Question

Prove that \[\vec A\left( {\vec A \times \vec B} \right) = 0\]

Answer 1

Hint: We know that there are two types of products in vectors. One is a dot product and the other is a cross product. Here a combination of both is given. So we will use properties of both the products. This will help to prove the answer.

Complete step by step solution:
We know that, in case of a cross product the angle between the vectors is defined by sin function. Thus the bracket can be written as,
\[\vec A \times \vec B = \left| a \right|\left| b \right|\sin \theta \]
This tells us that vector A is perpendicular to both A and B. and all these three vectors lie in the plane such that the \[\vec A \times \vec B\] is also a vector and that is perpendicular to vector \[\vec A.\] in the plane. Now the condition is like two vectors perpendicular to each other are in the dot product case.
Then we know that the dot product of two vectors that are perpendicular is equal to zero.
Such that,
\[\vec A.\left( {Perpendicular{\text{ }}to{\text{ }}vector{\text{ }}A} \right) = 0\]
Thus the statement above is proved.
So, the correct answer is “Option B”.

Note: Note that dot product of two identical vectors is zero.
\[\hat i,\hat j\& \hat k\] are said to be unit vectors.
the dot product of different unit vectors is zero. But that of the same unit vectors is 1.
 This is kept in consideration when we solve the problems of vectors.
The cross product in vectors is defined as, \[\vec A \times \vec B = \left| a \right|\left| b \right|\sin \theta \]
And the dot product is defined as \[\vec A.\vec B = \left| a \right|\left| b \right|\cos \theta \]
Cos and sin are the angles between the vectors.