Question

Under what condition will the equality: $\overrightarrow A \times \overrightarrow B = \overrightarrow A .\overrightarrow B$ hold good?

Answer 1

In this question we have to assume some angle between the vectors $\overrightarrow A$ and $\overrightarrow B$. Let that be $\theta$. So on angle $\theta$ the value of $\overrightarrow A \times \overrightarrow B$ and $\overrightarrow A .\overrightarrow B$ are equal. So use the formula to of $\overrightarrow A .\overrightarrow B$ and $\overrightarrow A \times \overrightarrow B$ after that equate both of them and from this equation we get the value of $\theta$. In this question We have to find the value of $\theta$.

Complete step-by-step solution:
Given,
This relation is given $\overrightarrow A \times \overrightarrow B = \overrightarrow A .\overrightarrow B$
To find,
Angle between the vector $\overrightarrow A$ and vector $\overrightarrow B$.
Let, the angle between the vectors $\overrightarrow A$ and $\overrightarrow B$is $\theta$.
Formula used:
Formula of cross product in terms of magnitudes of vector $\overrightarrow A$ and vector $\overrightarrow B$.
$\overrightarrow A \times \overrightarrow B = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\sin \theta$ ……(i)
Formula of dot product in terms of magnitudes of vector $\overrightarrow A$ and vector $\overrightarrow B$.
$\overrightarrow A .\overrightarrow B = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta$ ……(ii)
Both the equation (i) and equation (ii) are equal (given in the question)
$\overrightarrow A \times \overrightarrow B = \overrightarrow A .\overrightarrow B$ (given)……(iii)
Put the values in the equation (iii) from the equation (i) and equation (ii)
$\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\sin \theta = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta$
Here,
$\left| {\overrightarrow A } \right| = A$and $\left| {\overrightarrow B } \right| = B$
On putting the magnitude
$AB\sin \theta = AB\cos \theta$
On dividing $AB\cos \theta$ on both the side
$\dfrac{{AB\sin \theta }}{{AB\cos \theta }} = 1$
After canceling $AB$ from numerator and denominator
$\dfrac{{\sin \theta }}{{\cos \theta }} = 1$
$\tan \theta = 1$ ($\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta$)
$\tan \theta$ is equal to $1$ at an angle of ${45^ \circ }$
$\tan 45 = 1$
$\Rightarrow \theta = {45^ \circ }$
Final answer:
So the value of $\theta$ satisfying the equation $\overrightarrow A \times \overrightarrow B = \overrightarrow A .\overrightarrow B$ is
$\Rightarrow \theta = {45^ \circ }$

Note: On looking towards the question we have to make the equations and put all the values. If any of the variables are unknown then we assume some value of them. And then make an equation and on solving we are able to find the value of that unknown part. That is the answer. In order to get the answer, we must know the formula of the cross product and dot product and apply those formulas.