Question

If (vector) $\left| A \right|=2$and (vector) $\left| B \right|=2$ and $\left| A\times B \right|=8$, then what is $A.B$ equal to?
(1) $6$
(2) $2$
(3) $20$
(4) $8$

Answer 1

You should know the difference between the scalar product and vector product of two vectors. In a scalar product, direction has no use but in a vector product, it also gives us direction. Scalar product is represented by a dot and the vector product is represented by a cross.

Complete step-by-step solution:
The Scalar product of two vectors can be defined as the product of magnitudes of any vector and the cosine of the angle between those vectors. The Scalar product is also used to give the relation between energy and work. The Scalar product of any two vectors can be given by the formula below. Suppose A and B are any two vectors then their scalar product can be given as-
$A.B=AB\cos \theta$
If the angle $\theta$ between the vectors in the scalar product becomes $90{}^\circ <\theta \le 180{}^\circ$ then the scalar product comes out to be a negative value. $$$$
The Vector product of two vectors gives out the vector which is perpendicular to both of those vectors. The Vector product is obtained by multiplying the magnitude of both the vectors and the sine of angle between them. The right-hand thumb rule is used to find out the direction of the cross product. The Area of the parallelogram can be determined with the help of the cross product. Suppose A and B are two vectors then their cross product will be given as-
$A\times B=AB\sin \theta$
Also, a scalar product of two vectors is commutative while a cross product is not commutative.
So, we have gained much knowledge about scalar and vector products, now we will solve our question.
In the above question, it is given that,

& \left| A \right|=2 \\\ &\Rightarrow \left| B \right|=5 \\\ &\Rightarrow \left| A\times B \right|=8 \\\ \end{aligned}$$ Now we have to find the value of $$A.B$$ So according to the formula $${{\left| A\times B \right|}^{2}}+{{(A.B)}^{2}}={{\left| A \right|}^{2}}{{\left| B \right|}^{2}}$$ ……..eq(1) Now we will put the given values in eq(1) and we will find out the correct solution which is as follows. $$\begin{aligned} & {{\left| A\times B \right|}^{2}}+{{(A.B)}^{2}}={{\left| A \right|}^{2}}{{\left| B \right|}^{2}} \\\ &\Rightarrow {{8}^{2}}+{{(A.B)}^{2}}={{(2)}^{2}}{{(5)}^{2}} \\\ &\Rightarrow {{(A.B)}^{2}}=100-64 \\\ \end{aligned}$$ $$\Rightarrow {{(A.B)}^{2}}=36$$ $$\begin{aligned} &\Rightarrow A.B=\sqrt{36} \\\ &\Rightarrow A.B=6 \\\ \end{aligned}$$ So the scalar product of vectors A and B is $$6$$. **So the correct answer will be (1) $$6$$** **Note:** Scalar product is also known as dot product and vector product is also known as the cross product. If two vectors are parallel to each other then there will be an angle of zero degrees between them which in turn makes the cross product of two vectors equal to zero. On the other hand, if two vectors are perpendicular to each other, then their dot product will be equal to zero