If the position vectors of the points $A, B, C$ are $i, j, k$ respectively and $P$ is a point such that $\overrightarrow{AB} = \overrightarrow{CP},$ then the position vector of $P$ is

Let the position vectors of the vertices $A, B$ and $C$ of a triangle be $2 \hat{i}+2 \hat{j}+\hat{k}$,$\hat{i}+2 \hat{j}+2 \hat{k}$ and $2 \hat{i}+\hat{j}+2 \hat{k}$ respectively. Let $l_1, l_2$ and $l_3$ be the lengths of perpendiculars drawn from the orthocenter of the triangle on the sides $AB, BC$ and $CA$ respectively,then $l_1^2+l_2^2+l_3^2$ equals:

Let $ABC$ be a triangle. Let $u = \vec{AB}$ and $v = \vec{AC}$. If $D$ is the midpoint of $BC$,then $\vec{AD} =$

For any two vectors $\vec{a}$ and $\vec{b},$ we always have $|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|$ (triangle inequality).

The unit vector in the direction of the vector $(1, 0, 0)$ is $.......$

If $\vec{a}$ and $\vec{b}$ are non-zero vectors which are linearly dependent such that $\frac{|\vec{a} + \vec{b}|}{|\vec{a} - \vec{b}|} = 2$ and $|\vec{b}| > |\vec{a}|$,then: