If $a, b$ and $c$ are position vectors of the vertices of $\triangle ABC$,then $\frac{(a-c) \times (b-a)}{(b-a) \cdot (c-a)} = $

If $a \times b = b \times c \ne 0,$ where $a, b$ and $c$ are coplanar vectors,then for some scalar $k$

If $a$ and $b$ are two non-zero perpendicular vectors,then a vector $y$ satisfying the equations $a \cdot y = c$ (where $c$ is a scalar) and $a \times y = b$ is

Let the lines $L_1: \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}$ and $L_2: \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}$. The unit vector perpendicular to both $L_1$ and $L_2$ is:

Let $a, b$ and $c$ be three coplanar unit vectors. $A$ unit vector $d$ is perpendicular to them. If $(a \times b) \times (c \times d) = \frac{3}{26} i - \frac{2}{13} j + \frac{6}{13} k$ and the angle between $a$ and $b$ is $30^\circ$, then $c$ is equal to...

If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$, $\overrightarrow{b}=\hat{i}+\hat{j}$, $\overrightarrow{c}=\hat{i}$ and $(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}$, then $\lambda+\mu$ is equal to: