# Question 1 Exercise 3.4

Solutions of Question 1 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.

Find the cross product $\hat{j} \times(2 \hat{j}+3 \hat{k})$

Let \begin{align}\vec{a}=\hat{j}&=0 \hat{i}+\hat{j}+0 \hat{k}\\ \vec{b}&=0 \hat{i}+2 \hat{j}-3 \hat{k}\\ \vec{a} \times \vec{b}&=\hat{j} \times(2 \hat{j}+3 \hat{k})\\ &=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 0 \\ 0 & 2 & 3\end{array}\right|\\ \Rightarrow \vec{a} \times \vec{b}&=(3-0) \hat{i}=3 \hat{i}.\end{align}

Find the cross product $(2 \hat{i}-3 \hat{j}) \times \hat{k}$

Let \begin{align}\vec{a}&=2 \hat{i}-3 \hat{j}\\ \vec{b}&=\hat{k} \\ \therefore \vec{a} \times \vec{b}&=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 0 \\ 0 & 0 & 1 \end{array}\right| \\ \Rightarrow \vec{a} \times \vec{b}&=(-3-0) \hat{i}-(2-0) \hat{j} \\ \Rightarrow \vec{a} \times \vec{b}&=-3 \hat{i}-2 \hat{j} .\end{align}

Find the cross product $(2 \hat{i}-3 \hat{j}+5 \hat{k}) \times(6 \hat{i}+2 \hat{j}-3 \hat{k})$

Let \begin{align}\vec{a}&=2 \hat{i}-3 \hat{j}+5 \hat{k}\\ \vec{b}&=6 \hat{i}+2 \hat{j}-3 \hat{k} \\ \therefore \vec{a} \times \vec{b}&=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 5 \\ 6 & 2 & -3 \end{array}\right| \\ \Rightarrow \vec{a} \times \vec{b}&=(9-10) \hat{i}-(-6-30) \hat{j} +(4+18) \hat{k} \\ \Rightarrow \vec{a} \times \vec{b}&=-\hat{i}+36 \hat{j}+22 \hat{k} .\end{align}