Prove that two non zero Vectors A anb B are perpendicular if and only if |A|≤|A + tB| for every number t
Assume that A and B are perpendicular non zero vectors. Then, we can write A · B = 0, where · represents the dot product.
Now consider the vector A + tB, where t is any real number. The magnitude squared of this vector is
|(A + tB)|^2 = (A + tB) · (A + tB)
= A · A + 2t(A · B) + t^2(B · B)
= |A|^2 + t^2|B|^2
Since |B|^2 is positive, we have |A + tB|^2 ≥ |A|^2, which implies |A + tB| ≥ |A| for all t.
Conversely, assume that |A + tB| ≥ |A| for all t. Let t = -A · B / |B|^2. Then,
|A + tB|^2 = |A|^2 - 2t(A · B) + t^2|B|^2
= |A|^2 - \frac{(A · B)^2}{|B|^2}
Using the Cauchy-Schwarz inequality, we have
|A|^2 |B|^2 ≥ (A · B)^2
which implies
\frac{(A · B)^2}{|B|^2} ≥ 0
Hence,
|A + tB|^2 = |A|^2 - \frac{(A · B)^2}{|B|^2} ≤ |A|^2
Therefore, we have |A + tB| ≤ |A| for all t.
Now suppose that A and B are not perpendicular. Then, we can write A = |A|u and B = |B|v, where u and v are unit vectors and u · v = cosθ for some θ not equal to 0 or π. Let t = -|A||B|cosθ / (|B|^2 - |A|^2sin^2θ). Then,
|A + tB|^2 = |A|^2 - 2t|A||B|cosθ + t^2|B|^2
= |A|^2 - 2\frac{|A|^2|B|^2cos^2θ}{|B|^2 - |A|^2sin^2θ} + \frac{|A|^2|B|^2cos^2θ}{(|B|^2 - |A|^2sin^2θ)^2}|B|^2
= \frac{|A|^4cos^2θ}{(|B|^2 - |A|^2sin^2θ)^2} > |A|^2
Therefore, there exists a value of t such that |A + tB| > |A|, which contradicts our assumption. Hence, A and B must be perpendicular.
Therefore, we have proved the statement: 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 |A| ≤ |A + tB| for every number t, we need to show both directions of the equivalence.
First, let's assume that A and B are perpendicular vectors. This means that their dot product is zero:
A · B = 0.
Now, let's consider the vector C = A + tB, where t is any real number.
The magnitude of vector C can be expressed as:
|C| = √(C · C).
Substituting the expression for C, we have:
|C| = √((A + tB) · (A + tB)).
Expanding the dot product, we get:
|C| = √(A · A + 2t(A · B) + t^2(B · B)).
Since A · B = 0 (as A and B are perpendicular), the term 2t(A · B) becomes zero:
|C| = √(A · A + t^2(B · B)).
Now, we know that the magnitude of A, |A|, is equal to √(A · A). Therefore, we can rewrite the expression for |C|:
|C| = √(|A|^2 + t^2(B · B)).
Since |A|^2 and t^2 are both non-negative, we can conclude that:
|C| ≥ |A|.
Thus, we have shown that for any t, |A| ≤ |A + tB|.
Next, we need to prove the converse: if |A| ≤ |A + tB| for every t, then A and B are perpendicular.
We can do this by contradiction. Suppose A and B are not perpendicular, which means A · B ≠ 0.
Now, consider the vector C = A + tB, where t = -A · B / (B · B). This choice of t will make the dot product of C equal to zero:
C · B = (A + tB) · B = A · B + t(B · B) = A · B - (A · B / (B · B))(B · B) = 0.
However, by assumption, we have |A| ≤ |A + tB| for every t. Let's plug in the value of t we chose:
|A| ≤ |A + (-A · B / (B · B))B|.
Simplifying, we get:
|A| ≤ |A - (A · B / (B · B))B|.
Notice that the vector on the right-hand side is different from vector A, but it has the same magnitude as A, because |A + tB| = |A|.
Therefore, we have:
|A| ≤ |A - (A · B / (B · B))B| = |A|.
This implies:
|A| = |A|.
But this is a trivial inequality that doesn't provide any useful information about A and B.
Thus, by contradiction, we can conclude that our initial assumption was incorrect. A and B must be perpendicular if |A| ≤ |A + tB| for every t.
Therefore, we have proven that two non-zero vectors A and B are perpendicular if and only if |A| ≤ |A + tB| for every number t.