prove that two non zero vectors A and B are perpendicular if and only if, |A|<|A+tB| for every number t.
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.
To prove that two non-zero vectors A and B are perpendicular if and only if the inequality |A| < |A+tB| holds for every number t, we can proceed as follows:
First, let's assume that A and B are perpendicular. This means that the dot product of A and B is zero: A ⋅ B = 0.
Next, consider the vector C = A + tB, where t is any real number.
The magnitude (or length) of vector C can be calculated as:
|C| = sqrt((A + tB) ⋅ (A + tB))
Expanding and simplifying:
|C| = sqrt(A⋅A + 2tA⋅B + t^2B⋅B)
Since A⋅B = 0 (since A and B are perpendicular), the equation simplifies to:
|C| = sqrt(A⋅A + t^2B⋅B)
Now, let's compare the magnitude of vector C to the magnitude of vector A:
|C| < |A|
Squaring both sides of the inequality:
|C|^2 < |A|^2
Substituting the expression for |C| and squaring both sides again:
(A⋅A + t^2B⋅B) < A⋅A
Simplifying:
t^2(B⋅B) < 0
Since B⋅B is a non-zero value, the only way for this inequality to hold for every t is if t = 0. Thus, |A| < |A + 0⋅B|, which reduces to |A| < |A|.
Now, let's prove the other direction of the "if and only if" statement.
Assume that |A| < |A+tB| holds for every t.
For the sake of contradiction, let's assume that A and B are not perpendicular, which means that A⋅B ≠ 0.
Consider the vector C = A + tB, where t is chosen such that A⋅B ≠ 0. In this case, we can find a value of t such that (A + tB)⋅(A + tB) = 0, which leads to a contradiction.
Therefore, if the inequality holds for every t, A and B must be perpendicular.
Thus, we have proven that two non-zero vectors A and B are perpendicular if and only if the inequality |A| < |A+tB| holds for every number t.