Question

Find the cross product of two vectors \[\overrightarrow{A}=3i+2j-4k\]and $\overrightarrow{B}=2i-3j-6k$ . Also find the magnitude of $\overrightarrow{A}\times \overrightarrow{B}$ .

Answer 1

Hint: Here we have two vectors and we need to find their cross product. First we will find their cross product with the necessary formula. The cross product will be a vector so it will have both the magnitude and direction. Then we will find the magnitude of that vector with the necessary formula.
Formula used:
$i\times i=j\times j=k\times k=0,i\times j=k,j\times k=i,k\times i=j$ ,\[\overrightarrow{A}\times \overrightarrow{B}=-\overrightarrow{B}\times \overrightarrow{A},\left| A \right|=\sqrt{A_{1}^{2}+A_{2}^{2}+A_{3}^{2}}\]

Complete answer:
\[\overrightarrow{A}=3i+2j-4k\] and $\overrightarrow{B}=2i-3j-6k$,so using
$i\times i=j\times j=k\times k=0,i\times j=k,j\times k=i,k\times i=j$
We have
$\overrightarrow{A}\times \overrightarrow{B}=(3i+2j-4k)\times (2i-3j-6k)=-9k+18j-4k-12i-8j-12i=-24i+10j-13k$Thus
$\overrightarrow{A}\times \overrightarrow{B}=-24i+10j-13k$ .
Therefore
$\left| \overrightarrow{A}\times \overrightarrow{B} \right|=\sqrt{{{(-24)}^{2}}+{{(10)}^{2}}+{{(-13)}^{2}}}=\sqrt{576+100+169}=\sqrt{845}=29.07$

Additional information:
There are two types of products in vector, one is dot product and the other one is cross product. If we have two vectors $\overrightarrow{A}$ and$\overrightarrow{B}$ and the angle between the two vectors is$\theta $ ,
then their dot product is given by
$\overrightarrow{A}.\overrightarrow{B}=\left| A \right|\left| B \right|\cos \theta $ . The dot product gives a scalar quantity. One example is work. The cross product of two vectors is given by
$\overrightarrow{A}\times \overrightarrow{B}=\left| A \right|\left| B \right|\sin \theta \times \overrightarrow{n}$ . Where $\overrightarrow{n}$ is a unit vector perpendicular to both$\overrightarrow{A}$ and$\overrightarrow{B}$ .The cross product gives a vector quantity. One example is torque.
A unit vector is a vector whose magnitude is unity. If we divide a vector by its magnitude we get the unit vector along the direction of the vector. Thus if$\overrightarrow{a}$ is the unit vector along the vector $\overrightarrow{A}$ , we have
$\overrightarrow{a}=\dfrac{\overrightarrow{A}}{\left| A \right|}$

Note:
In this type of question we have to do the calculations very minutely as it requires rigorous work. In this problem we have used the formula for cross product. The problem could have been done by calculating the cross product in determinant form. In case of finding the magnitude, we should not forget to do the square root in the final step.