If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}| = 2$ and $|\vec{b}| = 3$, then the maximum value of $3 |(3\vec{a} + 2\vec{b})| + 4 |(3\vec{a} - 2\vec{b})|$ is:

If $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ are unit vectors such that $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1$ and $\vec{a} \cdot \vec{c} = \frac{1}{2}$,then :-

The vector $2\hat{i} + a\hat{j} + \hat{k}$ is perpendicular to the vector $2\hat{i} - \hat{j} - \hat{k},$ if $a = $

The projection of vector $\vec{a} = 2\hat{i} + 3\hat{j} - 2\hat{k}$ on the vector $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$ is:

Let $\vec{a}$ be a vector in the plane containing vectors $\vec{b}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{c}=2 \hat{i}-\hat{j}+\hat{k}$. If $\vec{a}$ is perpendicular to $\hat{i}+\hat{j}+3 \hat{k}$ and its projection on $\vec{b}$ is $3 \sqrt{6}$, then $|\vec{a}|^2=$

Using vectors,prove that $\cos (A-B)=\cos A \cos B+\sin A \sin B$.