Question

Find the moment about the point $i+2j-k$ of a force represented by $3i+k$ acting through the point $2i-j+3k$.
(A) $-3i+13j+9k$
(B) $3i-13j-9k$
(C) $3i-11j-9k$
(D) $-3i+11j+9k$

Answer 1

Hint: We solve this question by first considering the given vector of force and point of force and the point where the moment needs to be calculated. Then we consider the formula for the moment $\tau =\overrightarrow{r}\times \overrightarrow{F}$. Then we find the position vector of the point where force is exerted with respect to the point of force. Then we find the value of the cross product using the formula, the cross product of two vectors, $\overrightarrow{a}={{a}_{1}}i+{{a}_{2}}j+{{a}_{3}}k$ and $\overrightarrow{b}={{b}_{1}}i+{{b}_{2}}j+{{b}_{3}}k$ is $\overrightarrow{a}\times \overrightarrow{b}=\left| \begin{matrix}
   i & j & k \\
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
\end{matrix} \right|$. Then we find the determinant and find the value of moment.

Complete step by step answer:
We are given that a force represented by $3i+k$ is acting through the point $2i-j+3k$.
We need to find the moment about the point $i+2j-k$.
First let us consider the formula for the moment about a point due to a force.
Moment =$\tau =\overrightarrow{r}\times \overrightarrow{F}$
Here $\overrightarrow{r}=$ position vector of the point from the point of Force.
$\overrightarrow{F}=$ vector of force
So, from the above formula we can see that moment is the cross product of the position vector of the point and force exerted on it.
So, first let us find the vector of the point $i+2j-k$ with respect to the point of force that is $2i-j+3k$.
So, we get the value of $\overrightarrow{r}$ as,
$\begin{align}
  & \Rightarrow \overrightarrow{r}=\left( 2i-j+3k \right)-\left( i+2j-k \right) \\
 & \Rightarrow \overrightarrow{r}=2i-j+3k-i-2j+k \\
 & \Rightarrow \overrightarrow{r}=i-3j+4k \\
\end{align}$
So, we get the position vector as $\overrightarrow{r}=i-3j+4k$.
Now we need to find the cross product of the vectors $\overrightarrow{r}=i-3j+4k$ and force $3i+k$.
Now let us consider the formula for the cross product of two vectors, $\overrightarrow{a}={{a}_{1}}i+{{a}_{2}}j+{{a}_{3}}k$ and $\overrightarrow{b}={{b}_{1}}i+{{b}_{2}}j+{{b}_{3}}k$.
$\overrightarrow{a}\times \overrightarrow{b}=\left| \begin{matrix}
   i & j & k \\
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
\end{matrix} \right|$
So, using this formula we get the cross product of $\overrightarrow{r}=i-3j+4k$ and force $3i+k$ as,
$\begin{align}
  & \Rightarrow \overrightarrow{r}\times \overrightarrow{F}=\left| \begin{matrix}
   i & j & k \\
   1 & -3 & 4 \\
   3 & 0 & 1 \\
\end{matrix} \right| \\
 & \Rightarrow \overrightarrow{r}\times \overrightarrow{F}=\left| \begin{matrix}
   -3 & 4 \\
   0 & 1 \\
\end{matrix} \right|i-\left| \begin{matrix}
   1 & 4 \\
   3 & 1 \\
\end{matrix} \right|j+\left| \begin{matrix}
   1 & -3 \\
   3 & 0 \\
\end{matrix} \right|k \\
 & \Rightarrow \overrightarrow{r}\times \overrightarrow{F}=\left( -3-0 \right)i-\left( 1-12 \right)j+\left( 0-\left( -9 \right) \right)k \\
 & \Rightarrow \overrightarrow{r}\times \overrightarrow{F}=-3i-\left( -11 \right)j+9k \\
 & \Rightarrow \overrightarrow{r}\times \overrightarrow{F}=-3i+11j+9k \\
\end{align}$
So, we get the moment as $-3i+11j+9k$.

So, the correct answer is “Option D”.

Note: There is a possibility of one making a mistake while solving this problem by taking the cross product of position vector and force as, $\overrightarrow{F}\times \overrightarrow{r}$ and then get the value of moment as $3i-11j-9k$ and mark option C as the answer. But it is wrong because the formula for moment is $\tau =\overrightarrow{r}\times \overrightarrow{F}$ and cross product of vectors is not scalar product, interchanging the vectors changes the sign, that is direction.