Let $\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}$ be a vector such that $\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$ and $\vec{a} \cdot \vec{b}=3$. Then the projection of $\vec{b}$ on the vector $\vec{a}-\vec{b}$ is :-

If $\vec{a} = 4\hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} + \hat{j} + 2\hat{k}$ and $\vec{c} = 3\hat{i} + 4\hat{j} + 5\hat{k}$, then $(\vec{a} + \vec{b}) \cdot (\vec{b} + \vec{c}) =$

If $\vec{a}, \vec{b}$ and $\vec{c}$ are vectors with magnitudes $2, 3$ and $4$ respectively, then the best upper bound of $|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2$ among the given values is

The position vectors of points $A, B, C$ are given by $2\hat{i} - \hat{j} + \hat{k}$,$\hat{i} - 3\hat{j} - 5\hat{k}$,and $a\hat{i} - 3\hat{j} + \hat{k}$ respectively. If these points form a right-angled triangle with $\angle C = \pi/2$,find the value of $a$.

If $\vec{a}=\hat{i}+(\tan \theta) \hat{j}+\left(\frac{3}{\sqrt{\sin \frac{\theta}{2}}}\right) \hat{k}$ and $\vec{b}=\tan \theta(\hat{j}-\hat{i})-\left(2 \sqrt{\sin \frac{\theta}{2}}\right) \hat{k}$ are orthogonal vectors and $\vec{c}=(\sin 2 \theta) \hat{i}-2 \hat{j}+2 \hat{k}$ makes an obtuse angle with $X$-axis, then $\theta=$

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{i} - 3\hat{j} + 2\hat{k}$ and $\vec{c} = 3\hat{i} - 2\hat{k}$. If a vector $\vec{p}$ satisfies the conditions $\vec{p} \cdot \vec{c} = 0$ and $\vec{p} \times \vec{a} = \vec{b} \times \vec{a}$, then the value of $|\vec{p}|$ is: