If a=<1,3,4> b=<-2,2,1> the area of the parallelogram with 2 adjacent sides formed by vectors a & b will be?

Question

these are the possible solutions to choose from in the back of the book......


a. 12.4

b. 13.0

c. 15

d. 15.3

or none of these

Answer 1

b. 13.0

a.b = |a||b|cosx where x is the angle between vectors.
8 = sqrt(26).(3)cosx
x = 58.5 degrees

Area of half a parallelogram = area of triangle = (1/2)abSinC.

Area of parallelogram with unit vector sides
= 2 [(1/2)(sqrt26)(3)Sin58.5]
= 13.0

Answer 2

If a= <1,3,4> b= <-2,2,1> the area of the parallelogram with 2 adjacent sides formed by vectors a & b will be?

The area of a parallelogram for which we know the length of two sides, a and b, and the included angle θ is:

Area = absinθ.

The magnitude of the cross product of vectors a and b is:

| a X b | = || a || || b || sinθ

So we want the magnitude of the cross product. We don't know the angle θ directly but it is implicitly known since vectors have both magnitude and direction.

| a X b | = || <1, 3, 4> X <-2, 2, 1> || = || <-5, -9, 8> ||

| a X b | = √[(-5)² + (-9)² + 8²] = √(25 + 81 + 64) = √170

Area = √170 ≈ 13.0

The answer is b.

Answer 3

do you recognize what pass products are? the portion of the parallelogram is the scale of the pass made of the two vectors... ordinary to do by technique of determinants wish THIS WORKS A B C VECTOR a million D E F VECTOR 2 i j ok or on your case a million 3 4 -2 2 a million i j ok in case you do no longer submit to in concepts determinants they're (using the variables) (BF - CE )i - (AF-CD)j + (AE-BD) ok now take the sum of the squares of the coefficients of i, j, ok and then take the sq. root... right here we pass.. -5i -9j +8k or (the choice course) 5i+9j-8k So the section is sq.(5^2 + 9^2 + 8^2) = sq.(a hundred and seventy)

Answer 4

I am 100% sure it is d. 15.3 is correct