Show that if and only if A.B=0?
2. Let A1, A2, …, Ar be non-zero vectors such that Ai.Aj = o if i is different from j.
Let c1, c2, …., cr be numbers such that c1A1+ c2A2+…..+ crAr = 0. Show that ci = 0 for all i = 1,2, 3, …, r.
To prove that A.B = 0 if and only if A and B are orthogonal, we will show both directions of the statement.
Direction 1: If A.B = 0, then A and B are orthogonal.
Assume A.B = 0.
This means that the dot product of A and B is zero.
By the definition of dot product, we have A.B = ||A|| ||B|| cos(theta), where ||A|| and ||B|| are the magnitudes (or lengths) of A and B, and theta is the angle between A and B.
Since the dot product A.B is zero, we have ||A|| ||B|| cos(theta) = 0.
Since ||A|| and ||B|| are non-zero (given that A and B are non-zero vectors), we can conclude that cos(theta) must be zero.
Cos(theta) is equal to zero when theta is equal to 90 degrees or pi/2 radians.
Therefore, A and B are orthogonal.
Direction 2: If A and B are orthogonal, then A.B = 0.
Assume A and B are orthogonal.
This means that the angle between A and B is 90 degrees or pi/2 radians.
Using the definition of dot product, we have A.B = ||A|| ||B|| cos(theta), where ||A|| and ||B|| are the magnitudes (or lengths) of A and B, and theta is the angle between A and B.
Since cos(theta) is zero for a 90-degree angle, we have cos(theta) = 0.
Therefore, A.B = ||A|| ||B|| cos(theta) = ||A|| ||B|| (0) = 0.
Therefore, we have shown that A.B = 0 if and only if A and B are orthogonal.
Now, let's prove that ci = 0 for all i = 1,2, 3, ..., r when c1A1 + c2A2 + ... + crAr = 0.
Assume c1A1 + c2A2 + ... + crAr = 0.
Since Ai.Aj = 0 for i ≠ j, we can rewrite the above equation as:
c1A1.A1 + c2A2.A2 + ... + crAr.Ar = 0
c1 ||A1||^2 + c2 ||A2||^2 + ... + cr ||Ar||^2 = 0 (using the definition of dot product)
Since Ai ≠ 0 (given that Ai is non-zero), we can conclude that ||Ai||^2 ≠ 0 for all i = 1,2, 3, ..., r.
Therefore, in order for the above equation to hold true (c1 ||A1||^2 + c2 ||A2||^2 + ... + cr ||Ar||^2 = 0), all ci's must be zero.
Hence, ci = 0 for all i = 1,2, 3, ..., r when c1A1 + c2A2 + ... + crAr = 0.
Therefore, we have proven that ci = 0 for all i = 1,2, 3, ..., r when c1A1 + c2A2 + ... + crAr = 0.
To prove the statement, we will use the properties of dot product.
1. Assume A.B = 0:
- We need to prove that A and B are orthogonal vectors.
- By the definition of the dot product, A.B = |A||B|cosθ, where θ is the angle between A and B.
- Since A.B = 0, either |A| or |B| is zero or cosθ is zero.
- If |A| or |B| is zero, then A or B is the zero vector, which means they are orthogonal.
- If cosθ is zero, it means that the angle between A and B is 90 degrees, which again makes them orthogonal.
- Therefore, in both cases, A and B are orthogonal vectors.
2. Assume ci = 0 for all i = 1, 2, 3, ..., r:
- We need to prove that c1A1 + c2A2 + ... + crAr = 0.
- By the properties of the dot product, (c1A1 + c2A2 + ... + crAr).Ai = c1(A1.Ai) + c2(A2.Ai) + ... + cr(Ar.Ai).
- Since Ai.Aj = 0 for i ≠ j, all the terms in the above expression become zero except c1(A1.A1), c2(A2.A2), ..., cr(Ar.Ar).
- But since ci = 0 for all i, all these terms become zero, and we have (c1A1 + c2A2 + ... + crAr).Ai = 0.
- Since this holds for all i, it means that c1A1 + c2A2 + ... + crAr is orthogonal to each of Ai vectors.
- If a vector is orthogonal to all other vectors, it must be the zero vector.
- Therefore, c1A1 + c2A2 + ... + crAr = 0.
Hence, we have proven both directions. If A.B = 0, then A and B are orthogonal vectors, and if ci = 0 for all i, then c1A1 + c2A2 + ... + crAr = 0.