Question

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are two vectors, then $\left( 2\overrightarrow{a}+3\overrightarrow{b} \right)\times \left( 5\overrightarrow{a}+7\overrightarrow{b} \right)+\left( \overrightarrow{a}\times \overrightarrow{b} \right)$ is equal to?
(a) $0$
(b) $1$
(c) $\overrightarrow{a}\times \overrightarrow{b}$
(d) $\overrightarrow{b}\times \overrightarrow{a}$

Answer 1

Assume the given expression as E. Take the cross product $\left( 2\overrightarrow{a}+3\overrightarrow{b} \right)\times \left( 5\overrightarrow{a}+7\overrightarrow{b} \right)$ and simplify it using the relations $\overrightarrow{a}\times \overrightarrow{a}=0$ and $\overrightarrow{b}\times \overrightarrow{b}=0$. Further, use the relation $\overrightarrow{b}\times \overrightarrow{a}=-\overrightarrow{a}\times \overrightarrow{b}$ by considering the fact that the rotation of vectors in the two cases are in opposite direction and the vector obtained by the vector product is perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$. Add the term $\left( \overrightarrow{a}\times \overrightarrow{b} \right)$ present at the end to get the answer.

Complete step-by-step solution:
Here we have been provided with two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, we are asked to find the value of the expression $\left( 2\overrightarrow{a}+3\overrightarrow{b} \right)\times \left( 5\overrightarrow{a}+7\overrightarrow{b} \right)+\left( \overrightarrow{a}\times \overrightarrow{b} \right)$. Let us assume the expression as E, so we have,
$\Rightarrow E=\left( 2\overrightarrow{a}+3\overrightarrow{b} \right)\times \left( 5\overrightarrow{a}+7\overrightarrow{b} \right)+\left( \overrightarrow{a}\times \overrightarrow{b} \right)$
Now, first considering the cross product $\left( 2\overrightarrow{a}+3\overrightarrow{b} \right)\times \left( 5\overrightarrow{a}+7\overrightarrow{b} \right)$ we have,
\Rightarrow E=\left\\{ 10\left( \overrightarrow{a}\times \overrightarrow{a} \right)+14\left( \overrightarrow{a}\times \overrightarrow{b} \right)+15\left( \overrightarrow{b}\times \overrightarrow{a} \right)+21\left( \overrightarrow{b}\times \overrightarrow{b} \right) \right\\}+\left( \overrightarrow{a}\times \overrightarrow{b} \right)
We know that $\left( \overrightarrow{a}\times \overrightarrow{b} \right)=ab\sin \theta \left( {\hat{n}} \right)$, where a and b are the magnitudes of $\overrightarrow{a}$ and $\overrightarrow{b}$ respectively, $\theta$ is the angle between the two vectors and $\left( {\hat{n}} \right)$ is a unit vector perpendicular to the planes of both $\overrightarrow{a}$ and $\overrightarrow{b}$. Since the angle between two equal vectors is ${{0}^{\circ }}$ and we have $\sin {{0}^{\circ }}=0$, therefore $\left( \overrightarrow{a}\times \overrightarrow{a} \right)=0$ and $\left( \overrightarrow{b}\times \overrightarrow{b} \right)=0$.
\Rightarrow E=\left\\{ 14\left( \overrightarrow{a}\times \overrightarrow{b} \right)+15\left( \overrightarrow{b}\times \overrightarrow{a} \right) \right\\}+\left( \overrightarrow{a}\times \overrightarrow{b} \right)
In the cross product $\left( \overrightarrow{a}\times \overrightarrow{b} \right)$ we rotate $\overrightarrow{a}$ towards $\overrightarrow{b}$ and in cross product $\left( \overrightarrow{b}\times \overrightarrow{a} \right)$ we rotate $\overrightarrow{b}$ towards $\overrightarrow{a}$, so in the two cases the vectors obtained are perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$ but they are opposite in direction. If the unit vector along $\left( \overrightarrow{a}\times \overrightarrow{b} \right)$ is $\hat{n}$ then the unit vector along $\left( \overrightarrow{b}\times \overrightarrow{a} \right)$ will be $-\hat{n}$. Therefore, we have the relation $\overrightarrow{b}\times \overrightarrow{a}=-\overrightarrow{a}\times \overrightarrow{b}$, so we get,

& \Rightarrow E=\left\\{ 14\left( \overrightarrow{a}\times \overrightarrow{b} \right)-15\left( \overrightarrow{a}\times \overrightarrow{b} \right) \right\\}+\left( \overrightarrow{a}\times \overrightarrow{b} \right) \\\ & \Rightarrow E=-\left( \overrightarrow{a}\times \overrightarrow{b} \right)+\left( \overrightarrow{a}\times \overrightarrow{b} \right) \\\ & \therefore E=0 \\\ \end{aligned}$$ **Hence, option (a) is the correct answer.** **Note:** Note that the cross product results in a vector quantity and that is why it is also known as the vector product, while the dot product results in a scalar quantity and hence it is called the scalar product. In case of dot product we have the relation $$\overrightarrow{a}.\overrightarrow{b}=ab\cos \theta $$ and it is equal to 0 when the two vectors are perpendicular to each other.