Question

Consider points $A,B,C$ and $D$ with position vectors $(6,-4,7),(1,-6,10),(-1,-3,4)\text{and}(5,-1,5)$ respectively. Then quad $ABCD$ will have

• A. four sides equal
• B. three sides equal
• C. two sides equal
• D. all the sides unequal

Answer 1

The correct option is C (all the sides unequal)

We are given four points. The position vectors of four vectors in fact. Now looking at the options we can conclude that it can be solved by just finding the length of sides of the quadrilateral.
Since the position vectors of the vertices are given let’s write the coordinates of these points.

$A(6,-4,7)B(1,-6,10)C(-1,-3,4)\text{and}D(5,-1,5)$

The vector joining the two points is $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by $=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$
So, the vector joining the two points A (6,-4,7) ~B(1,-6,10) is $\overrightarrow{AB}=(1-6)\hat{i}+(-6+4)\hat{j}+(10-7)\hat{k}=-5\hat{i}-2\hat{j}+3\hat{k}$
Similar way, $\overrightarrow{BC}=-2\hat{i}+3\hat{j}-6\hat{k}$
$\overrightarrow{CD}=6\hat{i}+2\hat{j}+\hat{k}$
$\overrightarrow{DA}=\hat{i}-3\hat{j}+2\hat{k}$
Now finding the length of the sides, we get: $AB=\sqrt{(-5{)}^2+(-2{)}^2+3^2}=\sqrt{25+4+9}=\sqrt{38}$,
$BC=\sqrt{4+9+36}=\sqrt{49}=7$,
$CD=\sqrt{36+4+1}=\sqrt{41}$ and
$DA=\sqrt{1+9+4}=\sqrt{14}$
On observing, we see the length of all sides are unequal.