For the first question I really need all help I can get. Thanks!
1. Given a and b are unit vectors,
a) if the angle between them is 60 degrees, calculate (6a+b) . (a-2b)
b) if |a+b| = sqrt3, determine (2a-5b) . (b+3a)
2. The vectors a = 3i - 4j - k and b = 2i + 3j - 6k are the diagonals of a parallelogram. Show that this parallelogram is a rhombus, and determine the lengths of the sides and angles between the sides.
3. If a and b are perpendicular, show that |a|^2 + |b|^2 = |a + b|^2. What is the usual name of this result.
b) If a and b are not perpendicular, and a-b = c, express |c|^2 in terms of a and b. What is the usual name of this result?
You seem to post quite a few vector questions under the name of anonymous.
Are you the same person?
Please use a first name or some other nick to identify yourself.
Your first question uses the basic laws of vectors.
(2a-5b)∙(b+3a) = 2a∙b + 6│a│^2 - 5│b│^2 - 15a∙b
= 6│a│^2 - 5│b│^2 - 13a∙b
we know │a│ and │b│ are 1 each and
a∙b = │a││b│cos60º
a∙b = 1*1*1/2
so 6│a│^2 - 5│b│^2 - 13a∙b
= 6*1 - 5*1 - 13*1/2
= -11/2
for 1. b) make a diagram and find cosß using sides 1,1,√3
that way you can find a∙b and follow my example of a)
2. In a parallelogram the diagonals bisect each other, but in a rhombus (which is a parallelogram) they bisect each other at right angles.
so take the dot product, and see if you get zero. (you will)
then 1/2 of vector a + 1/2 of vector b will give you a side
etc.
Sure! I can help you with these questions.
1. a) To calculate (6a+b) . (a-2b):
- Use dot product rule: (6a+b) . (a-2b) = 6a . a + 6a . (-2b) + b . a + b . (-2b)
- Since a and b are unit vectors, their magnitudes are 1. Therefore, a . a = |a|^2 = 1 and b . b = |b|^2 = 1.
- Simplify the equation: 6a . a + (-12)b . a + b . a + (-2)b . b
- Substitute the values: 6(1) + (-12)(1) + 1 + (-2)(1)
- Calculate: 6 - 12 + 1 - 2 = -7
So, (6a+b) . (a-2b) = -7.
b) To determine (2a-5b) . (b+3a):
- Apply the dot product rule: (2a - 5b) . (b + 3a) = 2a . b + 2a . (3a) - 5b . b - 5b . (3a)
- Since a and b are unit vectors, their magnitudes are 1. Therefore, a . a = |a|^2 = 1 and b . b = |b|^2 = 1.
- Simplify the equation: 2a . b + 2a . (3a) - 5b . b - 5b . (3a)
- Substitute the values: 2(1) + 2(3) - 5(1) - 5(3)
- Calculate: 2 + 6 - 5 - 15 = -12
So, (2a-5b) . (b+3a) = -12.
2. To show that the parallelogram is a rhombus and determine the lengths of the sides and angles between the sides:
- Given the vectors a = 3i - 4j - k and b = 2i + 3j - 6k.
- Calculate the magnitudes of 'a' and 'b':
|a| = sqrt((3)^2 + (-4)^2 + (-1)^2) = sqrt(9 + 16 + 1) = sqrt(26)
|b| = sqrt((2)^2 + (3)^2 + (-6)^2) = sqrt(4 + 9 + 36) = sqrt(49) = 7
- The lengths of the sides are |a| = sqrt(26) and |b| = 7.
- To determine the angles between the sides, use the dot product:
cos(angle) = (a . b) / (|a| * |b|)
angle = arccos((a . b) / (|a| * |b|))
- Calculate the dot product of 'a' and 'b':
a . b = (3)(2) + (-4)(3) + (-1)(-6) = 6 - 12 + 6 = 0
- Calculate the angles:
angle = arccos(0 / (sqrt(26) * 7))
angle = arccos(0)
angle = 90 degrees
Since all angles between the sides are 90 degrees, the parallelogram is a rhombus. The lengths of the sides are |a| = sqrt(26) and |b| = 7, and the angles between the sides are all 90 degrees.
3. a) To show that |a|^2 + |b|^2 = |a + b|^2:
- Given that a and b are perpendicular unit vectors.
- Calculate the magnitude of 'a', 'b', and 'a + b':
|a| = 1
|b| = 1
|a + b| = sqrt((1 + 1)^2) = sqrt(4) = 2
- Simplify the equation: |a|^2 + |b|^2 = |a + b|^2
(1)^2 + (1)^2 = (2)^2
- Calculate: 1 + 1 = 2
So, |a|^2 + |b|^2 = |a + b|^2 is true. This result is known as the Pythagorean theorem.
b) To express |c|^2 in terms of 'a' and 'b':
- Given that a, b, and c are non-perpendicular vectors, and a - b = c.
- Use the magnitudes and dot product:
|c|^2 = (a - b) . (a - b)
- Apply the dot product rule:
(a - b) . (a - b) = a . a - a . b - b . a + b . b
- Substitute the values:
(|a|^2 - 2(a . b) + |b|^2)
So, |c|^2 = |a|^2 - 2(a . b) + |b|^2. This result is known as the Law of Cosines.