# Question 6 Exercise 3.4

Solutions of Question 6 of Exercise 3.4 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

A force $\vec{F}=3 \hat{i}-2 \hat{j}+5 \hat{k}$ acts on a particle at $(1,-2,2)$. Find the moment or torque of the force about the origin.

Let $\vec{r}$, the position vector of point $P(1,-2.2)$ relative to the origin that is $O(0,0,0)$, then
\begin{align}\vec{r}&=\overrightarrow{O P}\\ &=(1,-2,2)-(0,0,0) \\ \Rightarrow \vec{r}&=(1,-2,2).\\ \text { Hence } \vec{M}-\vec{r} \times \vec{F}&=\left|\begin{array}{ccc} i & j & k \\ 1 & -2 & 2 \\ 3 & -2 & 5 \end{array} \right| \\ \Rightarrow \vec{M}&=(-10+4) \hat{i}-(5-6) \hat{j}+(-2+6) \hat{k} \\ \Rightarrow \vec{M}&=-6 \hat{i}+\hat{j}+4 \hat{k}.\end{align}

A force $\vec{F}=3 \hat{i}-2 \hat{j}+5 \hat{k}$ acts on a particle at $(1,-2,2)$. Find the moment or torque of the force about the point $(1,2,1)$

Let $\vec{r}$ be the position vector of point $P(1,-2,2)$ relative to $A(1,2,1)$ then
\begin{align}\vec{r}&=\overrightarrow{A P}\\ &=(1,-2,2)-(1,2 ,1)\\ \Rightarrow \vec{r}&=(0 ,-4,1) \\ \vec{M}&=\vec{r} \times \vec{F}\\ &=\left |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 0 & -4 & 1 \\ 3 & -2 & 5 \end{array}\right| \\ \Rightarrow \vec{M}&=(-20+2) \hat{i}-(0-3) \hat{j}+(0+12) \hat{k} \\ \Rightarrow \vec{M}&=-18 \hat{i}+3 \hat{j}+12 \hat{k} .\end{align}