Prove that two non zero vectors A and B are perpendicular if and only if |A|< |A + tB| for every number t
To prove that two non-zero vectors A and B are perpendicular if and only if |A| < |A + tB| for every number t, we need to show both directions of the statement:
1) If two non-zero vectors A and B are perpendicular, then |A| < |A + tB| for every number t.
2) If |A| < |A + tB| for every number t, then two non-zero vectors A and B are perpendicular.
Proof:
1) If two non-zero vectors A and B are perpendicular, we can represent this as A · B = 0, where · denotes dot product.
Now, let's consider |A + tB|^2, which is the square of the magnitude of A + tB:
|A + tB|^2 = (A + tB) · (A + tB)
= (A · A) + 2t(A · B) + (tB · tB)
= |A|^2 + 2t(A · B) + t^2|B|^2 (by dot product properties)
= |A|^2 + t(2(A · B)) + t^2|B|^2
Since A · B = 0 (A and B are perpendicular), we can simplify the expression further:
|A + tB|^2 = |A|^2 + t(2(0)) + t^2|B|^2
= |A|^2 + t^2|B|^2
Now, taking the square root of both sides of the equation, we get:
|A + tB| = √(|A|^2 + t^2|B|^2)
Since |A|^2 and |B|^2 are non-negative, for any given t, the value of t^2|B|^2 will always be greater than or equal to zero. Therefore, we can conclude that:
|A + tB| ≥ |A| for every number t.
2) Now, let's prove the second direction of the statement, which is:
If |A| < |A + tB| for every number t, then two non-zero vectors A and B are perpendicular.
We can prove this by contradiction. Assume that A and B are not perpendicular, i.e., A · B ≠ 0. Since A and B are non-zero vectors, their magnitudes are also non-zero.
Now, let t = - (A · B) / (|B|^2).
Substituting this value of t into |A + tB|, we get:
|A + tB| = |A - (A · B) / (|B|^2) B|
= √(|A|^2 - 2(A · B)^2 / |B|^2 + (A · B)^2 / |B|^2)
Simplifying this expression further, we have:
|A + tB| = √(|A|^2 - (A · B)^2 / |B|^2)
Now, using the assumption that |A| < |A + tB| for every number t, we can write:
|A| < |A + tB|
|A|^2 < |A + tB|^2
|A|^2 < |A|^2 - (A · B)^2 / |B|^2
Since |A|^2 is positive, we can multiply through by |B|^2:
|A|^2|B|^2 < |A|^2|B|^2 - (A · B)^2
This simplifies to:
0 < -(A · B)^2
This implies that (A · B)^2 > 0, which contradicts our assumption that A · B ≠ 0.
Therefore, we can conclude that if |A| < |A + tB| for every number t, then two non-zero vectors A and B must be perpendicular.
Hence, we have proved both directions of the statement, and therefore, two non-zero vectors A and B are perpendicular if and only if |A| < |A + tB| for every number t.
To prove that two non-zero vectors A and B are perpendicular if and only if |A| < |A + tB| for every number t, we can follow these steps:
Step 1: Assume that A and B are perpendicular.
- Since A and B are perpendicular, the dot product of A and B is zero: A · B = 0.
- The magnitude (or length) of a vector A is denoted by |A|, which is equal to the square root of the sum of the squares of its components: |A| = √(A1^2 + A2^2 + ... + An^2).
- Therefore, the magnitude of A can be expressed as |A| = √(A · A).
Step 2: Calculate |A + tB|.
- The magnitude of the vector (A + tB) can be calculated similarly: |A + tB| = √((A + tB) · (A + tB)).
- Expanding the dot product, we get |A + tB| = √(A · A + 2t(A · B) + t^2(B · B)).
- Since A · B = 0 (from Step 1), the equation simplifies to |A + tB| = √(A · A + t^2(B · B)).
Step 3: Prove that |A| < |A + tB| for every number t.
- Substituting the expression for |A| from Step 1, we have |A| = √(A · A) and |A + tB| = √(A · A + t^2(B · B)).
- We need to show that |A| < |A + tB| for every t, which is equivalent to proving |A|^2 < |A + tB|^2.
- Squaring both sides requires comparing A · A with (A + tB) · (A + tB), which simplifies to A · A < A · A + t^2(B · B).
- Since A · A is a non-negative value, subtracting it from both sides gives 0 < t^2(B · B).
Step 4: Prove that 0 < t^2(B · B) for every t.
- Since B is non-zero, (B · B) is a positive value. Therefore, t^2(B · B) is also positive for any value of t, except t = 0.
- Note that the assumption states "for every number t," except t = 0, which means we can exclude t = 0 from the analysis.
- Therefore, for all non-zero vectors A and B, it can be concluded that |A| < |A + tB| for every number t.
Step 5: Conclude that A and B are perpendicular if and only if |A| < |A + tB| for every number t.
- Based on the above steps, we have proven that if A and B are perpendicular, then |A| < |A + tB| for every number t.
- To prove the converse, we can use the contrapositive: If |A| ≥ |A + tB| for some t, then A and B are not perpendicular.
- If |A| ≥ |A + tB| for some t, it implies that |A|^2 ≥ |A + tB|^2, which further implies that A · A ≥ A · A + t^2(B · B).
- Since (B · B) is positive, this inequality can only hold if A · A > A · A + t^2(B · B), which means A · A ≠ A · A + t^2(B · B).
- Therefore, A and B are not perpendicular.
- By proving the contrapositive, we have shown that if |A| ≥ |A + tB| for some t, then A and B are not perpendicular.
- Hence, it can be concluded that A and B are perpendicular if and only if |A| < |A + tB| for every number t.