Show that the scalar product of two vectors obeys the law of commutative.
Show that the scalar product of two vectors obeys the law of commutative.
If $\theta$ is the angle between $\vec{A}$ and $\vec{B}$, then scalar product
$\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}} =\mathrm{AB} \cos \theta$
$=\mathrm{BA} \cos \theta$
$\therefore \quad \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}$ $=\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{A}}$
$[\because \mathrm{AB}=\mathrm{BA}]$
Similar Questions
If $\overrightarrow A \times \overrightarrow B = \overrightarrow C + \overrightarrow D,$ then select the correct alternative-
If $\overrightarrow A \times \overrightarrow B = \overrightarrow C + \overrightarrow D,$ then select the correct alternative-
The angle between two vectors $4\hat i + 3\hat j + \hat k$ and $-3\hat i + 2\hat j + 6\hat k$ is ....... $^o$
The angle between two vectors $4\hat i + 3\hat j + \hat k$ and $-3\hat i + 2\hat j + 6\hat k$ is ....... $^o$
Given $A =3 \hat{ i }+4 \hat{ j }$ and $B =6 \hat{ i }+8 \hat{ j }$, which of the following statement is correct?
The angle between $(\overrightarrow A - \overrightarrow B )$ and $(\overrightarrow A \times \overrightarrow B )$ is $(\overrightarrow{ A } \neq \overrightarrow{ B })$
The angle between $(\overrightarrow A - \overrightarrow B )$ and $(\overrightarrow A \times \overrightarrow B )$ is $(\overrightarrow{ A } \neq \overrightarrow{ B })$
- [NEET 2017]
Find the angle between two vectors with the help of scalar product.
Find the angle between two vectors with the help of scalar product.