5. Given the vectors vec r 2 = 37 + 2j -5 vec k and vec r 2 =-2 vec l - j + vec k . a) Calculate the magnitude of the vector vec r_{1} + vec r 2

Question

b) Find the angle between the vectors overline r_{1} and vec r_{2} ^ +

Answer 1

a) To calculate the magnitude of the vector r₁ + r₂, we need to first find the individual vectors r₁ and r₂.

Given: r₁ = 37 + 2j - 5k
r₂ = -2l - j + k

Now, to find the magnitude, we use the formula:
|A| = sqrt(Ax² + Ay² + Az²)

For vector r₁:
|r₁| = sqrt((37)² + (2)² + (-5)²)
= sqrt(1369 + 4 + 25)
= sqrt(1398)

For vector r₂:
|r₂| = sqrt((-2)² + (-1)² + (1)²)
= sqrt(4 + 1 + 1)
= sqrt(6)

Now, let's calculate the resultant magnitude:
|r₁ + r₂| = sqrt((r₁x + r₂x)² + (r₁y + r₂y)² + (r₁z + r₂z)²)
= sqrt((37 - 2)² + (2 - 1)² + (-5 + 1)²)
= sqrt(35² + 1 + (-4)²)
= sqrt(1225 + 1 + 16)
= sqrt(1242)

So, the magnitude of vector r₁ + r₂ is sqrt(1242).

b) To find the angle between vectors r₁ and r₂, we use the dot product formula:
r₁ · r₂ = |r₁| |r₂| cos(θ)

Given: |r₁| = sqrt(1398)
|r₂| = sqrt(6)

Now, let's calculate the dot product:
r₁ · r₂ = (37)(-2) + (2)(-1) + (-5)(1)
= -74 - 2 - 5
= -81

Using the formula, we have:
-81 = sqrt(1398) * sqrt(6) * cos(θ)

Rearranging the equation, we get:
cos(θ) = -81 / (sqrt(1398) * sqrt(6))

Now, we can find the angle using inverse cosine:
θ = cos^(-1)(-81 / (sqrt(1398) * sqrt(6)))

Calculating θ, we get the angle between r₁ and r₂.