JEE Main
- Exam
- JEE Main
- Subject
- Physics
- Chapter
- Mathematics in Physics
JEE Main · Physics · Mathematics in Physics
The moment of the force, → F = 4 ^ i + 5 ^ j - 6 ^ k at ( 2, 0, - 3...
The moment of the force, \overset{→}{F} = 4\overset{^}{i} + 5\overset{^}{j} - 6\overset{^}{k} at about the point is given by
Options
- A(-7 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}-4 \widehat{\mathrm{k}})
- B(-4 \hat{\mathrm{i}}-\hat{\mathrm{j}}-8 \widehat{\mathrm{k}})
- C(-8 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-7 \hat{\mathrm{k}})
- D(-7 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-8 \widehat{\mathrm{k}})
Correct answer
D. (-7 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-8 \widehat{\mathrm{k}})
Step-by-step solution
\overset{→}{F} = 4\overset{^}{i} + 5\overset{^}{j} - 6\overset{^}{k}
\overset{→}{r} = \left(2 - 2\right)\overset{^}{i} + \left(0 - \left(- 2\right)\right)\overset{^}{j} + \left(- 3 - \left(- 2\right)\right)\overset{^}{k} = 2\overset{^}{j} - 1\overset{^}{k}
\overset{→}{r} \times \overset{→}{F} = \left(2\overset{^}{j} - 1\overset{^}{k}\right) \times \left(4\overset{^}{k} + 5\overset{^}{j} - 6\overset{^}{k}\right)
= \left|\begin{cases}\overset{^}{i}& \overset{^}{j} & \overset{^}{k} \\ 0 & 2 & - 1 \\ 4 & 5 & - 6\end{cases}\right| = - 7\overset{^}{i} - 4\overset{^}{j} - 8\overset{^}{k}