Let $m$ be the unit vector orthogonal to the vector $\hat{i}-\hat{j}+\hat{k}$ and coplanar with the vectors $2 \hat{i}+\hat{j}$ and $\hat{j}-\hat{k}$. If $a=\hat{i}-\hat{k}$,then the length of the perpendicular from the origin to the plane $r \cdot m=a \cdot m$ is

Let $\vec{c}$ and $\vec{d}$ be vectors such that $|\vec{c}+\vec{d}|=\sqrt{29}$ and $\vec{c}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{d}$. If $\lambda_1, \lambda_2$ $(\lambda_1 > \lambda_2)$ are the possible values of $(\vec{c}+\vec{d}) \cdot (-7\hat{i}+2\hat{j}+3\hat{k})$, then the equation $K^{2}x^{2}+(K^{2}-5K+\lambda_{1})xy+(3K+\frac{\lambda_{2}}{2})y^{2}-8x+12y+\lambda_{2}=0$ represents a circle, for $K$ equal to:

Let $P, Q, R$ and $S$ be the points on the plane with position vectors $-2 \hat{i}-\hat{j}, 4 \hat{i}, 3 \hat{i}+3 \hat{j}$ and $-3 \hat{i}+2 \hat{j}$ respectively. The quadrilateral $PQRS$ must be a

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 $\vec{OA} = 2\hat{i} + 2\hat{j} + \hat{k}$,$\vec{OB} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ and the length of the internal bisector of $\angle BOA$ of triangle $AOB$ is $k$,then $9k^2 =$

If $a \neq 0, b \neq 0$ and $|a + b| = |a - b|,$ then the vectors $a$ and $b$ are