If the position vectors of the vertices $A, B, C, D$ of a quadrilateral are $7 \hat{i}-4 \hat{j}+7 \hat{k}, \hat{i}-6 \hat{j}+10 \hat{k}, -\hat{i}-3 \hat{j}+4 \hat{k}$,and $5 \hat{i}-\hat{j}+5 \hat{k}$ respectively,then $ABCD$ is

If $a = 2i + 2j - k$ and $|xa| = 1$,then $x =$

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:

Show that each of the given three vectors is a unit vector:
$\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k}), \frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k}), \frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})$
Also,show that they are mutually perpendicular to each other.

$ABCD$ is a parallelogram such that $L$ is the mid-point of $BC$. Then,$\vec{AL}$ is equal to:

Represent graphically a displacement of $40 \, km$,$30^{\circ}$ east of north.