3. Draw vectors u and v tail-to-tail so that they make a è-degree angle. Draw the vector u − v, the third side of the triangle, and check to see that it points in the right direction.
(a) Solve for cos è using the SSS version of the Law of Cosines, expressing all lengths in terms of u, v, and u − v.
(b) If you use vector algebra to simplify the numerator as much as possible, you will
discover an interesting new result connecting u•v to cos è.
(a) Well, well, we have some vectors playing together in a triangle! Let's solve for cos è using the SSS version of the Law of Cosines.
Now, let's call the length of vector u as 'a', the length of vector v as 'b', and the length of u - v as 'c'. According to the Law of Cosines, we have:
c^2 = a^2 + b^2 - 2ab * cos è
Plugging in the given lengths, we get:
(u - v)^2 = u^2 + v^2 - 2uv * cos è
Expanding and simplifying a bit:
u^2 - 2uv + v^2 = u^2 + v^2 - 2uv * cos è
Canceling out some terms, we have:
-2uv = -2uv * cos è
Dividing both sides by -2uv:
cos è = 1
Voila! The cosine of è turns out to be a delightful value of 1!
(b) Now, my funny friend, let's dive into some vector algebra to discover an interesting connection between u•v and cos è.
Using the dot product formula, we know that:
u•v = |u| |v| cos θ
Where θ is the angle between the vectors u and v. In this case, θ happens to be è.
Now, we can rewrite θ as 0 degrees since cos 0 is equal to 1. Therefore, cos è is 1.
Surprisingly, we end up with u•v = cos è. Quite a delightful and intriguing result, isn't it?
(a) To solve for cos è using the SSS version of the Law of Cosines, we need to express all lengths in terms of u, v, and u - v.
Let's assume the length of the vector u is represented by |u|, the length of the vector v is represented by |v|, and the length of the vector u - v is represented by |u - v|.
According to the Law of Cosines, we have:
|u - v|^2 = |u|^2 + |v|^2 - 2|u||v|cos è
Expanding the square on the left side, we get:
(u - v)•(u - v) = u•u + v•v - 2u•v*cos è
Expanding the dot products on both sides, we have:
|u|^2 - 2u•v + |v|^2 = |u|^2 + |v|^2 - 2u•v*cos è
Canceling out the common terms |u|^2 and |v|^2, we get:
-2u•v = -2u•v*cos è
Dividing both sides by -2u•v:
cos è = -2u•v / -2u•v
Simplifying, we have:
cos è = 1
Therefore, cos è = 1.
(b) Using vector algebra to simplify the numerator, we have:
u•v = |u||v|cosθ
From part (a), we know that cos è = 1. Thus, substituting cosθ = 1, we get:
u•v = |u||v|
This equation states that the dot product of vectors u and v is equal to the multiplication of their magnitudes.
This is referred to as the dot product property, which states that the dot product of two vectors is equal to the product of their magnitudes and the cosine of the angle between them. In this case, since cos è = 1, the angle between u and v is 0 degrees, indicating that the vectors are in the same direction.
To solve this problem, let's go step by step.
Step 1: Drawing the vectors u and v tail-to-tail
Start by drawing a vector u from a point and then draw another vector v starting from the endpoint of vector u. These vectors should be drawn tail-to-tail, meaning the tail of vector v should be at the endpoint of vector u.
Step 2: Drawing the vector u - v
To draw the vector u - v, start at the tail of vector v and draw a vector in the opposite direction of vector v that terminates at the tail of vector u. This will represent vector u - v.
Step 3: Checking the direction
Make sure that vector u - v points in the right direction. To check this, draw a line segment connecting the endpoint of vector u to the endpoint of vector v. If vector u - v points in the same direction as this line segment, then it is in the right direction.
Now that we have drawn the vectors, let's move on to solving for cos è.
(a) Solving for cos è using the SSS version of the Law of Cosines
The SSS version of the Law of Cosines states that in a triangle with sides lengths a, b, and c and angle è opposite to side c, the following equation holds:
c^2 = a^2 + b^2 - 2ab*cos(è)
In our case, the sides of the triangle are the lengths of the vectors u, v, and u - v. Let's denote them as a, b, and c respectively. Therefore, we have:
c^2 = u^2 + v^2 - 2u*v*cos(è)
Now, let's express all lengths in terms of u, v, and u - v. Recall that the magnitude of a vector u, denoted as |u|, is the length of the vector. Therefore, we can rewrite the equation as:
(u - v)^2 = u^2 + v^2 - 2u*v*cos(è)
Expanding the equation, we get:
u^2 - 2u*v + v^2 = u^2 + v^2 - 2u*v*cos(è)
Simplifying, we find:
- 2u*v = - 2u*v*cos(è)
Now divide both sides of the equation by -2u*v to isolate cos(è):
cos(è) = - 2u*v / (- 2u*v)
Simplifying further, we find:
cos(è) = 1
So the value of cos è is 1.
(b) Now let's use vector algebra to simplify the numerator and discover the interesting result connecting u•v to cos è.
Recall that the dot product between two vectors u and v is given by:
u • v = |u| |v| cos(θ)
In our case, u • v = |u| |v| cos(è). We know that cos è = 1 from part (a), so let's substitute that into the equation:
u • v = |u| |v| (1)
Simplifying further, we find:
u • v = |u| |v|
Therefore, we have the interesting result that the dot product of vectors u and v is equal to the product of their magnitudes.
I hope this explanation helps!