Name: Resolving Vectors Into Components
Uploaded: 2019-07-19
Duration: 5 min 37 s
Description: A vector can be resolved into a number of components according to convenience. Watch this video to learn how this can make our lives easier while solving pro...

Question

Prove that scalar and vector product of two vectors are distributive in nature.

Accepted Answer

Consider the vector diagrams given below representing the above vectors.

From the diagram, $\vec{A} . \vec{B} = A (B \cos θ_{1})$

Putting the values, we get, $\vec{A} . \vec{B} = (O P) (O S)$

Similarly,

$\begin{array}{rcl} \vec{A} . \vec{C} & = & A (C \cos θ_{2}) \\ = & (O P) (S T) \end{array}$

Thus,

$\begin{array}{rcl} \vec{A} . (\vec{B} + \vec{C}) & = & O P (O T) \\ = & O P . (O S + S T) \\ = & O P . O S + O P . S T \\ = & \vec{A} . \vec{B} + \vec{A} . \vec{C} \end{array}$

This proves the distributive property of dot product of two vectors.ature.

Again,

$\begin{array}{rcl} \vec{A} X \vec{B} & = & A (B \sin θ_{1}) \\ = & (O P) (N S) \end{array}$

Similarly, we have,

$\begin{array}{rcl} \vec{A} X \vec{C} & = & A (C \sin θ_{2}) \\ = & (O P) (Q T) \end{array}$

Thus,

$\begin{array}{rcl} \vec{A} X (\vec{B} + \vec{C}) & = & (O P) (Q T) \\ = & (O P) (T M + M Q) \\ = & (O P) (T M) + (O P) (M Q) \\ = & \vec{A} X \vec{B} + \vec{A} X \vec{C} \end{array}$

This proves the distributive property of cross product of two vectors

Prove that scalar and vector product of two vectors are distributive in nature.

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