prove that two non zero vectors A and B are perpendicular if and only if, |A|<|A+tB| for every number t.
A+tB is the diagonal of a parallelogram, right?
That diagonal has zero length only if A and B are perpendicular.
I don't think so
I didn't know
To see the step of work
Well, this is a classic case of "orthogonal humor"! Let's see if we can come up with a proof filled with laughter:
To prove that two non-zero vectors A and B are perpendicular if and only if |A| < |A + tB| for every number t, just imagine that A and B are standing at a party.
In one scenario, they are perpendicular, which means they both have awkward dance moves and keep bumping into each other. Now, if we take A and add tB to it, it's like trying to add a dance partner who is exactly the opposite of A's style. They will blend even less harmoniously and dance with even more awkwardness. So, it's safe to say that the magnitude of A will be less than the magnitude of A + tB for every t.
On the other hand, if A and B are not perpendicular, it means they have similar dance moves, and they go along well. Adding tB to A is like inviting a dance partner who has equivalent moves to A. As they start syncing and coordinating their moves, the magnitude of A + tB will become larger than the magnitude of A.
So, in conclusion, if A and B are perpendicular, they will keep crashing into each other and never share a perfect dance, resulting in |A| < |A + tB| for every t. But if they are not perpendicular, they will dance in sync, resulting in |A| ≥ |A + tB| for some values of t.
Remember, math and humor can dance together too!
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 approach it by breaking it down into two parts: the forward direction and the backward direction.
Forward Direction:
Assume that A and B are perpendicular vectors. We need to show that |A| < |A + tB| for every value of t.
Let's consider the vector A + tB. The magnitude of this vector, |A + tB|, represents the length of the vector. This length is the distance between the initial point and the terminal point of the vector.
Notice that A is a non-zero vector, which means it has a non-zero length. And since A and B are perpendicular, it implies that A and B are in different directions or are orthogonal to each other.
Now, let's consider the expression |A + tB|. We can rewrite it as |(1/t)A + B|. Since A and B are perpendicular, the expression (1/t)A + B represents a vector resulting from scaling vector A by (1/t) and adding vector B.
For every value of t, the vector (1/t)A + B represents the terminal point obtained by moving in the direction of B from the initial point of A. Notice that for any value of t, the magnitude of the vector (1/t)A + B is greater than or equal to |A|.
Hence, we can conclude that |A| < |A + tB| for every value of t when A and B are perpendicular.
Backward Direction:
Now assume that |A| < |A + tB| for every value of t. We need to show that A and B are perpendicular.
Let's consider the case where t = 1. According to the given condition, |A| < |A + B|. This tells us that the length of A is less than the length of the vector obtained by moving in the direction of B from A's initial point.
Now, consider the resulting vector A + B. The length of A + B represents the distance between the initial point of A and the terminal point obtained by moving in the direction of B.
Given |A| < |A + B|, it implies that the length of A is strictly less than the length of A + B. This condition is only possible if A and B are orthogonal or perpendicular to each other.
Hence, we can conclude that if |A| < |A + tB| for every value of t, then A and B are perpendicular.
Therefore, we have proven both directions of the statement, and it follows that two non-zero vectors A and B are perpendicular if and only if |A| < |A + tB| for every number t.