Answer
Verified
480.9k+ views
Hint: In this question first find out the position vectors of the adjacent sides, later on apply the property of cross product, so, using these concepts we can reach the solution of the question.
Position vectors of the rectangle having vertices A, B, C, D are given as
$
\vec O\vec A = - \hat i + \dfrac{1}{2}\hat j + 4\hat k \\
\vec O\vec B = \hat i + \dfrac{1}{2}\hat j + 4\hat k \\
\vec O\vec C = \hat i - \dfrac{1}{2}\hat j + 4\hat k \\
\vec O\vec D = - \hat i - \dfrac{1}{2}\hat j + 4\hat k \\
$
The adjacent sides of rectangle ABCD is given as (AB, BC)
So, the position vectors of these sides is given as
$\left( {\vec A\vec B} \right) = \left( {\vec O\vec B} \right) - \left( {\vec O\vec A} \right)$
Now substitute the above values in above equation we have,
$
\Rightarrow \left( {\vec A\vec B} \right) = \left( {\hat i + \dfrac{1}{2}\hat j + 4\hat k} \right) - \left( { - \hat i + \dfrac{1}{2}\hat j + 4\hat k} \right) \\
\Rightarrow \left( {\vec A\vec B} \right) = \left( {1 + 1} \right)\hat i + \left( {\dfrac{1}{2} - \dfrac{1}{2}} \right)\hat j + \left( {4 - 4} \right)\hat k = 2\hat i \\
$
And, $\left( {\vec B\vec C} \right) = \left( {\vec O\vec C} \right) - \left( {\vec O\vec B} \right)$
Now substitute the above values in above equation we have,
$
\Rightarrow \left( {\vec B\vec C} \right) = \left( {\hat i - \dfrac{1}{2}\hat j + 4\hat k} \right) - \left( {\hat i + \dfrac{1}{2}\hat j + 4\hat k} \right) \\
\Rightarrow \left( {\vec B\vec C} \right) = \left( {1 - 1} \right)\hat i + \left( { - \dfrac{1}{2} - \dfrac{1}{2}} \right)\hat j + \left( {4 - 4} \right)\hat k = - \hat j \\
$
Now we all know that the area of the rectangle whose adjacent sides are $\vec a$ and $\vec b$ is the modulus of cross product of these vectors which is $\left| {\left( {\vec a \times \vec b} \right)} \right|$.
So the area of the rectangle having adjacent sides $\left( {\vec A\vec B} \right)$ and $\left( {\vec B\vec C} \right)$ is $\left| {\left( {\left( {\vec A\vec B} \right) \times \left( {\vec B\vec C} \right)} \right)} \right|$
Now first calculate cross product of these vectors we have
$\left( {\vec A\vec B} \right) \times \left( {\vec B\vec C} \right) = \left| {\begin{array}{*{20}{c}}
{\hat i}&{\vec j}&{\vec k} \\
2&0&0 \\
0&{ - 1}&0
\end{array}} \right| = \left( {0 - 0} \right)\hat i - \left( {0 - 0} \right)\vec j + \left( { - 2 - 0} \right)\vec k = - 2\vec k$
Now the modulus of above cross product is
$\left| {\left( {\left( {\vec A\vec B} \right) \times \left( {\vec B\vec C} \right)} \right)} \right| = \left| { - 2\hat k} \right| = \sqrt {{{\left( { - 2} \right)}^2}} = 2$
So, the area of the rectangle is 2 sq. units.
Hence, option (c) is correct.
Note: In such types of questions the key concept we have to remember is that always recall that the area of rectangle whose adjacent sides are $\vec a$ and $\vec b$ is the modulus of cross product of these vectors which is$\left| {\left( {\vec a \times \vec b} \right)} \right|$, so first calculate the adjacent sides of the rectangle as above then apply this concept, we will get the required area of the rectangle.
Position vectors of the rectangle having vertices A, B, C, D are given as
$
\vec O\vec A = - \hat i + \dfrac{1}{2}\hat j + 4\hat k \\
\vec O\vec B = \hat i + \dfrac{1}{2}\hat j + 4\hat k \\
\vec O\vec C = \hat i - \dfrac{1}{2}\hat j + 4\hat k \\
\vec O\vec D = - \hat i - \dfrac{1}{2}\hat j + 4\hat k \\
$
The adjacent sides of rectangle ABCD is given as (AB, BC)
So, the position vectors of these sides is given as
$\left( {\vec A\vec B} \right) = \left( {\vec O\vec B} \right) - \left( {\vec O\vec A} \right)$
Now substitute the above values in above equation we have,
$
\Rightarrow \left( {\vec A\vec B} \right) = \left( {\hat i + \dfrac{1}{2}\hat j + 4\hat k} \right) - \left( { - \hat i + \dfrac{1}{2}\hat j + 4\hat k} \right) \\
\Rightarrow \left( {\vec A\vec B} \right) = \left( {1 + 1} \right)\hat i + \left( {\dfrac{1}{2} - \dfrac{1}{2}} \right)\hat j + \left( {4 - 4} \right)\hat k = 2\hat i \\
$
And, $\left( {\vec B\vec C} \right) = \left( {\vec O\vec C} \right) - \left( {\vec O\vec B} \right)$
Now substitute the above values in above equation we have,
$
\Rightarrow \left( {\vec B\vec C} \right) = \left( {\hat i - \dfrac{1}{2}\hat j + 4\hat k} \right) - \left( {\hat i + \dfrac{1}{2}\hat j + 4\hat k} \right) \\
\Rightarrow \left( {\vec B\vec C} \right) = \left( {1 - 1} \right)\hat i + \left( { - \dfrac{1}{2} - \dfrac{1}{2}} \right)\hat j + \left( {4 - 4} \right)\hat k = - \hat j \\
$
Now we all know that the area of the rectangle whose adjacent sides are $\vec a$ and $\vec b$ is the modulus of cross product of these vectors which is $\left| {\left( {\vec a \times \vec b} \right)} \right|$.
So the area of the rectangle having adjacent sides $\left( {\vec A\vec B} \right)$ and $\left( {\vec B\vec C} \right)$ is $\left| {\left( {\left( {\vec A\vec B} \right) \times \left( {\vec B\vec C} \right)} \right)} \right|$
Now first calculate cross product of these vectors we have
$\left( {\vec A\vec B} \right) \times \left( {\vec B\vec C} \right) = \left| {\begin{array}{*{20}{c}}
{\hat i}&{\vec j}&{\vec k} \\
2&0&0 \\
0&{ - 1}&0
\end{array}} \right| = \left( {0 - 0} \right)\hat i - \left( {0 - 0} \right)\vec j + \left( { - 2 - 0} \right)\vec k = - 2\vec k$
Now the modulus of above cross product is
$\left| {\left( {\left( {\vec A\vec B} \right) \times \left( {\vec B\vec C} \right)} \right)} \right| = \left| { - 2\hat k} \right| = \sqrt {{{\left( { - 2} \right)}^2}} = 2$
So, the area of the rectangle is 2 sq. units.
Hence, option (c) is correct.
Note: In such types of questions the key concept we have to remember is that always recall that the area of rectangle whose adjacent sides are $\vec a$ and $\vec b$ is the modulus of cross product of these vectors which is$\left| {\left( {\vec a \times \vec b} \right)} \right|$, so first calculate the adjacent sides of the rectangle as above then apply this concept, we will get the required area of the rectangle.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Change the following sentences into negative and interrogative class 10 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE