Prove that two non zero vectors A and B are perpendicular if and only if |A|< |A + tB| for every number t

Answer 1

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.