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
The angle between vectors $(\vec{M} \times \vec{N})$ and $(\bar{N} \times \vec{M})$ is ................
The angle between vectors $(\vec{M} \times \vec{N})$ and $(\bar{N} \times \vec{M})$ is ................
If the projection of $2 \hat{i}+4 \hat{j}-2 \hat{k}$ on $\hat{i}+2 \hat{j}+\alpha \hat{k}$ is zero. Then, the value of $\alpha$ will be.
If the projection of $2 \hat{i}+4 \hat{j}-2 \hat{k}$ on $\hat{i}+2 \hat{j}+\alpha \hat{k}$ is zero. Then, the value of $\alpha$ will be.
- [JEE MAIN 2022]
Two vectors $\overrightarrow A $ and $\overrightarrow B $ are at right angles to each other, when
Two vectors $\overrightarrow A $ and $\overrightarrow B $ are at right angles to each other, when
- [AIIMS 1987]
Let $\vec{A}=2 \hat{i}-3 \hat{j}+4 \hat{k}$ and $\vec{B}=4 \hat{i}+j+2 \hat{k}$ then $|\vec{A} \times \vec{B}|$ is equal to ...................
Let $\vec{A}=2 \hat{i}-3 \hat{j}+4 \hat{k}$ and $\vec{B}=4 \hat{i}+j+2 \hat{k}$ then $|\vec{A} \times \vec{B}|$ is equal to ...................
Why the product of two vectors is not commutative ?
Why the product of two vectors is not commutative ?