Question

Let a and b be vectors, satisfying |a|=|b|=5 and (a,b) =45°,find the area of the triangle having a-2b and 3a+2b as two of its sides

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given,

|a|=|b|=5

The angle between a and b(θ) = 45°

To Find,

The area of a triangle formed by these vectors =?

Solution,

We know from the are of a triangle that,

Area = (Base * Height) / 2

Putting the values of the vectors in the equation,

Area = |(a x b) x (3a+2b )| / 2

Area = |(-2a x b) - (a x b )| / 2

Area = |4a x b| / 2

Area = 2 * |a x b|

Putting the values of  a and b and using formula axb = |a| * |b| * Sin θ,

Area = 2 * 5 * 5 * Sin 45°

Area = 50 * (1 /√2 )

Area = 25√2 sq. units

Hence, the area of a triangle formed by these vectors is 25√2 sq. units.