Math Problem Statement

If a and b are two unit vectors and x, y are the minimum and maximum values of (3/2)*|a+b| + 2*|a-b| respectively, then which of the following options are correct: a) x+y=0, b) xy=-25, c) x+y=8, d) xy=15?

Solution

The problem involves two unit vectors, a\vec{a} and b\vec{b}, and asks about the minimum (xx) and maximum (yy) values of the expression:

32a+b+2ab\frac{3}{2}|\vec{a} + \vec{b}| + 2|\vec{a} - \vec{b}|

Let’s analyze this step by step.

Step 1: Understanding the Magnitudes of Vector Expressions

Given that a\vec{a} and b\vec{b} are unit vectors, the magnitudes of the expressions a+b|\vec{a} + \vec{b}| and ab|\vec{a} - \vec{b}| depend on the angle θ\theta between the two vectors:

  • a+b=2+2cosθ|\vec{a} + \vec{b}| = \sqrt{2 + 2\cos\theta}
  • ab=22cosθ|\vec{a} - \vec{b}| = \sqrt{2 - 2\cos\theta}

The values of these magnitudes will range depending on θ\theta, with:

  • cosθ=1\cos\theta = 1 when a\vec{a} and b\vec{b} are parallel (minimum angle, giving maximum magnitude),
  • cosθ=1\cos\theta = -1 when a\vec{a} and b\vec{b} are opposite (maximum angle, giving minimum magnitude).

Step 2: Identifying Maximum and Minimum Values

Maximum Value:

When a\vec{a} and b\vec{b} are parallel (cosθ=1\cos\theta = 1):

  • a+b=4=2|\vec{a} + \vec{b}| = \sqrt{4} = 2
  • ab=0=0|\vec{a} - \vec{b}| = \sqrt{0} = 0

Substitute into the expression: 32(2)+2(0)=3\frac{3}{2}(2) + 2(0) = 3 This gives the maximum value y=3y = 3.

Minimum Value:

When a\vec{a} and b\vec{b} are opposite (cosθ=1\cos\theta = -1):

  • a+b=0=0|\vec{a} + \vec{b}| = \sqrt{0} = 0
  • ab=4=2|\vec{a} - \vec{b}| = \sqrt{4} = 2

Substitute into the expression: 32(0)+2(2)=4\frac{3}{2}(0) + 2(2) = 4 This gives the minimum value x=4x = 4.

Step 3: Solving the Given Equation

Now we need to check which of the given equations corresponds to the relation between the maximum and minimum values.

We know:

  • x=4x = 4
  • y=3y = 3

Testing the options:

  • Option (a): x+y=0x + y = 04+304 + 3 \neq 0
  • Option (b): xy=25xy = -254×3=12254 \times 3 = 12 \neq -25
  • Option (c): x+y=8x + y = 84+3=784 + 3 = 7 \neq 8
  • Option (d): xy=15xy = 154×3=12154 \times 3 = 12 \neq 15

None of these options are correct based on the calculated values of x=4x = 4 and y=3y = 3. However, there may be a slight variation in calculation, or perhaps the question has an error.

Do you want to discuss the next steps or alternative strategies?

Follow-Up Questions:

  1. What would change in the calculations if a\vec{a} and b\vec{b} were not unit vectors?
  2. How would the result differ if the coefficients in the given expression were altered?
  3. What other methods can be used to find the maximum and minimum values of vector magnitudes?
  4. Could you verify the calculation of the maximum and minimum values by another approach, such as dot product analysis?
  5. How would the problem differ if the vectors were in three-dimensional space?

Tip: When working with vector magnitudes, always double-check assumptions like vector lengths and angles, as they significantly affect the results.

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Math Problem Analysis

Mathematical Concepts

Vectors
Magnitude of Vectors
Unit Vectors
Vector Addition and Subtraction
Cosine Rule

Formulas

|a + b| = sqrt(2 + 2*cos(theta))
|a - b| = sqrt(2 - 2*cos(theta))
Expression: (3/2) * |a+b| + 2 * |a-b|

Theorems

Cosine Law for Vectors
Trigonometric Identity for Vector Addition/Subtraction

Suitable Grade Level

Grades 10-12