The angle between two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ with magnitudes $\sqrt{3}$ and 4, respectively, and $\vec{\text{a}}\cdot\vec{\text{b}}=2\sqrt{3}$ is:
- $\frac{\pi}{6}$
- $\frac{\pi}{3}$
- $\frac{\pi}{2}$
- $\frac{5\pi}{2}$
Solution:
Here, $|\vec{\text{a}}|=\sqrt{3},|\vec{\text{b}}|=4$ and $\vec{\text{a}}\cdot\vec{\text{b}}=2\sqrt{3}$ [given]
We know that, $\vec{\text{a}}\cdot\vec{\text{b}}=|\vec{\text{a}}||\vec{\text{b}}|\cos\theta$
$\Rightarrow2\sqrt{3}=\sqrt{3}.4.\cos\theta$
$\Rightarrow\cos\theta=\frac{2\sqrt{3}}{4\sqrt{3}}=\frac{1}{2}$
$\therefore\theta=\frac{\pi}{3}$
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$ x-2 y+3 z=-1 $ ; $ -x+y-2 z=k $ ; $ x-3 y+4 z=1$
$STATEMENT -1$ : The system of equations has no solution for $\mathrm{k} \neq 3$. and
$STATEMENT - 2$ : The determinant $\left|\begin{array}{ccc}1 & 3 & -1 \\ -1 & -2 & \mathrm{k} \\ 1 & 4 & 1\end{array}\right| \neq 0$, for $\mathrm{k} \neq 3$.