Question

If the angle between the vectors $\vec A$ and $\vec B$ is $\theta $, then the value of the product $(\vec B \times \vec A) \cdot \vec A$ equals
A) $B{A^2}\sin \theta $
B) $B{A^2}\cos \theta \sin \theta $
C) $B{A^2}\cos \theta $
D) zero

Answer 1

Hint: The cross product of two vectors gives us a vector that is perpendicular to both the vectors. The dot product of two vectors is scalar in nature and is equal to 0 if the two vectors are perpendicular to each other.

Formula used:
$\Rightarrow \vec B \times \vec A = BA\hat n\sin \theta $ where $\hat n$ is a unit vector that is perpendicular to both $\vec A$ and $\vec B$.

Complete step by step solution:
To evaluate the expression $(\vec B \times \vec A) \cdot \vec A$, let’s start with the term inside the bracket in the expression. i.e. $(\vec B \times \vec A)$ which can be determined as:
$\Rightarrow \vec B \times \vec A = BA\hat n\sin \theta $
In the above expression, we know that $\hat n$ is a unit vector that is perpendicular to both $\vec A$ and$\vec B$. Hence, the expression in question can be written as:
$\Rightarrow (\vec B \times \vec A) \cdot \vec A = (BA\sin \theta \hat n) \cdot \vec A$
Now, the dot product of two vectors $\vec M$ and $\vec N$ can be written in the form of
$\vec M \cdot \vec N = MN\cos \theta $ where $\theta $ is the angle between these two vectors. We notice here that if $\vec M$ and $\vec N$ are perpendicular to each other i.e. $\theta = 90^\circ $, $\cos 90^\circ = 0$ and hence, the dot product of two vectors will be zero.
Similarly, in the expression $(\vec B \times \vec A) \cdot \vec A = (BA\sin \theta \hat n) \cdot \vec A$ , we know that the vector $\hat n$ is a unit vector that is perpendicular to both $\vec A$ and$\vec B$.
Since $\hat n$ is perpendicular to $\vec A$, $(\vec B \times \vec A) \cdot \vec A = 0$ which corresponds to option (D).

Hence, the correct option is option (D).

Note:
We must be familiar with vector properties such as the cross and the dot product to approach such problems with ease. Option (B) also seems very tempting however we must remember that the cross product of vectors is a vector that has a direction in itself and must be taken into account when calculating the dot product after the cross product $(\vec B \times \vec A)$.