if vectors a+2b and 5a-4b are perpendicular to each other and a and b are unit vectors. Find the angle between a and b.

Let θ be the angle between a and b. We know that cosθ = (a∙b) / (|a||b|). Since a and b are unit vectors, |a|=|b|=1. Therefore we need to find a∙b.

a+2b and 5a-4b are perpendicular, so

(a+2b)∙(5a-4b)=0

Multiply this out to obtain

5a∙a + 6a∙b - 8b∙b = 0

which results in

6a∙b = 8b∙b - 5a∙a

Since a and b are unit vectors,

6a∙b = 8-5 = 3

Therefore, cos θ = 3/6. θ=π/3

Good!

If two vectors are perpendicular their dot product will equal zero.

(a+2b)•(5a-4b)=0
5a•a+10a•b-4a•b-8b•b=0
6a•b=8b•b-5a•a
6a•b=8-5=3
cosθ=3/6.
θ=π/3

To find the angle between vectors a and b, we can use the dot product formula and the fact that two vectors are perpendicular to each other if and only if their dot product is zero.

Given vectors a+2b and 5a-4b are perpendicular, their dot product is zero:

(a+2b) · (5a-4b) = 0

Expanding the dot product gives us:

5(a · a) + 10(a · b) - 4(b · a) - 8(b · b) = 0

Since both a and b are unit vectors, a · a = ||a||^2 = 1, and b · b = ||b||^2 = 1. Thus, the equation simplifies to:

5 + 10(a · b) - 4(b · a) - 8 = 0

Rearranging the terms gives:

10(a · b) - 4(b · a) = 3

Using the commutative property of the dot product, we can rewrite the equation as:

(10a · b) - (4a · b) = 3

Simplifying, we get:

6(a · b) = 3

Dividing both sides by 6:

(a · b) = 1/2

The dot product of two vectors can also be expressed as the product of their magnitudes and the cosine of the angle between them:

(a · b) = ||a|| ||b|| cosθ

Substituting the given information that a and b are unit vectors, we have:

1/2 = 1 * 1 * cosθ

cosθ = 1/2

To find the angle θ, we can take the inverse cosine (arccos) of both sides:

θ = arccos(1/2)

Calculating the value of arccos(1/2) gives us:

θ ≈ 60°

Therefore, the angle between vectors a and b is approximately 60 degrees.

Good