Answer
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Hint: 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 $ - \widehat i + \dfrac{1}{2}\widehat j + 4\widehat k$. So its coordinate will be $\left( { - 1,\dfrac{1}{2},4} \right)$.
For vertex B,
Position vector is $\widehat i + \dfrac{1}{2}\widehat j + 4\widehat k$. So its coordinate will be $\left( {1,\dfrac{1}{2},4} \right)$.
For vertex C,
Position vector is $\widehat i - \dfrac{1}{2}\widehat j + 4\widehat k$.So its coordinate will be$\left( {1, - \dfrac{1}{2},4} \right)$.
For vertex D,
Position vector is $ - \widehat i - \dfrac{1}{2}\widehat j + 4\widehat k$.So its coordinate will be$\left( { - 1, - \dfrac{1}{2},4} \right)$.
Distance formula between points $({x_1},{y_1},{z_1})$ and \[({x_2},{y_2},{z_2})\] is $\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
=$\sqrt {{{(1 - ( - 1))}^2} + {{(\dfrac{1}{2} - \dfrac{1}{2})}^2} + {{(4 - 4)}^2}} $
=$\sqrt {{2^2}} $
=$2$
Similarly,
BC= Distance between point B and point C
=$\sqrt {{{(1 - 1)}^2} + {{(\dfrac{1}{2} - ( - \dfrac{1}{2}))}^2} + {{(4 - 4)}^2}} $
=$\sqrt {{1^2}} $
=$1$
Area of rectangle = length of AB $ \times $ length of BC
= $2 \times 1$
=$2$
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.
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 $ - \widehat i + \dfrac{1}{2}\widehat j + 4\widehat k$. So its coordinate will be $\left( { - 1,\dfrac{1}{2},4} \right)$.
For vertex B,
Position vector is $\widehat i + \dfrac{1}{2}\widehat j + 4\widehat k$. So its coordinate will be $\left( {1,\dfrac{1}{2},4} \right)$.
For vertex C,
Position vector is $\widehat i - \dfrac{1}{2}\widehat j + 4\widehat k$.So its coordinate will be$\left( {1, - \dfrac{1}{2},4} \right)$.
For vertex D,
Position vector is $ - \widehat i - \dfrac{1}{2}\widehat j + 4\widehat k$.So its coordinate will be$\left( { - 1, - \dfrac{1}{2},4} \right)$.
Distance formula between points $({x_1},{y_1},{z_1})$ and \[({x_2},{y_2},{z_2})\] is $\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
=$\sqrt {{{(1 - ( - 1))}^2} + {{(\dfrac{1}{2} - \dfrac{1}{2})}^2} + {{(4 - 4)}^2}} $
=$\sqrt {{2^2}} $
=$2$
Similarly,
BC= Distance between point B and point C
=$\sqrt {{{(1 - 1)}^2} + {{(\dfrac{1}{2} - ( - \dfrac{1}{2}))}^2} + {{(4 - 4)}^2}} $
=$\sqrt {{1^2}} $
=$1$
Area of rectangle = length of AB $ \times $ length of BC
= $2 \times 1$
=$2$
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.
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