Question

The area of parallelogram represented by the vectors A=2i^ + 3j^ and B= i^ + 4j^ is :- (A) 14 units (B) 7.5 unit (C) 10 unit (D)5 unit. Please explain as well if you can

Answer 1

Given A = 2i+3j and B = i+4j .

Area of paralleogram A x B = (2i+3j+0k) x(i+4j+0k) 

For  i component we have,

(3 * 0) -  (4 * 0) = 0

For j component we have,

(0 * 2) - (1 * 0) = 0

For k component we have,

(2 * 4) - (1 * 3) = 5.

Now take the magnitude of this vector to find the area of the parallelogram:

A * B = Whole root of 0^2 + 0^2 + 5^2

         = 0 + 0 + 5(Square and root will be cancelled)

         = 5 units.


Hope this helps!

Answer 2

Answer:

Area of parallelogram formed by two vectors A and B is 5units.

Explanation:

Given $\vec A =2\hat{i}+3\hat{j}$ and $\vec A =\hat{i}+4\hat{j}$

Area of parallelogram is $\big| \vec A \times\vec B\big|$

    $\vec A \times\vec B=\left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\2&3&0\\1&4&0\end{array}\right|$

               $=\big[ (3\times0)-(4\times0)\big]\hat{i}- \big[ (2\times0)-(1\times0)\big]\hat{j}+ \big[ (2\times4)-(1\times3)\big]\hat{k}$

               $= \big[8-3\big]\hat{k}=5\hat{k}$

Magnitude of $\vec A \times\vec B$ is   $\big| \vec A \times\vec B\big|=\sqrt{0^{2} +0^{2}+ 5^{2} }$

                                                    $=\sqrt{ 5^{2} }=5units$

Therefore, area of parallelogram formed by the given vectors is 5units.