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The torque due to force $\mathbf{F}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ acting at a…
The torque due to force $\mathbf{F}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ acting at a point $\mathbf{r}=8 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ is
$-4 \hat{\mathbf{i}}-17 \hat{\mathbf{j}}+22 \hat{\mathbf{k}}$ $4 \hat{\mathbf{i}}+17 \hat{\mathbf{j}}-22 \hat{\mathbf{k}}$ $8 \hat{\mathbf{i}}-17 \hat{\mathbf{j}}+22 \hat{\mathbf{k}}$ $-4 \hat{\mathbf{i}}-17 \hat{\mathbf{j}}-22 \hat{\mathbf{k}}$ Solution
Given,
$
\begin{aligned}
F & =3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{r}=8 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\
\tau & =\mathbf{v} \times \mathbf{F}=(8 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \times(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \\
& =\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
8 & 2 & 3 \\
3 & -2 & -1
\end{array}\right| \\
& =\hat{\mathbf{i}}(-2+6)-\hat{\mathbf{j}}(-8-9)+\hat{\mathbf{k}}(-16-6) \\
& =4 \hat{\mathbf{i}}+17 \hat{\mathbf{j}}-22 \hat{\mathbf{k}}
\end{aligned}
$
Asked in: AP EAMCET 2021 (25 Aug Shift 1)
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