Show that the magnitude of a vector is equal to the square root of the scalar product of the vector with itself.
Show that the magnitude of a vector is equal to the square root of the scalar product of the vector with itself.
If $\overrightarrow{\mathrm{A}} \| \overrightarrow{\mathrm{B}}$, then $\theta=0^{\circ}$
$\therefore \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=\mathrm{AB} \cos \theta=\mathrm{AB} \text { and } \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{A}}=|\overrightarrow{\mathrm{A}}||\overrightarrow{\mathrm{A}}|=\mathrm{A}^{2}$
$\therefore|\overrightarrow{\mathrm{A}}|=\sqrt{\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{A}}}$
hence the magnitude of a vector is equal to the square root of the scalar product of the vector with itself.
Magnitude of vector is equal to the square root of the scalar product of the vector with itself.
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-
Why the product of two vectors is not commutative ?
Why the product of two vectors is not commutative ?
If $\theta$ is the angle between two vectors $A$ and $B$, then match the following two columns.
colum $I$
colum $II$
$(A)$ $A \cdot B =| A \times B |$
$(p)$ $\theta=90^{\circ}$
$(B)$ $A \cdot B = B ^2$
$(q)$ $\theta=0^{\circ}$ or $180^{\circ}$
$(C)$ $|A+B|=|A-B|$
$(r)$ $A=B$
$(D)$ $|A \times B|=A B$
$(s)$ None
| colum $I$ | colum $II$ |
| $(A)$ $A \cdot B =| A \times B |$ | $(p)$ $\theta=90^{\circ}$ |
| $(B)$ $A \cdot B = B ^2$ | $(q)$ $\theta=0^{\circ}$ or $180^{\circ}$ |
| $(C)$ $|A+B|=|A-B|$ | $(r)$ $A=B$ |
| $(D)$ $|A \times B|=A B$ | $(s)$ None |
If $\overrightarrow{ P }=3 \hat{ i }+\sqrt{3} \hat{ j }+2 \hat{ k }$ and $\overrightarrow{ Q }=4 \hat{ i }+\sqrt{3} \hat{ j }+2.5 \hat{ k }$ then, The unit vector in the direction of $\overrightarrow{ P } \times \overrightarrow{ Q }$ is $\frac{1}{x}(\sqrt{3} \hat{i}+\hat{j}-2 \sqrt{3} \hat{k})$. The value of $x$ is
If $\overrightarrow{ P }=3 \hat{ i }+\sqrt{3} \hat{ j }+2 \hat{ k }$ and $\overrightarrow{ Q }=4 \hat{ i }+\sqrt{3} \hat{ j }+2.5 \hat{ k }$ then, The unit vector in the direction of $\overrightarrow{ P } \times \overrightarrow{ Q }$ is $\frac{1}{x}(\sqrt{3} \hat{i}+\hat{j}-2 \sqrt{3} \hat{k})$. The value of $x$ is
- [JEE MAIN 2023]