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
b) Find the angle between the vectors overline r_{1} and vec r_{2} ^ +
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.
To calculate the magnitude of vector vec r_{1} + vec r_{2}, you first need to determine the values of vec r_{1} and vec r_{2}.
You mentioned two different vectors: vec r 2 and vec r 2. I assume that one of them is vec r_{1}. Please clarify this.
Next, once you have the values for the vectors vec r_{1} and vec r_{2}, you can add them together by simply adding their corresponding components.
a) To find the magnitude of the vector vec r_{1} + vec r_{2}, you can use the Pythagorean theorem, which states that the magnitude of a vector with components (a,b,c) is given by sqrt(a^2 + b^2 + c^2).
So, first, add the corresponding components of vec r_{1} and vec r_{2} to find vec r_{1} + vec r_{2}.
Then, calculate the magnitude of the resultant vector by using the formula mentioned above.
b) To find the angle between vectors vec r_{1} and vec r_{2}, you can use the dot product between the two vectors.
The dot product of two vectors vec A = ai + bj + ck and vec B = di + ej + fk is given by A dot B = ad + be + cf.
In this case, we need to find the dot product between vec r_{1} and vec r_{2}. Let's call the dot product r_dot.
Once you have calculated the dot product r_dot, you can use the formula: cos(theta) = r_dot / (||r_1|| * ||r_2||), where theta is the angle between the vectors.
To find the angle, you can take the inverse cosine (cos^(-1)) of the value obtained from the above calculation.
Note: It's important to ensure that the vectors are expressed in the same coordinate system and that the units are consistent before performing any calculations.
a) To calculate the magnitude of the vector vec r_{1} + vec r 2, we first need to find the individual vectors and then add them together.
Given vec r_{1} = 37 + 2j - 5 vec k
and vec r_{2} = -2 vec l - j + vec k
Adding the corresponding components gives us:
vec r_{1} + vec r_{2} = (37 - 2) + (2 - 1) + (-5 + 1) = 35 + 1 - 4 = 32 + 1j - 4k
Now, we can calculate the magnitude of vec r_{1} + vec r_{2} using the formula sqrt(a^2 + b^2 + c^2), where a, b, and c are the components of the vector.
Magnitude = sqrt((32)^2 + (1)^2 + (-4)^2)
= sqrt(1024 + 1 + 16)
= sqrt(1041)
≈ 32.25
Therefore, the magnitude of the vector vec r_{1} + vec r 2 is approximately 32.25.
b) To find the angle between the vectors overline r_{1} and vec r_{2} ^, we can use the dot product formula and the magnitude formula. The dot product formula is given by:
vec r_{1} · vec r_{2} = |vec r_{1}| × |vec r_{2}| × cos(theta)
where |vec r_{1}| and |vec r_{2}| are the magnitudes of the vectors and theta is the angle between them. Rearranging the formula gives us:
cos(theta) = (vec r_{1} · vec r_{2}) / (|vec r_{1}| × |vec r_{2}|)
Now, let's calculate the dot product and magnitudes.
Dot product: vec r_{1} · vec r_{2} = (37)(-2) + (2)(-1) + (-5)(1) = -74 - 2 - 5 = -81
Magnitudes: |vec r_{1}| = sqrt((37)^2 + (2)^2 + (-5)^2) = sqrt(1369 + 4 + 25) = sqrt(1398) ≈ 37.39
|vec r_{2}| = sqrt((-2)^2 + (-1)^2 + (1)^2) = sqrt(4 + 1 + 1) = sqrt(6) ≈ 2.45
Now, substituting these values into the formula:
cos(theta) = (-81) / (37.39 × 2.45)
≈ -0.877
Using the inverse cosine function, we can find the angle:
theta = arccos(-0.877)
≈ 150.36°
Therefore, the angle between the vectors overline r_{1} and vec r_{2} ^ is approximately 150.36 degrees.