Question

The angle between two adjacent sides\[\left| {\overrightarrow a } \right|\] and \[\left| {\overrightarrow b } \right|\] of parallelogram is \[\dfrac{\pi }{2}\]. If\[\left| {\overrightarrow a } \right| = \left( {2, - 2,1} \right)\] and\[\left| {\overrightarrow b } \right| = 2\left| {\overrightarrow a } \right|\], then the area of this parallelogram is ………….
A.$9$
B.$18$
C.\[\dfrac{9}{2}\]
D.\[\dfrac{3}{4}\]

Answer 1

Hint: The area of parallelogram is given by expression \[\left| {\overrightarrow a } \right| \times \left| {\overrightarrow b } \right|\],where \[\left| {\overrightarrow a } \right|\] and \[\left| {\overrightarrow b } \right|\] are two vectors and \[\theta \] is the angle between two given vectors? Here it is given that \[\theta \] is equal to \[\dfrac{\pi }{6}\] . We can expand\[\left| {\overrightarrow a } \right| \times \left| {\overrightarrow b } \right|\] in the form of \[\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin (\dfrac{\pi }{6})\] and substitute the value with\[\left| {\overrightarrow a } \right| = 3\], and using the relation\[\left| {\overrightarrow b } \right| = 2\left| {\overrightarrow a } \right|\], we will get the value of \[\left| {\overrightarrow b } \right|\] as $6$.

Complete step-by-step answer:
Given: \[\left| {\overrightarrow a } \right| = \left( {2, - 2,1} \right)\] and\[\left| {\overrightarrow b } \right| = 2\left| {\overrightarrow a } \right|\]and angle between two adjacent sides\[\left| {\overrightarrow a } \right|\] and \[\left| {\overrightarrow b } \right|\] of parallelogram is \[\dfrac{\pi }{6}\]
Suppose\[\left| {\overrightarrow b } \right| = a\widehat i + b\widehat j + c\widehat k\]
 The angle between two vectors is \[\dfrac{\pi }{6}\]
\[
   \Rightarrow \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin (\dfrac{\pi }{6}) = \left| {\overrightarrow a } \right| \times \left| {\overrightarrow b } \right| \\
   \Rightarrow \left| {\overrightarrow a } \right| = \sqrt {{2^2} + {{( - 2)}^2} + {1^2}} = 3 \\
   \Rightarrow 3 \times 6 \times \dfrac{1}{2} = \left| {\overrightarrow a } \right| \times \left| {\overrightarrow b } \right| \\
 \]
Area of parallelogram is given by \[\left| {\overrightarrow a } \right| \times \left| {\overrightarrow b } \right|\]
Hence the area is \[9unit{s^2}\]

So, the correct answer is “Option A”.

Note: Students sometimes use\[\dfrac{1}{2}\left| {\overrightarrow a } \right| \times \left| {\overrightarrow b } \right|\] which is wrong for parallelogram, as it is the formula for area of triangle. Here you will see \[\left| {\overrightarrow b } \right|\]that is unknown and we will calculate the value of \[\left| {\overrightarrow b } \right|\] from the other given relations that are mentioned in our question. Keep in mind that everything is not given in the question. You must calculate it using an indirect method. Also be careful while placing the values given in the formula. For competitive exams and board exams, try to memorize the formula of area of figures in the vector form. Vectors are an important part of mathematics and you should spend some time playing around with vectors. Also, know the difference between scalar product and vector product and understand their application in solving such problems.