Let $\bar{a}$ and $\bar{b}$ be two vectors such that $|\bar{a}|=|\bar{b}|$ and $|\bar{a}+2 \bar{b}|=|2 \bar{a}-\bar{b}|$. If $\bar{c}$ is a vector parallel to $\bar{a}$,then the angle between $\bar{b}$ and $\bar{c}$ is (in $^{\circ}$)

If $a, b, c$ are three non-zero,non-coplanar vectors and $b_1 = b - \frac{b \cdot a}{|a|^2} a$,$b_2 = b + \frac{b \cdot a}{|a|^2} a$,$c_2 = c - \frac{c \cdot a}{|a|^2} a - \frac{c \cdot b_1}{|b_1|^2} b_1$,$c_3 = c - \frac{c \cdot a}{|a|^2} a - \frac{c \cdot b_2}{|b_2|^2} b_2$,and $c_4 = a - \frac{c \cdot a}{|a|^2} a$. Then which of the following is a set of mutually orthogonal vectors?

The set of all $\alpha$,for which the vectors $\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k}$ and $\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k}$ are inclined at an obtuse angle for all $t \in R$ is:

The vector$(s)$ which is/are coplanar with vectors $\hat{i}+\hat{j}+2\hat{k}$ and $\hat{i}+2\hat{j}+\hat{k}$,and perpendicular to the vector $\hat{i}+\hat{j}+\hat{k}$ is/are:
$(A) \hat{j}-\hat{k}$
$(B) -\hat{i}+\hat{j}$
$(C) \hat{i}-\hat{j}$
$(D) -\hat{j}+\hat{k}$

If $4\hat{i}+\hat{j}-\hat{k}$ and $3\hat{i}+m\hat{j}+2\hat{k}$ are at a right angle,then $m = $

In quadrilateral $ABCD$,$\overrightarrow{AB}=\vec{a}$,$\overrightarrow{BC}=\vec{b}$,$\overrightarrow{DA}=\vec{a}-\vec{b}$. $M$ is the midpoint of $BC$ and $X$ is a point on $DM$ such that $\overrightarrow{DX}=\frac{4}{5} \overrightarrow{DM}$. Then the points $A, X$ and $C$: