + 5 \hat{k}$ and the angular velocity of a particle is $\vec{ω} = 3 \hat{i} - 4 \hat{k}$ . Then the linear velocity of the particle is

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$2 \hat{i} + 3 \hat{j} + \hat{k}$

$15 \hat{i} + 8 \hat{j} - 6 \hat{k}$

$15 \hat{i} - 8 \hat{j} + 6 \hat{k}$

$- 23 \hat{j}$

answer is D.

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$\bar{r} = 2 \hat{i} + 5 \hat{k} \bar{ω} = 3 \hat{i} - 4 \hat{k} \bar{v} = \bar{ω} \times \bar{r} \bar{v} = |\begin{array}{ccc} i & j & k \\ 3 & 0 & - 4 \\ 2 & 0 & 5 \end{array}| = i (0) - j (15 + 8) + k (0) = - 23 \hat{j}$

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The radius vector r→=2i^+5k^ and the angular velocity of a particle is ω→=3i^−4k^. Then the linear velocity of the particle is

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The radius vector $\vec{r} = 2 \hat{i} + 5 \hat{k}$ and the angular velocity of a particle is $\vec{ω} = 3 \hat{i} - 4 \hat{k}$ . Then the linear velocity of the particle is