Solveeit Logo

Question

Question: Area of a rectangle having vertices A, B, C and D with position vectors \[ - \widehat i + \dfrac{1...

Area of a rectangle having vertices A, B, C and D with position vectors
i^+12j^+4k,^i^+12j^+4k,^i^12j^+4k,^- \widehat i + \dfrac{1}{2}\widehat j + 4\widehat {k,}\widehat i + \dfrac{1}{2}\widehat j + 4\widehat {k,}\widehat i - \dfrac{1}{2}\widehat j + 4\widehat {k,} and i^12j^+4k^ - \widehat i - \dfrac{1}{2}\widehat j + 4\widehat k respectively is
(A) 12\dfrac{1}{2}
(B) 11
(C) 22
(D) 44

Explanation

Solution

This is a vector algebra based problem. We have been given with the position vectors of all four vertices of a rectangle ABCD. Here, we may involve coordinate geometry methods to find the distance between points and hence to find the area of the rectangle.

Complete step-by-step answer:
First we will find the coordinates of all four vertices with the help of their position vectors.
For vertex A,
Position vector is i^+12j^+4k^ - \widehat i + \dfrac{1}{2}\widehat j + 4\widehat k. So its coordinate will be (1,12,4)\left( { - 1,\dfrac{1}{2},4} \right).
For vertex B,
Position vector is i^+12j^+4k^\widehat i + \dfrac{1}{2}\widehat j + 4\widehat k. So its coordinate will be (1,12,4)\left( {1,\dfrac{1}{2},4} \right).
For vertex C,
Position vector is i^12j^+4k^\widehat i - \dfrac{1}{2}\widehat j + 4\widehat k.So its coordinate will be(1,12,4)\left( {1, - \dfrac{1}{2},4} \right).
For vertex D,
Position vector is i^12j^+4k^ - \widehat i - \dfrac{1}{2}\widehat j + 4\widehat k.So its coordinate will be(1,12,4)\left( { - 1, - \dfrac{1}{2},4} \right).
Distance formula between points (x1,y1,z1)({x_1},{y_1},{z_1}) and (x2,y2,z2)({x_2},{y_2},{z_2}) is (x1x2)2+(y1y2)2+(z1z2)2\sqrt {{{({x_1} - {x_2})}^2} + {{({y_1} - {y_2})}^2} + {{({z_1} - {z_2})}^2}}
Now, we can find the length of sides of the rectangle ABCD , let us suppose AB and BC.
AB = Distance between point A and point B
=(1(1))2+(1212)2+(44)2\sqrt {{{(1 - ( - 1))}^2} + {{(\dfrac{1}{2} - \dfrac{1}{2})}^2} + {{(4 - 4)}^2}}
=22\sqrt {{2^2}}
=22
Similarly,
BC= Distance between point B and point C
=(11)2+(12(12))2+(44)2\sqrt {{{(1 - 1)}^2} + {{(\dfrac{1}{2} - ( - \dfrac{1}{2}))}^2} + {{(4 - 4)}^2}}
=12\sqrt {{1^2}}
=11
Area of rectangle = length of AB ×\times length of BC
= 2×12 \times 1
=22
Thus option D is correct.

Note: Vector is an object which has magnitude and direction. This problem is a good example of a geometry related question where coordinates of the points are playing an important role for the computation of other relevant terms of some given shape. Here we have used a distance formula for finding the length and breadth and hence area of the rectangle.