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 this statement, we will demonstrate both the forward and backward directions.

First, we will prove the forward direction: if vectors A and B are perpendicular, then |A| < |A+tB| for every number t.

Assume that A and B are perpendicular vectors, meaning their dot product is zero: A • B = 0.

Now, consider the vector A + tB. To find the magnitude of this vector, we square each component, sum them, and take the square root:

|A + tB|^2 = (A + tB) • (A + tB) = (A • A) + 2t(A • B) + (tB • tB)

= |A|^2 + 2t(A • B) + t^2|B|^2
(Notice that the cross term 2t(A • B) disappears since A • B = 0.)

Since |A|^2 and t^2|B|^2 are always positive, the magnitude of A + tB is minimized when t = 0.

Thus, |A + tB| ≥ |A| for every t, with equality holding only when t = 0.

Therefore, we can conclude that if A and B are perpendicular vectors, then |A| < |A+tB| for every t.

Now, we will prove the backward direction: if |A| < |A+tB| for every number t, then vectors A and B are perpendicular.

Assume that |A| < |A+tB| for every t.

Now, let's consider the vector A + tB again. By applying the triangle inequality, we have:

|A + tB| ≤ |A| + |tB|

Since |A| < |A + tB| for every t, we can rewrite the inequality as:

|A| + |tB| > |A + tB|

Simplifying this inequality, we get:

|tB| > 0 (since |tB| is always positive)

Dividing both sides by |t| (note that |t| > 0), we obtain:

|B| > 0

Therefore, we can conclude that |B| > 0, which means B is a non-zero vector.

Next, consider the case when t = -1. We have:

|A + (-1)B| = |A - B|

Since |A| < |A + tB| for every t, we know that |A| < |A - B|.

Squaring both sides of this inequality, we get:

|A|^2 < |A - B|^2

Expanding the square on the right side, we have:

|A|^2 < (A - B) • (A - B) = |A|^2 - 2(A • B) + |B|^2

Canceling out |A|^2 from both sides, we obtain:

0 < -2(A • B) + |B|^2

Rearranging this inequality, we get:

2(A • B) < |B|^2

Dividing both sides by 2 (note that 2 > 0), we have:

(A • B) < |B|^2/2

Since |B|^2/2 is always positive, (A • B) must be negative. This implies that the angle between A and B is obtuse, which further implies that they are perpendicular vectors.

Therefore, we have shown that if |A| < |A+tB| for every number t, then vectors A and B are perpendicular.

By proving both the forward and backward directions, we have demonstrated that two non-zero vectors A and B are perpendicular if and only if |A| < |A+tB| for every number t.