Question

Consider points A, B, C and D with position vectors$<1,2,0>,<1,2,2>,<1,5,2>and<1,5,0>$ respectively. Then ABCD is a

• A. square
• B. trapezium
• C. rectangle

Answer 1

The correct option is C rectangle

The four points given enclose a plane. Hence it forms a quadrilateral.

Also, we can find the nature of the quadrilateral by finding the side lengths.

$AB=\sqrt{(1-1{)}^2+(2-2{)}^2+(0-2{)}^2}=\sqrt{4}=2$

$BC=\sqrt{(1-1{)}^2+(5-2{)}^2+(2-2{)}^2}=\sqrt{9}=3$

$CD=\sqrt{(1-1{)}^2+(5-5{)}^2+(2-0{)}^2}=\sqrt{4}=2$

$DA=\sqrt{(1-1{)}^2+(5-2{)}^2+(0-0{)}^2}=\sqrt{9}=3$

Here,AB = CD and AD = BC. Opposite sides are equal.

Hence, the quadrilateral could be a parallelogram or a rectangle.

Now the angle between AB and BC is ${90}^{\circ }$. This is because AB is parallel to the XY plane and BC is parallel to the XZ plane. These planes, and hence, AB and BC, are perpendicular to each other.

Hence, ABCD is a rectangle.