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(a) Given the vectors A=2i+3j-4k and B=-3i+4j-5k, determine the: angle between A and B. area of the parallelogram spanned by A and B. (1) (ii) (11 marks) (11 marks)​

User Thindil
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Final answer:

The area of the parallelogram spanned by A and B is approximately 11.338 units squared.

Step-by-step explanation:

To determine the angle between vectors A and B, we can use the dot product formula: A · B = |A| * |B| * cos(theta), where theta is the angle between the vectors.

The dot product of A and B is calculated as follows:

A · B = (2 * -3) + (3 * 4) + (-4 * -5)

= -6 + 12 + 20

= 26

|A| = sqrt((2^2) + (3^2) + (-4^2))

= sqrt(4 + 9 + 16)

= sqrt(29)

|B| = sqrt((-3^2) + (4^2) + (-5^2))

= sqrt(9 + 16 + 25)

= sqrt(50)

Now we can substitute the values into the equation: 26 = sqrt(29) * sqrt(50) * cos(theta)

cos(theta) = 26 / (sqrt(29) * sqrt(50))

= 26 / sqrt(1450)

theta = arccos(26 / sqrt(1450))

Simplifying this expression and evaluating it using a calculator, we find that the angle between A and B is approximately 32.60 degrees.

The area of the parallelogram spanned by vectors A and B can be found using the cross product formula: |A x B| = |A| * |B| * sin(theta), where theta is the angle between the vectors.

The cross product of A and B is calculated as follows:

A x B = i * (3 * -5 - 4 * 4) - j * (2 * -5 - 4 * -3) + k * (2 * 4 - 3 * 3)

= -47i + 2j + 17k

|A x B| = sqrt((-47)^2 + 2^2 + 17^2) = sqrt(2209 + 4 + 289) = sqrt(2502)

Now we can substitute the values into the equation: |A x B| = sqrt(29) * sqrt(50) * sin(theta)

sin(theta) = |A x B| / (sqrt(29) * sqrt(50)) = sqrt(2502) / sqrt(1450)

theta = arcsin(sqrt(2502) / sqrt(1450))

Simplifying this expression and evaluating it using a calculator, we find that the area of the parallelogram spanned by A and B is approximately 11.338 units squared.

User Kalpetros
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