Question

If $\overline a $ and $\overline b $ are vectors, satisfying $\left| {\overline a } \right| = \left| {\overline b } \right| = 5$ and $\left( {\overline a ,\overline b } \right) = 45^\circ $ , then the area of the triangle constructed with vectors $\overline a - 2\overline b $ and $3\overline a + 2\overline b $ is
A. $50\sqrt 2 $ sq. units
B. $5\sqrt 5 $ sq. units
C. $4\sqrt 5 $ sq. units
D. $3\sqrt 5 $ sq. units

Answer 1

Hint: The area of triangle formed by any two vectors is given by $\dfrac{1}{2} \times $ | cross product of two vectors |.
Now, do the cross product of the two given vectors and take the magnitude of the cross product.
After that, divide the above result by 2 to get the answer.

Complete step-by-step answer:
We are given that, $\overline a $ and $\overline b $ are vectors, satisfying $\left| {\overline a } \right| = \left| {\overline b } \right| = 5$ and $\left( {\overline a ,\overline b } \right) = 45^\circ $ .
Now, it is said that, the vectors $\overline a - 2\overline b $ and $3\overline a + 2\overline b $ form a triangle.
The area of triangle formed by vectors is $\dfrac{1}{2} \times $ | cross product of two vectors |
 $\therefore \Delta = \dfrac{1}{2}\left| {\left( {\overline a - 2\overline b } \right) \times \left( {3\overline a + 2\overline b } \right)} \right|$
  $ = \dfrac{1}{2} \times \left| {\overline a \times 3\overline a + \overline a \times 2\overline b - 2\overline b \times 3\overline a - 2\overline b \times 2\overline b } \right|$
Applying the property $\overline a \times \overline a = 0$ ,
 \[\therefore \Delta = \dfrac{1}{2}\left| {\overline a \times 2\overline b - 2\overline b \times 3\overline a } \right|\]
Also, $\overline a \times \overline b = - \overline b \times \overline a $
 \[\therefore \Delta = \dfrac{1}{2}\left| {\overline a \times 2\overline b + 3\overline a \times 2\overline b } \right|\]
Applying $\overline a \times \overline b = ab\sin \theta $
 $\therefore \Delta = \dfrac{1}{2}\left| {2ab\sin 45^\circ + 6ab\sin 45^\circ } \right|$
        $
   = \dfrac{1}{2}\left| {8ab\sin 45^\circ } \right| \\
   = \dfrac{1}{2}\left| {8ab \times \dfrac{1}{{\sqrt 2 }}} \right| \\
 $
Now, $\left| {\overline a } \right| = \left| {\overline b } \right| = 5$
 $\therefore \Delta = \dfrac{1}{2} \times 8 \times 5 \times 5 \times \dfrac{1}{{\sqrt 2 }}$
         $
   = \dfrac{{100}}{{\sqrt 2 }} \\
   = 50\sqrt 2 \\
 $
Thus, the area of the triangle constructed with vectors $\overline a - 2\overline b $ and $3\overline a + 2\overline b $ is $50\sqrt 2 $ sq. units.
So, Option (A) is correct.

Note: We have written some properties of cross product, which will be useful to us while solving such types of questions.
1. $\overline a \times \overline a = 0$
2.$\overline a \times \overline b = - \overline b \times \overline a $
3. $\left( {p\overline a } \right) \times \overline b = \overline a \times \left( {p\overline b } \right) = p\left( {\overline a \times \overline b } \right)$
4. $\left| {\overline a \times \overline b } \right| = \left| {\overline a } \right|\left| {\overline b } \right|\sin \theta $
5. $\overline a \cdot \left( {\overline b \times \overline c } \right) = \left( {\overline a \times \overline b } \right) \cdot \overline c $
6. $\overline a \times \left( {\overline b \times \overline c } \right) = \left( {\overline a \cdot \overline c } \right)\overline b - \left( {\overline a \cdot \overline b } \right)\overline c $ .