Prove that two non-zero vectors A and B are perpendicular if and only if for every number t.
To prove that two non-zero vectors A and B are perpendicular if and only if for every number t, we need to show two separate conditions:
1. If A and B are perpendicular, then for every number t:
A · (tB) = 0
2. If for every number t, A · (tB) = 0, then A and B are perpendicular.
We will prove each condition separately:
Condition 1:
If A and B are perpendicular, then for every number t:
A · (tB) = 0
Proof:
If A and B are perpendicular, it means that the dot product of A and B is zero:
A · B = 0
Now, consider the vector tB. This vector has a magnitude of t times the magnitude of B and has the same direction as B.
The dot product of A and tB can be expressed as:
A · (tB) = |A| |tB| cosθ
where |A| represents the magnitude of A, |tB| represents the magnitude of tB, and θ represents the angle between A and tB.
Since |tB| = t|B|, we can rewrite the expression as:
A · (tB) = |A| t|B| cosθ
Now, we know that |A| and |B| are (non-zero) constants, and since A and B are perpendicular, the angle θ between them is 90 degrees. Therefore, cosθ = 0.
Substituting cosθ = 0 in the expression, we get:
A · (tB) = |A| t|B| 0
A · (tB) = 0
Therefore, if A and B are perpendicular, then for every number t:
A · (tB) = 0
Condition 2:
If for every number t, A · (tB) = 0, then A and B are perpendicular.
Proof:
Let's assume that for every number t, A · (tB) = 0.
Consider the case when t = 1. Then we have:
A · (1B) = 0
A · B = 0
This shows that A and B are perpendicular.
Therefore, if for every number t, A · (tB) = 0, then A and B are perpendicular.
Since we have proven both conditions, we have proved that two non-zero vectors A and B are perpendicular if and only if for every number t:
A · (tB) = 0
To prove that two non-zero vectors A and B are perpendicular if and only if their dot product is equal to zero for every value of t, we need to demonstrate both the "if" and "only if" directions of the statement.
First, we will prove the "if" direction. Suppose A and B are perpendicular. This means that the angle between A and B is 90 degrees. The dot product of two vectors is given by the formula:
A · B = |A| |B| cos(θ)
where |A| and |B| are the magnitudes (lengths) of vectors A and B, and θ is the angle between them. In this case, since A and B are perpendicular (θ = 90 degrees), cos(θ) = 0. Therefore, the dot product of A and B is zero:
A · B = |A| |B| cos(90) = 0
This proves the "if" direction.
Now, we will prove the "only if" direction. Suppose the dot product of A and B is zero for every t. Using the same formula for dot product as above, we have:
A · B = |A| |B| cos(θ)
Since the dot product is zero for every t, it means that cos(θ) = 0 for every t. The only way for cos(θ) to be zero for every t is if the angle θ between A and B is consistently 90 degrees, or in other words, if A and B are perpendicular.
This completes the proof of the "only if" direction.
Therefore, we have proven that two non-zero vectors A and B are perpendicular if and only if their dot product is equal to zero for every value of t.