The projection of the vector $\hat{i} + \hat{j} + \hat{k}$ on the line $\vec{r} = 3\hat{i} - \hat{j} + \lambda(\hat{i} + 2\hat{j} + 3\hat{k})$ is:

If $(a \times b)^{2} + (a \cdot b)^{2} = 144$ and $|a| = 4$,then $|b|$ is equal to

Show that the points $A, B$ and $C$ with position vectors $\vec{a}=3 \hat{i}-4 \hat{j}-4 \hat{k}$,$\vec{b}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-3 \hat{j}-5 \hat{k}$ respectively form the vertices of a right-angled triangle.

If $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$,$\vec{b} = \hat{i} + \hat{j} - 2 \hat{k}$,and $\vec{c} = \hat{i} + 3 \hat{j} - \hat{k}$ are given vectors. If $\vec{a}$ is perpendicular to $\lambda \vec{b} + \vec{c}$,then $\lambda = . . . . . .$.

The vectors $2\,i + 3\,j - 4\,k$ and $a\,i + b\,j + c\,k$ are perpendicular,when

If $\theta$ is the angle between the vectors $\vec{a}$ and $\vec{b}$ and $|\vec{a} \times \vec{b}| = \vec{a} \cdot \vec{b}$,then $\theta = $