The angle between the vectors $a = 2i + 3j + k$ and $b = 2i - j - k$ is

Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that $\vec{a} \cdot \vec{b} = 0$. For some $x, y \in \mathbb{R}$,let $\vec{c} = x\vec{a} + y\vec{b} + (\vec{a} \times \vec{b})$. If $|\vec{c}| = 2$ and the vector $\vec{c}$ is inclined at the same angle $\alpha$ to both $\vec{a}$ and $\vec{b}$,then the value of $8 \cos^2 \alpha$ is . . . . .

Let the point $A$ divide the line segment joining the points $P(-1, -1, 2)$ and $Q(5, 5, 10)$ internally in the ratio $r : 1$ $(r > 0)$. If $O$ is the origin and $(\overrightarrow{OQ} \cdot \overrightarrow{OA}) - \frac{1}{5}|\overrightarrow{OP} \times \overrightarrow{OA}|^2 = 10$,then the value of $r$ is:

Let $a = i + 2j + k$, $b = i - j + k$, and $c = i + j - k$. $A$ vector in the plane of $a$ and $b$ has a projection of $\frac{1}{\sqrt{3}}$ on $c$. Then, one such vector is:

Consider two vectors $\overrightarrow{u} = 3\hat{i} - \hat{j}$ and $\overrightarrow{v} = 2\hat{i} + \hat{j} - \lambda\hat{k}$,where $\lambda > 0$. The angle between them is given by $\cos^{-1}\left(\frac{\sqrt{5}}{2\sqrt{7}}\right)$. Let $\vec{v} = \vec{v}_1 + \vec{v}_2$,where $\vec{v}_1$ is parallel to $\overrightarrow{u}$ and $\vec{v}_2$ is perpendicular to $\overrightarrow{u}$. Then the value $|\vec{v}_1|^2 + |\vec{v}_2|^2$ is equal to

Suppose $\overrightarrow{a}=\lambda \hat{i}-7 \hat{j}+3 \hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}+\hat{j}+2 \lambda \hat{k}$. If the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is greater than $90^{\circ}$, then $\lambda$ satisfies the inequality: