A force \(\vec F = 3\hat i + 4\;\hat j - 3\;\hat k\) is applied at the point P, whose position vector is \(\vec r = 2\hat i - 2\hat j - 3\hat k\). What is the magnitude of the moment of the force about the origin?

Concept:

Moment = \(\vec r \times \vec F\), Where r is position vector and F is applied force.

Cross product of two vectors:

\(\begin{array}{l} \vec{a}=x_{1} \hat{1}+y_{1} \hat{j}+z_{1} \hat{k}\\ \vec{b}=x_{2} \hat{1}+y_{2} \hat{j}+z_{2} \hat{k} \\ \vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & j & \hat{k} \\ x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \end{array}\right| \end{array}\)

Calculation: 

\(\vec F = 3\hat i + 4\;\hat j - 3\;\hat k\) and \(\vec r = 2\hat i - 2\hat j - 3\hat k\)

Moment = \(\vec r \times \vec F\)

\(\begin{aligned} &=\left|\begin{array}{ccc} i & j & \hat{k} \\ 2 & -2 & -3 \\ 3 & 4 & -3 \end{array}\right|\\ &\begin{array}{l} = \hat{i}[(-2)(-3)-(-3)(4)]-\hat{j}[(-3)(2)-(3)(3)] +\hat{k}[(2)(4)-(-2)(3)] \end{array}\\ &=\hat{i}(6+12)-\hat{j}(-6+9)+\hat{k}(8+6)\\ &=18 \hat{i}-3 \hat{j}+14 \hat{k} \end{aligned}\)

Now, magnitude \(|\vec r \times \vec F|\) = √((18)2 + (-3)2  + (14)2)

= √(324 + 9 + 196)

= √529

= 23  

Hence, option (1) is correct.