Math Problem Statement

1.     Use the definition of scalar product, aG ⋅bG = ab cos θ, and the fact that aG ⋅bG = axbx + ayby + azbz , calculate the angle between the two vectors given by aG= 3.0iˆ+3.0ˆj +3.0kˆ    and    bG= 2.0iˆ+1.0ˆj +3.0 .kˆ

Solution

We can calculate the angle between the two vectors a\vec{a} and b\vec{b} using the definition of the scalar (dot) product:

ab=abcosθ\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta

Where:

  • a=3.0i^+3.0j^+3.0k^\vec{a} = 3.0\hat{i} + 3.0\hat{j} + 3.0\hat{k}
  • b=2.0i^+1.0j^+3.0k^\vec{b} = 2.0\hat{i} + 1.0\hat{j} + 3.0\hat{k}
  • θ\theta is the angle between the vectors.

Step 1: Calculate the dot product ab\vec{a} \cdot \vec{b}

The dot product of two vectors a=axi^+ayj^+azk^\vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k} and b=bxi^+byj^+bzk^\vec{b} = b_x \hat{i} + b_y \hat{j} + b_z \hat{k} is:

ab=axbx+ayby+azbz\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z

For a=3.0i^+3.0j^+3.0k^\vec{a} = 3.0\hat{i} + 3.0\hat{j} + 3.0\hat{k} and b=2.0i^+1.0j^+3.0k^\vec{b} = 2.0\hat{i} + 1.0\hat{j} + 3.0\hat{k}:

ab=(3.0)(2.0)+(3.0)(1.0)+(3.0)(3.0)\vec{a} \cdot \vec{b} = (3.0)(2.0) + (3.0)(1.0) + (3.0)(3.0)

ab=6.0+3.0+9.0=18.0\vec{a} \cdot \vec{b} = 6.0 + 3.0 + 9.0 = 18.0

Step 2: Calculate the magnitudes a|\vec{a}| and b|\vec{b}|

The magnitude of a vector a=axi^+ayj^+azk^\vec{a} = a_x \hat{i} + a_y \hat{j} + a_z \hat{k} is given by:

a=ax2+ay2+az2|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}

For a=3.0i^+3.0j^+3.0k^\vec{a} = 3.0\hat{i} + 3.0\hat{j} + 3.0\hat{k}:

a=(3.0)2+(3.0)2+(3.0)2=9.0+9.0+9.0=27.0=33|\vec{a}| = \sqrt{(3.0)^2 + (3.0)^2 + (3.0)^2} = \sqrt{9.0 + 9.0 + 9.0} = \sqrt{27.0} = 3\sqrt{3}

Similarly, for b=2.0i^+1.0j^+3.0k^\vec{b} = 2.0\hat{i} + 1.0\hat{j} + 3.0\hat{k}:

b=(2.0)2+(1.0)2+(3.0)2=4.0+1.0+9.0=14.0|\vec{b}| = \sqrt{(2.0)^2 + (1.0)^2 + (3.0)^2} = \sqrt{4.0 + 1.0 + 9.0} = \sqrt{14.0}

Step 3: Calculate the angle θ\theta

From the dot product formula:

ab=abcosθ\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta

We can solve for cosθ\cos \theta:

cosθ=abab=18.0(33)(14)\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{18.0}{(3\sqrt{3})(\sqrt{14})}

Simplifying:

cosθ=18.0342=642\cos \theta = \frac{18.0}{3\sqrt{42}} = \frac{6}{\sqrt{42}}

Now, calculate cosθ\cos \theta:

cosθ=64266.48070.9258\cos \theta = \frac{6}{\sqrt{42}} \approx \frac{6}{6.4807} \approx 0.9258

Finally, solve for θ\theta:

θ=cos1(0.9258)\theta = \cos^{-1}(0.9258)

θ22.4\theta \approx 22.4^\circ

Conclusion:

The angle between the two vectors a\vec{a} and b\vec{b} is approximately 22.422.4^\circ.


Would you like any additional details or have further questions?

Here are 5 questions related to this problem:

  1. How does the magnitude of a vector affect the dot product?
  2. Can the dot product be negative, and if so, what does that signify about the angle between the vectors?
  3. How would the calculation change if one of the vectors was a unit vector?
  4. What are the properties of dot products in higher dimensions?
  5. How is the dot product used in physics applications such as work or energy?

Tip: The dot product of two perpendicular vectors is always zero.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Dot Product
Trigonometry

Formulas

Dot product formula: a ⋅ b = axbx + ayby + azbz
Magnitude of vector: |a| = √(ax² + ay² + az²)
Cosine rule for dot product: a ⋅ b = |a| |b| cos(θ)

Theorems

Dot Product Theorem
Cosine Rule

Suitable Grade Level

Grades 10-12