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.
Prove that two non zero Vectors A anb B are perpendicular if and only if |A|≤|A + tB| for every number t
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