Let a, b, and c be unit vectors, such that a+b+c=0. Show that the angle between any two of these vectors is 120 degrees.
I proved this geometrically, but now my teacher wants a proof using vectors, and I don't know what to do.
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To prove that the angle between any two of these vectors, let's say a and b, is 120 degrees, we can use vector algebra.
Given:
a + b + c = 0
We are required to prove:
The angle between vectors a and b is 120 degrees.
To begin, let's focus on vectors a and b. To find the angle between them using vector algebra, we can use the dot product:
a · b = |a| |b| cos(theta)
Where a · b represents the dot product of vectors a and b, |a| and |b| represent their magnitudes, and theta represents the angle between them.
We can simplify the given equation using c = -a - b:
a + b + c = 0
a + b + (-a - b) = 0
a + (-a) + b + (-b) = 0
0 = 0
From the equation above, we can see that both sides are equal to 0. This implies that the dot product of any two of these vectors must also be 0:
a · b + a · c + b · c = 0
Since a · c = |a| |c| cos(theta_ac) and b · c = |b| |c| cos(theta_bc), and |a|, |b|, and |c| are all unit vectors (with magnitude 1), the equation becomes:
a · b + cos(theta_ac) + cos(theta_bc) = 0
Now, let's consider the dot product a · b. We'll use the fact that the dot product is defined as:
a · b = |a| |b| cos(theta_ab)
Since both a and b are unit vectors, their magnitudes are both equal to 1:
a · b = 1 * 1 * cos(theta_ab)
= cos(theta_ab)
Now, substituting back into the equation from before, we have:
cos(theta_ab) + cos(theta_ac) + cos(theta_bc) = 0
To prove that the angle between any two vectors is 120 degrees, we must show that each term in this equation is equal to -1/2.
Let's consider the angle between vectors a and b. Since the sum of the angles in a triangle is 180 degrees, we know that the angle between vectors a and c is 180 - 120 = 60 degrees. Similarly, the angle between vectors b and c is also 60 degrees.
Using the trigonometric identity cos(180 - x) = -cos(x), we can calculate:
cos(theta_ac) = cos(60) = 1/2
cos(theta_bc) = cos(60) = 1/2
Substituting these values back into the equation:
cos(theta_ab) + cos(theta_ac) + cos(theta_bc) = cos(theta_ab) + 1/2 + 1/2 = cos(theta_ab) + 1 = 0
Solving for cos(theta_ab):
cos(theta_ab) = -1
Since the range of the cosine function is [-1, 1], and the angle between vectors a and b is 120 degrees, we have successfully proven that the angle between any two vectors a, b, and c is 120 degrees.