Question

If $\vec{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}-{{a}_{3}}\hat{k}$ and $\vec{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}-{{b}_{3}}\hat{k}$ then find the value of $\vec{a}\times \vec{b}$

Answer 1

Hint: We have to do the cross product of two given vectors the cross product of two vectors is given by:
$\vec{A}\times \vec{B}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ {{x}_{1}} & {{y}_{1}} & {{z}_{1}} \\\ {{x}_{2}} & {{y}_{2}} & {{z}_{2}} \\\ \end{matrix} \right|$

Complete step-by-step answer:
When the vectors are $\vec{A}={{x}_{1}}\hat{i}+{{y}_{1}}\hat{j}+{{z}_{1}}\hat{k}$ and $\vec{B}={{x}_{2}}\hat{i}+{{y}_{2}}\hat{j}+{{z}_{2}}\hat{k}$
In the question given above, we have to do the cross product or vector product of two given vectors $\vec{a}$ and $\vec{b}$ . The cross product or vector product is a binary operation on two vectors in three-dimensional space and is denoted by the symbol (X). The cross product of the two vectors is a vector perpendicular to both the vectors and thus the resultant vector will be normal to the plane of vectors containing them. The cross product of any two vectors will be given by the determinant:
$\vec{A}\times \vec{B}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ {{x}_{1}} & {{y}_{1}} & {{z}_{1}} \\\ {{x}_{2}} & {{y}_{2}} & {{z}_{2}} \\\ \end{matrix} \right|$
When the vectors are $\vec{A}={{x}_{1}}\hat{i}+{{y}_{1}}\hat{j}+{{z}_{1}}\hat{k}$ and $\vec{B}={{x}_{2}}\hat{i}+{{y}_{2}}\hat{j}+{{z}_{2}}\hat{k}$
Here, $\hat{i}$ is the unit vector in x – direction, $\hat{j}$ is the unit vector in y direction and $\hat{k}$ is the unit vector in z direction. Thus, the cross product of the vectors given in question is:
$\vec{a}\times \vec{b}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ {{a}_{1}} & {{a}_{2}} & -{{a}_{3}} \\\ {{b}_{1}} & {{b}_{2}} & -{{b}_{3}} \\\ \end{matrix} \right|$
The above determinant of order 3 can be solved as shown below:

{{a}_{2}} & -{{a}_{3}} \\\ {{b}_{2}} & -{{b}_{3}} \\\ \end{matrix} \right|-\hat{j}\left| \begin{matrix} {{a}_{1}} & -{{a}_{3}} \\\ {{b}_{1}} & -{{b}_{3}} \\\ \end{matrix} \right|+\hat{k}\left| \begin{matrix} {{a}_{1}} & {{a}_{2}} \\\ {{b}_{1}} & {{b}_{2}} \\\ \end{matrix} \right|$$ Now, we can solve these determinants of order 2 by simply cross multiplying and subtracting the product. Thus, it is done as shown: $\vec{a}\times \vec{b}-=\hat{i}\left[ \left( {{a}_{2}} \right)\left( -{{b}_{3}} \right)-\left( -{{a}_{3}} \right)\left( {{b}_{2}} \right) \right]-\hat{j}\left[ \left( {{a}_{1}} \right)\left( -{{b}_{3}} \right)-\left( -{{a}_{3}} \right)\left( {{b}_{1}} \right) \right]+\hat{k}\left[ \left( {{a}_{1}} \right)\left( {{b}_{2}} \right)-\left( {{a}_{2}} \right)\left( {{b}_{1}} \right) \right]$ $$\begin{aligned} & \Rightarrow \vec{a}\times \vec{b}=\hat{i}\left[ -{{a}_{2}}{{b}_{3}}+{{a}_{3}}{{b}_{2}} \right]-\hat{j}\left[ -{{a}_{1}}{{b}_{3}}+{{a}_{3}}{{b}_{1}} \right]+\hat{k}\left[ {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}} \right] \\\ & \Rightarrow \vec{a}\times \vec{b}=\hat{i}\left[ {{a}_{3}}{{b}_{2}}-{{a}_{2}}{{b}_{3}} \right]+\hat{j}\left[ {{a}_{3}}{{b}_{1}}-{{a}_{1}}{{b}_{3}} \right]+\hat{k}\left[ {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}} \right] \\\ & \Rightarrow \vec{a}\times \vec{b}=\left( {{a}_{3}}{{b}_{2}}-{{a}_{2}}{{b}_{3}} \right)\hat{i}+\left( {{a}_{3}}{{b}_{1}}-{{a}_{1}}{{b}_{3}} \right)\hat{j}+\left( {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}} \right)\hat{k} \\\ \end{aligned}$$ Thus, above is the required cross product we wanted to calculate. The above cross product is perpendicular to the plane containing vectors $\vec{a}$ and $\vec{b}$ . Note: We could have also used the formula of cross product to solve the question. If $$\vec{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$$ and $$\vec{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$$ then the cross product is given by $\vec{a}\times \vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\sin \theta \hat{n}$ , where $\left| {\vec{a}} \right|$ and $\left| {\vec{b}} \right|$ are the length of vectors $\vec{a}$ and $\vec{b}$ and $\theta $ is the angle between the two vectors. $\hat{n}$ is the vector perpendicular to both the vectors $\vec{a}$ and $\vec{b}$