Given |a|=10, |b|=15, and |a+b|=20, find |a-b|
Draw the parallelogram with sides a and b, where the angle from a to be is θ. Then |a+b| is the length of one diagonal
|a+b|^2 = a^2 + b^2 + 2ab cosθ
plug in your numbers, and you can see that cosθ = 1/4
The length of the other diagonal is
|a-b|^2 = a^2 + b^2 - 2ab cosθ = 250
so |a-b| = 5√10
This question is about vector.
To find |a-b|, we can use the properties of absolute value and the given information.
We know that |a+b| = 20, which means the magnitude of the vector sum of a and b is 20. Using the triangle inequality, we can write:
|a-b| ≤ |a| + |b|
|a-b| ≤ 10 + 15 (since |a| = 10 and |b| = 15)
|a-b| ≤ 25
Now, to determine the minimum possible value of |a-b|, we need to consider the worst-case scenario. In this case, a and b would be pointing in opposite directions.
If a and b are pointing in opposite directions, then the magnitude of their vector sum (a+b) would be the difference between their magnitudes. Therefore:
|a+b| = |a| - |b|
Substituting the given values, we have:
|a+b| = 10 - 15
20 = -5
However, this is not possible since the magnitude of a vector cannot be negative. Therefore, we conclude that the minimum value of |a-b| is zero.
Thus, |a-b| = 0.
To find the magnitude of the vector difference, |a-b|, we can use the properties of vectors. The magnitude of the vector difference is equal to the absolute value of the difference between the magnitudes of the two vectors.
First, let's label the given information:
|a| = 10
|b| = 15
|a+b| = 20
Now, we can use the properties of vectors to derive the equation:
|a+b| = |a| + |b| + 2|a||b|cos(theta)
In this equation, theta represents the angle between vectors a and b. Since we are given the magnitudes, we need to find the value of cos(theta).
Rearranging the equation, we have:
|a| + |b| + 2|a||b|cos(theta) = |a+b|
Substituting the known magnitudes, we have:
10 + 15 + 2(10)(15)cos(theta) = 20
Simplifying the equation gives:
25 + 300cos(theta) = 20
Rearranging again, we have:
300cos(theta) = 20 - 25
300cos(theta) = -5
Finally, we solve for the value of cos(theta):
cos(theta) = -5/300
Using a calculator or trigonometric table, we find that the value of cos(theta) is approximately -0.01667.
Now, we can substitute the value of cos(theta) back into the original equation to solve for |a-b|:
|a - b| = sqrt(|a|^2 + |b|^2 - 2|a||b|cos(theta))
Substituting the given magnitudes and the value of cos(theta), we have:
|a - b| = sqrt(10^2 + 15^2 - 2(10)(15)(-0.01667))
Simplifying this expression will give us the magnitude of the vector difference, |a - b|.