7.3 Q-1

Question Statement

Find the cosine of the angle between the following pairs of vectors: (i) $\underline{\mathbf{U}} = 3\underline{\mathbf{i}} + \underline{\mathbf{j}} - \underline{\mathbf{k}}, \quad \underline{\mathbf{V}} = 2\underline{\mathbf{i}} - 3\underline{\mathbf{j}} + \underline{\mathbf{k}}$

(ii) $\underline{\mathbf{U}} = \underline{\mathbf{i}} - 3\underline{\mathbf{j}} + 4\underline{\mathbf{k}}, \quad \underline{\mathbf{V}} = 4\underline{\mathbf{i}} - \underline{\mathbf{j}} + 3\underline{\mathbf{k}}$

(iii) $\underline{\mathbf{U}} = [-3, 5], \quad \underline{\mathbf{V}} = [6, -2]$

(iv) $\underline{\mathbf{U}} = [2, -3, 1], \quad \underline{\mathbf{V}} = [2, 4, 1]$

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Background and Explanation

To solve this problem, we need to use the formula for the cosine of the angle between two vectors $\underline{\mathbf{U}}$ and $\underline{\mathbf{V}}$:

$$\cos \theta = \frac{\underline{\mathbf{U}} \cdot \underline{\mathbf{V}}}{|\underline{\mathbf{U}}| |\underline{\mathbf{V}}|}$$

Where:

• $\underline{\mathbf{U}} \cdot \underline{\mathbf{V}}$ is the dot product of the two vectors.
• $|\underline{\mathbf{U}}|$ and $|\underline{\mathbf{V}}|$ are the magnitudes (lengths) of the vectors.

The dot product $\underline{\mathbf{U}} \cdot \underline{\mathbf{V}}$ for vectors $\underline{\mathbf{U}} = [u_1, u_2, u_3]$ and $\underline{\mathbf{V}} = [v_1, v_2, v_3]$ is computed as:

$$\underline{\mathbf{U}} \cdot \underline{\mathbf{V}} = u_1v_1 + u_2v_2 + u_3v_3$$

The magnitude of a vector $\underline{\mathbf{U}} = [u_1, u_2, u_3]$ is:

$$|\underline{\mathbf{U}}| = \sqrt{u_1^2 + u_2^2 + u_3^2}$$

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Solution

(i) $\underline{\mathbf{U}} = 3\underline{\mathbf{i}} + \underline{\mathbf{j}} - \underline{\mathbf{k}}, \quad \underline{\mathbf{V}} = 2\underline{\mathbf{i}} - 3\underline{\mathbf{j}} + \underline{\mathbf{k}}$

First, calculate the magnitudes of $\underline{\mathbf{U}}$ and $\underline{\mathbf{V}}$:

$$|\underline{\mathbf{U}}| = \sqrt{(3)^2 + (1)^2 + (-1)^2} = \sqrt{9 + 1 + 1} = \sqrt{11}$$ $$|\underline{\mathbf{V}}| = \sqrt{(2)^2 + (-3)^2 + (1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$$

Next, compute the dot product:

$$\underline{\mathbf{U}} \cdot \underline{\mathbf{V}} = (3)(2) + (1)(-3) + (-1)(1) = 6 - 3 - 1 = 2$$

Now, substitute these values into the cosine formula:

$$\cos \theta = \frac{2}{\sqrt{11} \times \sqrt{14}} = \frac{2}{\sqrt{154}}$$

(ii) $\underline{\mathbf{U}} = \underline{\mathbf{i}} - 3\underline{\mathbf{j}} + 4\underline{\mathbf{k}}, \quad \underline{\mathbf{V}} = 4\underline{\mathbf{i}} - \underline{\mathbf{j}} + 3\underline{\mathbf{k}}$

Calculate the magnitudes:

$$|\underline{\mathbf{U}}| = \sqrt{(1)^2 + (-3)^2 + (4)^2} = \sqrt{1 + 9 + 16} = \sqrt{26}$$ $$|\underline{\mathbf{V}}| = \sqrt{(4)^2 + (-1)^2 + (3)^2} = \sqrt{16 + 1 + 9} = \sqrt{26}$$

Now, compute the dot product:

$$\underline{\mathbf{U}} \cdot \underline{\mathbf{V}} = (1)(4) + (-3)(-1) + (4)(3) = 4 + 3 + 12 = 19$$

Substitute these values into the cosine formula:

$$\cos \theta = \frac{19}{\sqrt{26} \times \sqrt{26}} = \frac{19}{26}$$

(iii) $\underline{\mathbf{U}} = [-3, 5], \quad \underline{\mathbf{V}} = [6, -2]$

First, calculate the magnitudes:

$$|\underline{\mathbf{U}}| = \sqrt{(-3)^2 + (5)^2} = \sqrt{9 + 25} = \sqrt{34}$$ $$|\underline{\mathbf{V}}| = \sqrt{(6)^2 + (-2)^2} = \sqrt{36 + 4} = \sqrt{40}$$

Now, compute the dot product:

$$\underline{\mathbf{U}} \cdot \underline{\mathbf{V}} = (-3)(6) + (5)(-2) = -18 - 10 = -28$$

Substitute these values into the cosine formula:

$$\cos \theta = \frac{-28}{\sqrt{34} \times \sqrt{40}} = \frac{-28}{\sqrt{1360}} = \frac{-28}{2\sqrt{340}}$$

(iv) $\underline{\mathbf{U}} = [2, -3, 1], \quad \underline{\mathbf{V}} = [2, 4, 1]$

First, calculate the magnitudes:

$$|\underline{\mathbf{U}}| = \sqrt{(2)^2 + (-3)^2 + (1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$$ $$|\underline{\mathbf{V}}| = \sqrt{(2)^2 + (4)^2 + (1)^2} = \sqrt{4 + 16 + 1} = \sqrt{21}$$

Now, compute the dot product:

$$\underline{\mathbf{U}} \cdot \underline{\mathbf{V}} = (2)(2) + (-3)(4) + (1)(1) = 4 - 12 + 1 = -7$$

Substitute these values into the cosine formula:

$$\cos \theta = \frac{-7}{\sqrt{14} \times \sqrt{21}} = \frac{-7}{\sqrt{294}} = \frac{-7}{7\sqrt{6}} = \frac{-1}{\sqrt{6}}$$

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Key Formulas or Methods Used

• Cosine of angle between vectors:

$$\cos \theta = \frac{\underline{\mathbf{U}} \cdot \underline{\mathbf{V}}}{|\underline{\mathbf{U}}| |\underline{\mathbf{V}}|}$$

• Dot product of vectors:

$$\underline{\mathbf{U}} \cdot \underline{\mathbf{V}} = u_1v_1 + u_2v_2 + u_3v_3$$

• Magnitude of a vector:

$$|\underline{\mathbf{U}}| = \sqrt{u_1^2 + u_2^2 + u_3^2}$$

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Summary of Steps

1. Calculate the magnitudes of both vectors $\underline{\mathbf{U}}$ and $\underline{\mathbf{V}}$.
2. Compute the dot product of the vectors.
3. Substitute the dot product and magnitudes into the cosine formula $\cos \theta = \frac{\underline{\mathbf{U}} \cdot \underline{\mathbf{V}}}{|\underline{\mathbf{U}}| |\underline{\mathbf{V}}|}$.
4. Simplify the expression to find $\cos \theta$.