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
Prove that two non-zero vectors A and B are perpendicular if and only if for every number t.
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