For a triangle $ABC$ with vertices $A(1, 0, 0)$,$B(0, 1, 0)$,and $C(0, 0, 1)$,the angle $A = \dots$

If $\bar{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}$,$\bar{b}=-\hat{i}+2 \hat{j}+\hat{k}$ and $\bar{c}=3 \hat{i}+\hat{j}$ are vectors such that $\bar{a}+\lambda \bar{b}$ is perpendicular to $\bar{c}$,then the value of $\lambda$ is

Given that $\vec{a}, \vec{b}, \vec{p}$,and $\vec{q}$ are four vectors such that $\vec{a} + \vec{b} = \mu \vec{p}$,$\vec{b} \cdot \vec{q} = 0$,and $|\vec{b}|^2 = 1$,then $|(\vec{a} \cdot \vec{q}) \vec{p} - (\vec{p} \cdot \vec{q}) \vec{a}|$ is equal to:

If $\theta$ is the angle between the unit vectors $a$ and $b$,then $a - \sqrt{2}b$ will be a unit vector if $\theta = $

If $\bar{a}=\hat{i}-2 \hat{j}+3 \hat{k}$ and $\bar{b}=2 \hat{i}+3 \hat{j}-\hat{k}$,then the angle between the vectors $(2 \bar{a}+\bar{b})$ and $(\bar{a}+2 \bar{b})$ is

The vector $2i + j - k$ is perpendicular to $i - 4j + \lambda k,$ if $\lambda = $