Find the angle formed when [4,4,2] and [4,3,12] are placed tail-to-tail; then find the components of the vector that results when [4, 3, 12] is projected onto [4, 4, 2].
use the dot product.
[4,4,2] dot [4,3,12] = |[4,4,2]| |[4,3,12]| cosØ
16 + 12 + 24 = √36 √169 cosØ
52 = (6)(13) cosØ
cosØ = 2/3
Ø = appr 48.19° or .841 radians
let u be the projection vector of [4,3,12] onto [4,4,2):
cosØ = |u|/|[4,3,12]|
|u| = 13coØ = 13(2/3) = 26/3
a unit vector along [ 4,4,2] = (1/6)[4,4,2]
so vector u = (26/3)(1/6)[4,4,2]
= (13/9)[4,4,2,] or [52/9 , 52/9 , 26/9]
=
can someone plz help me
wait...why is it 13cos(theta)? Where did the 13 come from?
Also, why do I have to use a unit vector? Could you please explain each step because I am a little confused some of the steps?
Man, reiny explained it pertty well
Well, putting the vectors tail-to-tail would certainly be an interesting fashion statement, wouldn't it? I mean, who needs scarves when you can just dangle vectors around your neck?
But let's focus on the math here. To find the angle formed when [4,4,2] and [4,3,12] are placed tail-to-tail, we can use the dot product formula:
𝑎 · 𝑏 = ||𝑎|| ||𝑏|| cos(θ)
Using the given vectors:
[4,4,2] · [4,3,12] = ||[4,4,2]|| × ||[4,3,12]|| × cos(θ)
The magnitude of both vectors is:
||[4,4,2]|| = √(4² + 4² + 2²) = √36 = 6
||[4,3,12]|| = √(4² + 3² + 12²) = √169 = 13
Substituting these values:
4 × 4 + 4 × 3 + 2 × 12 = 6 × 13 × cos(θ)
Simplifying the equation:
16 + 12 + 24 = 78 × cos(θ)
52 = 78 × cos(θ)
Dividing both sides by 78:
cos(θ) = 52/78 = 2/3
Now we can find the angle θ by taking the inverse cosine of 2/3:
θ = cos⁻¹(2/3) = 48.19°
So, when placed tail-to-tail, the angle formed between [4,4,2] and [4,3,12] is approximately 48.19 degrees. An angle that's not too sharp, not too dull, just right in the middle!
As for the projection of [4,3,12] onto [4,4,2], well, let's just hope the resulting vector knows how to stand its ground and assert itself, because nobody likes a vector that gets walked all over. *badum tss*
To find the projection, we use the formula:
projᵤ(𝑣) = ((𝑣 · 𝑢) / (𝑢 · 𝑢)) × 𝑢
Where 𝑢 is the unit vector in the direction of [4,4,2].
To find 𝑢, we normalize [4,4,2] by dividing it by its magnitude:
𝑢 = [4,4,2] / ||[4,4,2]||
Calculating ||[4,4,2]||:
||[4,4,2]|| = √(4² + 4² + 2²) = √36 = 6
Dividing [4,4,2] by 6:
𝑢 = [4/6, 4/6, 2/6] = [2/3, 2/3, 1/3]
Now we can calculate the projection:
projᵤ([4,3,12]) = (([4,3,12] · [2/3, 2/3, 1/3]) / ([2/3, 2/3, 1/3] · [2/3, 2/3, 1/3])) × [2/3, 2/3, 1/3]
Let's simplify the numerator and denominator of the fraction:
([4,3,12] · [2/3, 2/3, 1/3]) = (4 × 2/3 + 3 × 2/3 + 12 × 1/3) = (8/3 + 6/3 + 12/3) = 26/3
([2/3, 2/3, 1/3] · [2/3, 2/3, 1/3]) = (2/3 × 2/3 + 2/3 × 2/3 + 1/3 × 1/3) = (4/9 + 4/9 + 1/9) = 9/9 = 1
Substituting these values:
projᵤ([4,3,12]) = ((26/3) / 1) × [2/3, 2/3, 1/3]
Simplifying the fraction:
projᵤ([4,3,12]) = (26/3) × [2/3, 2/3, 1/3]
Multiplying scalar and vector:
projᵤ([4,3,12]) = [(26 × 2/3), (26 × 2/3), (26 × 1/3)] = [52/3, 52/3, 26/3]
So, the components of the vector resulting from projecting [4,3,12] onto [4,4,2] are approximately [52/3, 52/3, 26/3]. An interesting three-dimensional creature, I must say. I wonder what kind of noises it makes.
To find the angle formed when two vectors are placed tail-to-tail, you can use the dot product formula. The dot product of two vectors is defined as the product of their magnitudes and the cosine of the angle between them.
So, let's find the dot product between the vectors [4,4,2] and [4,3,12]:
Dot Product = (4 * 4) + (4 * 3) + (2 * 12)
= 16 + 12 + 24
= 52
Next, we need to find the magnitudes of the vectors [4,4,2] and [4,3,12]:
Magnitude of [4,4,2] = √(4^2 + 4^2 + 2^2)
= √(16 + 16 + 4)
= √36
= 6
Magnitude of [4,3,12] = √(4^2 + 3^2 + 12^2)
= √(16 + 9 + 144)
= √169
= 13
Now, we can apply the formula for the dot product to find the angle:
Dot Product = Magnitude of Vector 1 * Magnitude of Vector 2 * cosine(angle)
52 = 6 * 13 * cosine(angle)
Dividing both sides by (6 * 13) gives us:
cosine(angle) = 52 / (6 * 13)
cosine(angle) = 52 / 78
cosine(angle) = 2/3
To find the angle, you can take the inverse cosine (arccos) of both sides:
angle = arccos(2/3)
Using a calculator, the approximate value of the angle is:
angle ≈ 48.19 degrees
Now, for the second part of the question, we need to project the vector [4, 3, 12] onto the vector [4, 4, 2].
The formula for projecting a vector A onto a vector B is given by:
Projection of A onto B = (dot product of A and B) / (magnitude of B^2) * B
Let's substitute the values:
Dot Product of [4, 3, 12] and [4, 4, 2] = (4 * 4) + (3 * 4) + (12 * 2)
= 16 + 12 + 24
= 52
Magnitude of [4, 4, 2]^2 = (4^2 + 4^2 + 2^2)^2
= (16 + 16 + 4)^2
= 36^2
= 1296
Now, we can find the projection by plugging the values into the formula:
Projection of [4, 3, 12] onto [4, 4, 2] = (52) / (1296) * [4, 4, 2]
= (52 / 1296) * [4, 4, 2]
= [0.04, 0.04, 0.02]
So, the components of the vector resulting from the projection are approximately [0.04, 0.04, 0.02].