Question

What is the value of \[{\left( {A \times B} \right)^2} + {\left( {A \cdot B} \right)^2}\] ?

Answer 1

Hint: To solve this question, one must know about the concept of cross (vector) and dot (scalar) products.Here, in this question firstly we have calculated both the values of \[{\left( {A \times B} \right)^2} + {\left( {A \cdot B} \right)^2}\] separately and at last added both the obtained value and hence in this way we got our required solution.

Complete step by step answer:
Two vectors $A$ and $B$ are given above. And to solve two vector product we do,
$A \times B = \left| A \right|\left| B \right|\sin \theta $
So, calculating, \[{\left( {A \times B} \right)^2}\] we will get,
${A^2}{B^2}{\sin ^2}\theta $
And two vectors $A$ and $B$ are given above. And to solve two scalar product we do,
$A \cdot B = \left| A \right|\left| B \right|\cos \theta $
So, calculating, \[{\left( {A \cdot B} \right)^2}\] we will get,
${A^2}{B^2}{\cos ^2}\theta $

Now, as per question we have to add both the obtained value, so adding we will get,
${A^2}{B^2}{\sin ^2}\theta + {A^2}{B^2}{\cos ^2}\theta $
Now on further solving,
${A^2}{B^2}{\sin ^2}\theta + {A^2}{B^2}{\cos ^2}\theta = {A^2}{B^2}\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) \\ $
And we know that the value of ${\sin ^2}\theta + {\cos ^2}\theta = 1$
${A^2}{B^2}\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = {A^2}{B^2} \\ $

Therefore, the value of \[{\left( {A \times B} \right)^2} + {\left( {A \cdot B} \right)^2}\] is ${A^2}{B^2}$.

Note: Remember that the value of ${\sin ^2}\theta + {\cos ^2}\theta = 1$ .
When two vectors are scalarized, you get a number or a scalar. When it comes to identifying energy and work relationships, scalar products come in handy. The labour done by a Force (which is a vector) in displacing (a vector) an object is represented by the scalar product of Force and Displacement vectors, which is an example of a scalar product. The vector product, also known as the cross product of two vectors, is a new vector with a magnitude equal to the sum of the magnitudes of the two vectors plus the sine of the angle between them.According to the right-hand screw rule or right-hand thumb rule, the product vector is perpendicular to the plane containing the two vectors.