91Ó°ÊÓ

$\stackrel{â‡\P Ä}{a}.\stackrel{â‡\P Ä}{b}=\stackrel{â‡\P Ä}{a}.\stackrel{â‡\P Ä}{c}$

The problem deals with the scalar product. It is also called dot product which is the multiplication operation on two vectors. It is the sum of the product of the corresponding components of the vectors.

Formula:

$$\begin{gathered} \stackrel{â‡\P Ä}{a}.\stackrel{â‡\P Ä}{b}=ab\cos x \\ \stackrel{â‡\P Ä}{a}.\stackrel{â‡\P Ä}{c}=ac\cos y \end{gathered}$$

According to the property of the scalar product, the scalar product of two vectors is equivalent to the product of the magnitude of one vector with the component of the other, in the direction of the first.

It x is the angle between vectors $\stackrel{â‡\P Ä}{a}\mathrm{and}\stackrel{â‡\P Ä}{b}$and y is the angle between vectors role="math" localid="1660894029946" $\stackrel{â‡\P Ä}{a}\mathrm{and}\stackrel{â‡\P Ä}{c}$then,

$$\begin{gathered} \stackrel{â‡\P Ä}{a}.\stackrel{â‡\P Ä}{b}=ab\cos x \\ \stackrel{â‡\P Ä}{a}.\stackrel{â‡\P Ä}{b}=a(b\cos x) \\ \stackrel{â‡\P Ä}{a}.\stackrel{â‡\P Ä}{c}=ac\cos y \\ \stackrel{â‡\P Ä}{a}.\stackrel{â‡\P Ä}{c}=a(c\cos y) \end{gathered}$$

(i)

From equations (i) and (ii), since $\stackrel{â‡\P Ä}{a}.\stackrel{â‡\P Ä}{b}=\stackrel{â‡\P Ä}{a}.\stackrel{â‡\P Ä}{c}$,

$$\begin{gathered} a\left(b\cos x\right)=a\left(c\cos y\right) \\ b\cos x=c\cos y \end{gathered}$$ (ii)

This does not mean that$\stackrel{â‡\P Ä}{b}=\stackrel{â‡\P Ä}{c}$

But, if the vectoris zero, then we can have,

$$\begin{gathered} \stackrel{â‡\P Ä}{a}.\stackrel{â‡\P Ä}{b}=0 \\ \mathrm{And}, \\ \stackrel{â‡\P Ä}{\mathrm{a}}.\stackrel{â‡\P Ä}{\mathrm{c}}=0 \end{gathered}$$

In this situation, we may have$\stackrel{â‡\P Ä}{b}≠\stackrel{â‡\P Ä}{c}$