Question

A force \(\vec F = 3\hat i + 4\;\hat j - 3\;\hat k\) is applied at the point P, which lies on 1 unit along the positive x-axis. What is the magnitude of the moment of the force about the point Q which lies on 1 unit along the negative x-axis?
1. 12
2. 5
3. 10
4. 3

Answer 1

Correct Answer - Option 3 : 10

Concept:

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

Cross product of two vectors:

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

Calculation:

Given that

\(\vec F = 3̂ i + 4\;̂ j - 3\;̂ k\)

The position vector of points P & Q is î and -î respectively. 

\(\Rightarrow\ \vec{PQ} = -\vec{i} -\hat{i}\ =-2\ \hat{i}\)

We know that, moment M = \(\vec r \times \vec F\)

\(\Rightarrow\ \vec{M}= \left| {\begin{array}{*{20}{c}} {\widehat i}&{\widehat j}&{\widehat k}\\ { - 2}&0&{ 0}\\ 3&4&-3 \end{array}} \right|\)

\(\Rightarrow\ \vec{M}= \hat{i}(0\ -\ 0)\ -\ ​​​​\hat{j}(6\ -\ 0)\ +\ \hat{k}(-8\ -\ 0)\)

\(\Rightarrow\ \vec{M}= -​​​​6\hat{j}\ -\ 8\hat{k}\)

Now, magnitude

 \(|\vec M |\ =\ \sqrt{6^2\ +\ 8^2}\) 

\(|\vec M |\ =\ 10\)

Hence, option (3) is correct.