Question

What is the area of the parallelogram formed by the vectors. (Hint A=one over two (absin teta) A=_3i +3j and B=_2i _2j

Answer 1

Dot Product:

|A|=√13, |B|=√17 {Magnitude of vectors}

I will represent vectors with a prefix of #.

General formula of “dot product”:

(#A).(#B)=|A|.|B|.cos ∆ {where, ∆ is the angle between two vectors}

Putting values, we get

cos ∆= 14/√221.

As, the vectors, have different magnitudes, they will be adjacent in the parallelogram.

Area of parallelogram is, (A).(B).(sin ∆).

Converting​ value cos∆, to sin ∆, we get,

sin ∆= 5/√221.

Putting values in the formula of area of parallelogram, we get

Area= A.B.sin∆=√221*(5/√221) unit square = 5 unit square.

Cross Product:

It's, second approach, and a short one!

As, vectors are in it's component form, we can use “cross product” directly, to multiply it's sides to obtain area of parallelogram.

AxB=(2i+3j+0k)x(i+4j+0k).

AxB= 5k.

Magnitude of AxB= √(5^2)=5 unit square.

This, the area of the parallelogram is 5 unit square.