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 = $

The scalar product of vectors $\bar{a}=\hat{i}+2 \hat{j}+\hat{k}$ and a unit vector along the sum of vectors $\bar{b}=2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\bar{c}=\lambda \hat{i}+2 \hat{j}-3 \hat{k}$ is $1$. Then the value of $\lambda$ is:

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}|=\sqrt{14}$,$|\vec{b}|=\sqrt{6}$ and $|\vec{a} \times \vec{b}|=\sqrt{48}$. Then $(\vec{a} \cdot \vec{b})^2$ is equal to $...........$.

Let $a, b, c$ be three vectors such that $a \neq 0$,$a \times b = 2a \times c$,$|a| = |c| = 1$,$|b| = 4$,and $|b \times c| = \sqrt{15}$. If $b - 2c = \lambda a$,then $\lambda$ equals:

Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $1$ and $2$ respectively,given that $\vec{a} \cdot \vec{b} = 1$.

If the coordinates of $A, B, C, D$ are $(2, 3, -1), (3, 5, -3), (1, 2, 3)$ and $(3, 5, 7)$ respectively,then what is the projection of $AB$ on $CD$?