Prove that two non - zero vectors AandB are perpendicular if and only if|A|<=|A+tB| for every number t
1425
1425 can be factored as 3 x 5 x 5 x 19.
To find the prime factorization of 1425, we can start by dividing it by the smallest prime number, 2. However, 1425 is an odd number and is not divisible by 2.
Next, we can try dividing it by 3. 1425 ÷ 3 = 475, which tells us that 3 is a factor of 1425. We can now focus on finding the prime factorization of 475.
Dividing 475 by 5 gives us 95, which is not divisible by 5 any further.
Dividing 95 by 5 gives us 19, which is a prime number.
Therefore, the prime factorization of 1425 is 3 x 5 x 5 x 19.
To prove the statement, we will need to show two implications:
1. If A and B are perpendicular, then |A| <= |A + tB| for every number t.
2. If |A| <= |A + tB| for every number t, then A and B are perpendicular.
Let's proceed with the proof:
Proof:
1. If A and B are perpendicular, then |A| <= |A + tB| for every number t.
Given that A and B are perpendicular, we know that their dot product is zero: A · B = 0.
Now, let's calculate the magnitude of A + tB:
|A + tB|² = (A + tB) · (A + tB) [Using the definition of magnitude: |A + tB|² = (A + tB) · (A + tB)]
Expanding the dot product, we get:
|A + tB|² = (A + tB) · (A + tB)
= A · A + t(A · B) + t(B · A) + t²(B · B) [Using the distributive property and dot product properties]
= |A|² + 2t(A · B) + t²|B|² [Since A · B = B · A and A · A = |A|²]
Since A · B = 0 (since A and B are perpendicular), the equation simplifies to:
|A + tB|² = |A|² + t²|B|²
Since squares are always non-negative, we have |A + tB|² >= |A|² for any value of t. Now, taking the square root of both sides, we have:
|A + tB| >= |A|
Therefore, we have shown that if A and B are perpendicular, then |A| <= |A + tB| for every number t.
2. If |A| <= |A + tB| for every number t, then A and B are perpendicular.
Suppose |A| <= |A + tB| for every number t.
Let's take t = 1 in the given inequality:
|A| <= |A + B|
Squaring both sides of the inequality, we get:
|A|² <= |A + B|²
= (A + B) · (A + B)
= A · A + 2(A · B) + B · B
= |A|² + 2(A · B) + |B|²
Canceling out |A|² from both sides of the inequality, we have:
0 <= 2(A · B) + |B|²
Since squares are always non-negative, we have |B|² >= 0. Therefore, 2(A · B) + |B|² >= 0.
Since the inequality holds for every t, it must hold for t = -1. Substituting t = -1, we have:
|A| <= |A - B|
Squaring both sides of the inequality, we get:
|A|² <= |A - B|²
= (A - B) · (A - B)
= A · A - 2(A · B) + B · B
= |A|² - 2(A · B) + |B|²
Canceling out |A|² from both sides of the inequality, we have:
0 <= -2(A · B) + |B|²
Since the left-hand side of the inequality is non-negative, we have:
0 <= |B|² - 2(A · B)
Simplifying, we get:
2(A · B) <= |B|²
Dividing both sides of the inequality by 2, we have:
(A · B) <= (1/2)|B|²
Since |B|² is non-negative, (1/2)|B|² is also non-negative. Therefore, (A · B) <= (1/2)|B|² implies that (A · B) = 0.
Hence, we have shown that if |A| <= |A + tB| for every number t, then A and B are perpendicular.
Therefore, we have proved both implications, which demonstrates that 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 consider both directions of the statement.
First, let's assume that A and B are perpendicular.
To prove that |A| ≤ |A + tB| for every number t, we need to show that the length of vector A is less than or equal to the length of vector A + tB for any value of t.
The length of vector A, denoted as |A|, can be calculated using the formula:
|A| = sqrt(Ax^2 + Ay^2 + Az^2)
Similarly, the length of vector A + tB, denoted as |A + tB|, can be calculated as:
|A + tB| = sqrt((Ax + tBx)^2 + (Ay + tBy)^2 + (Az + tBz)^2)
Now, let's compare the squares of the lengths of both vectors to avoid square root calculations:
|A|^2 = Ax^2 + Ay^2 + Az^2
|A + tB|^2 = (Ax + tBx)^2 + (Ay + tBy)^2 + (Az + tBz)^2
= Ax^2 + 2tAxBx + t^2Bx^2 + Ay^2 + 2tAyBy + t^2By^2 + Az^2 + 2tAzBz + t^2Bz^2
Since A and B are perpendicular, the dot product of A and B is zero:
A · B = Ax*Bx + Ay*By + Az*Bz = 0
Now, consider the difference:
|A + tB|^2 - |A|^2 = 2t(Ax*Bx + Ay*By + Az*Bz) + t^2(Bx^2 + By^2 + Bz^2)
As A · B = 0, it follows that:
|A + tB|^2 - |A|^2 = t^2(Bx^2 + By^2 + Bz^2)
Since t^2 is non-negative, we can conclude that:
|A + tB|^2 - |A|^2 ≥ 0
Taking the square root of both sides, we have:
|A + tB| ≥ |A|
Thus, we have shown that |A| ≤ |A + tB| for every number t, when A and B are perpendicular.
Now, let's consider the reverse direction:
Assume that |A| ≤ |A + tB| for every number t.
We need to prove that A and B are perpendicular.
If A and B are not perpendicular, then their dot product A · B is non-zero.
Let us consider the case where A and B are not parallel, so A · B ≠ 0.
Now, we select a value of t such that t = -(A · B) / |B|^2.
Substituting this value of t into the inequality |A| ≤ |A + tB|, we get:
|A| ≤ |A + tB|
|A| ≤ |A + (-A · B / |B|^2)B|
|A| ≤ |A - (A · B / |B|^2)B|
Expanding the modulus, we have:
|A|^2 ≤ (A - (A · B / |B|^2)B) · (A - (A · B / |B|^2)B)
Simplifying further:
|A|^2 ≤ A · A - (A · B)^2 / |B|^2
Now, recall that A · A is the square of the length of A, |A|^2, and we also know that |A|^2 = A · A.
Substituting this, we can rewrite the inequality as:
|A|^2 ≤ |A|^2 - (A · B)^2 / |B|^2
Here, we observe that both sides of the inequality are equal. But this implies that:
|A|^2 = |A|^2 - (A · B)^2 / |B|^2
Subtracting |A|^2 from both sides, we get:
0 = -(A · B)^2 / |B|^2
Since the square of a number is zero if and only if the number itself is zero, we have:
A · B = 0
This implies that A and B are perpendicular.
Therefore, we have now proven both directions:
1. If A and B are perpendicular, then |A| ≤ |A + tB| for every number t.
2. If |A| ≤ |A + tB| for every number t, then A and B are perpendicular.
Hence, two non-zero vectors A and B are perpendicular if and only if |A| ≤ |A + tB| for every number t.