Math Problem Statement

Given that | vec a |=1 , | vec b |= sqrt(3) and that overline a and vec b are perpendicular, evaluat (a) |( vec a + vec b )( vec a - vec b )| (b) |(2 vec a + vec b )( vec a -2 vec b )|

Solution

Given that a=1|\vec{a}| = 1, b=3|\vec{b}| = \sqrt{3}, and a\vec{a} and b\vec{b} are perpendicular, we need to evaluate:

(a) (a+b)(ab)|(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b})|
(b) (2a+b)(a2b)|(2\vec{a} + \vec{b}) \cdot (\vec{a} - 2\vec{b})|

Solution:

Part (a):

First, expand the dot product: (a+b)(ab)=aaab+babb(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = \vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b}

Since ab=0\vec{a} \cdot \vec{b} = 0 (they are perpendicular): (a+b)(ab)=aabb(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = \vec{a} \cdot \vec{a} - \vec{b} \cdot \vec{b}

Using a2=aa=1|\vec{a}|^2 = \vec{a} \cdot \vec{a} = 1 and b2=bb=3|\vec{b}|^2 = \vec{b} \cdot \vec{b} = 3: (a+b)(ab)=13=2(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 1 - 3 = -2

So, the magnitude is: (a+b)(ab)=2=2|(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b})| = |-2| = 2

Part (b):

Expand the dot product: (2a+b)(a2b)=2aa4ab+ba2bb(2\vec{a} + \vec{b}) \cdot (\vec{a} - 2\vec{b}) = 2\vec{a} \cdot \vec{a} - 4\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} - 2\vec{b} \cdot \vec{b}

Again, using ab=0\vec{a} \cdot \vec{b} = 0: (2a+b)(a2b)=2aa2bb(2\vec{a} + \vec{b}) \cdot (\vec{a} - 2\vec{b}) = 2\vec{a} \cdot \vec{a} - 2\vec{b} \cdot \vec{b}

Substitute the magnitudes: (2a+b)(a2b)=2(1)2(3)=26=4(2\vec{a} + \vec{b}) \cdot (\vec{a} - 2\vec{b}) = 2(1) - 2(3) = 2 - 6 = -4

So, the magnitude is: (2a+b)(a2b)=4=4|(2\vec{a} + \vec{b}) \cdot (\vec{a} - 2\vec{b})| = |-4| = 4

Final Answers:

  • (a) (a+b)(ab)=2|(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b})| = 2
  • (b) (2a+b)(a2b)=4|(2\vec{a} + \vec{b}) \cdot (\vec{a} - 2\vec{b})| = 4

Would you like any additional details or explanations?


Here are some related questions to consider:

  1. What are the properties of dot products for perpendicular vectors?
  2. How does the magnitude of vectors affect the dot product?
  3. Can dot products be used to determine the angle between two vectors?
  4. What happens to the dot product if one of the vectors is multiplied by a scalar?
  5. How can we extend this problem to three dimensions?
  6. What is the geometric interpretation of the result in part (a)?
  7. How would the answers change if a\vec{a} and b\vec{b} were not perpendicular?
  8. How does vector addition affect the overall magnitude of the resulting vector?

Tip: Always check the angle between vectors when dealing with dot products, as this can simplify the calculation significantly.

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Math Problem Analysis

Mathematical Concepts

Vectors
Dot Product
Magnitude of Vectors

Formulas

Dot product formula: \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \)

Theorems

Properties of dot products for perpendicular vectors

Suitable Grade Level

Advanced High School