91Ó°ÊÓ

The given figure $\stackrel{â‡\P Ä}{A}$is going along the positive axis. The four vectors as $\stackrel{â‡\P Ä}{B},\stackrel{â‡\P Ä}{C},\stackrel{â‡\P Ä}{D}\mathrm{and}\stackrel{â‡\P Ä}{\mathrm{E}}$ have the same magnitude but different orientations as shown in the above figure.

The scalar product of two vectors is a scalar quantity. Therefore, if the vectors are the same in magnitude and make the same angle as the third vector, then the scalar product of each of these two vectors with the third vector will be the same. In this case, the direction of the vectors will not matter.

The scalar product of two vectors is given by the following equation.

$\stackrel{â‡\P Ä}{A}.\stackrel{â‡\P Ä}{B}=AB\cos \mathrm{θ}$

According to the figure, $\stackrel{â‡\P Ä}{A}$is going along the positive x-axis. We have given $B=C=D=E$and they are orientated in different directions as shown in the figure.

According to the expression of the scalar product of two vectors$\stackrel{â‡\P Ä}{A}\mathrm{and}\stackrel{â‡\P Ä}{\mathrm{B}}$is,

$\stackrel{â‡\P Ä}{A}.\stackrel{â‡\P Ä}{B}=AB\cos \mathrm{θ}$

Similarly, scalar product of $\stackrel{â‡\P Ä}{A}\mathrm{and}\stackrel{â‡\P Ä}{C}$is,

$\stackrel{â‡\P Ä}{A}.\stackrel{â‡\P Ä}{C}=AC\cos \mathrm{θ}$

From the figure, role="math" localid="1660897218718" $B\cos \mathrm{θ}\mathrm{and}C\cos \mathrm{θ}$are the same because they are acting along the same direction as along the positive x-axis. Hence

$AB\cos \mathrm{θ}=AC\cos \mathrm{θ}$

Therefore,

$\stackrel{â‡\P Ä}{A}.\stackrel{â‡\P Ä}{B}=\stackrel{â‡\P Ä}{A}.\stackrel{â‡\P Ä}{C}$

According to the expression of the scalar product of two vectors $\stackrel{â‡\P Ä}{A}\mathrm{and}\stackrel{â‡\P Ä}{\mathrm{D}}$is,

$\stackrel{â‡\P Ä}{A}.\stackrel{â‡\P Ä}{D}=-AD\cos \mathrm{θ}$

And the scalar product of $\stackrel{â‡\P Ä}{A}\mathrm{and}\stackrel{â‡\P Ä}{C}$is,

$\stackrel{â‡\P Ä}{A}.\stackrel{â‡\P Ä}{E}=-AE\cos \mathrm{θ}$

From the figure,$D\cos \mathrm{θ}\mathrm{and}E\cos \mathrm{θ}$ are acting along the same direction as the negative x axis so they are equal.

$AD\cos \mathrm{θ}=AE\cos \mathrm{θ}$

Therefore, we found

$\stackrel{â‡\P Ä}{A}.\stackrel{â‡\P Ä}{D}=\stackrel{â‡\P Ä}{A}.\stackrel{â‡\P Ä}{E}$

Thus,$\stackrel{â‡\P Ä}{A}.\stackrel{â‡\P Ä}{B}=\stackrel{â‡\P Ä}{A}.\stackrel{â‡\P Ä}{D}=\stackrel{â‡\P Ä}{A}.\stackrel{â‡\P Ä}{E}$ have the same dot product with $\stackrel{â‡\P Ä}{A}$.

The vectors $\stackrel{â‡\P Ä}{D}\mathrm{and}\stackrel{â‡\P Ä}{E}$makes an angle with negative x axis as shown in figure, hence their x components or components acting along the $\stackrel{â‡\P Ä}{A}$are negative. They are oppositely oriented.

Hence scalar product of$\stackrel{â‡\P Ä}{D}\mathrm{with}\stackrel{â‡\P Ä}{A}$ is negative.

role="math" localid="1660897716103" $\stackrel{â‡\P Ä}{A}.\stackrel{â‡\P Ä}{D}=-AD\cos \mathrm{θ}$

The scalar product of $\stackrel{â‡\P Ä}{E}\mathrm{with}\stackrel{â‡\P Ä}{A}$is negative.

$\stackrel{â‡\P Ä}{A}.\stackrel{â‡\P Ä}{E}=-AE\cos \mathrm{θ}$

Thus, $\stackrel{â‡\P Ä}{A}.\stackrel{â‡\P Ä}{D}\mathrm{and}\stackrel{â‡\P Ä}{\mathrm{A}}.\stackrel{â‡\P Ä}{\mathrm{E}}$have a negative dot product with $\stackrel{â‡\P Ä}{A}$.