The angle between the two vectors $\vec A = 3\hat i + 4\hat j + 5\hat k$ and $\vec B = 3\hat i + 4\hat j - 5\hat k$ will be....... $^o$

If $\vec{P} = b \hat{i} + 6 \hat{j} + \hat{k}$ and $\vec{Q} = \hat{i} - a \hat{j} + 4 \hat{k}$ are perpendicular to each other,and $3b - a = 5$,find the values of $a$ and $b$.

$\vec A$ and $\vec B$ are two vectors and $\theta$ is the angle between them. If $|\vec A \times \vec B| = \sqrt{3}(\vec A \cdot \vec B)$,the value of $\theta$ is ......... $^\circ$.

If $\vec{A}$ and $\vec{B}$ are two vectors satisfying the relation $\vec{A} \cdot \vec{B} = |\vec{A} \times \vec{B}|$,then the value of $|\vec{A} - \vec{B}|$ will be:

For three vectors $\vec{A} = (-x \hat{i} - 6 \hat{j} - 2 \hat{k})$,$\vec{B} = (-\hat{i} + 4 \hat{j} + 3 \hat{k})$ and $\vec{C} = (-8 \hat{i} - \hat{j} + 3 \hat{k})$,if $\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$,then the value of $x$ is:

Three particles $P, Q$ and $R$ are moving along the vectors $\vec{A}=\hat{i}+\hat{j}, \vec{B}=\hat{j}+\hat{k}$ and $\vec{C}=-\hat{i}+\hat{j}$ respectively. They strike a point and start to move in different directions. Now,particle $P$ is moving normal to the plane containing vectors $\vec{A}$ and $\vec{B}$. Similarly,particle $Q$ is moving normal to the plane containing vectors $\vec{A}$ and $\vec{C}$. The angle between the directions of motion of $P$ and $Q$ is $\cos^{-1}\left(\frac{1}{\sqrt{x}}\right)$. Then the value of $x$ is ...... .