7.2 Q-11

Question Statement

Find the direction cosines for the following vectors:

(i) $\underline{\mathbf{V}} = 3 \underline{\mathbf{i}} - \underline{\mathbf{j}} + 2 \underline{\mathbf{k}}$
(ii) $\underline{\mathbf{V}} = 6 \underline{\mathbf{i}} - 2 \underline{\mathbf{j}} + \underline{\mathbf{k}}$
(iii) $\mathbf{P} = (2, 1, 5)$, $\mathbf{Q} = (1, 3, 1)$ (find the direction cosines of the vector $\overrightarrow{PQ}$)

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Background and Explanation

The direction cosines of a vector represent the cosines of the angles that the vector makes with the $x$-, $y$-, and $z$-axes. To find the direction cosines for any vector $\mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} + v_z \mathbf{k}$, we use the following formula: $\cos \alpha = \frac{v_x}{|\mathbf{v}|}, \quad \cos \beta = \frac{v_y}{|\mathbf{v}|}, \quad \cos \gamma = \frac{v_z}{|\mathbf{v}|}$

Where:

• $|\mathbf{v}|$ is the magnitude of the vector.
• $\alpha$, $\beta$, and $\gamma$ are the angles between the vector and the $x$-, $y$-, and $z$-axes, respectively.

The magnitude of a vector $\mathbf{v} = v_x \mathbf{i} + v_y \mathbf{j} + v_z \mathbf{k}$ is calculated using: $|\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$

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Solution

(i) Find the direction cosines of $\underline{\mathbf{V}} = 3 \underline{\mathbf{i}} - \underline{\mathbf{j}} + 2 \underline{\mathbf{k}}$

1. Find the magnitude of $\underline{\mathbf{V}}$: $|\underline{\mathbf{V}}| = \sqrt{(3)^2 + (-1)^2 + (2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$

2. Find the direction cosines: Using the formula for the direction cosines: $\cos \alpha = \frac{3}{\sqrt{14}}, \quad \cos \beta = \frac{-1}{\sqrt{14}}, \quad \cos \gamma = \frac{2}{\sqrt{14}}$

Thus, the direction cosines of $\underline{\mathbf{V}}$ are: $\left[ \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}}, \frac{2}{\sqrt{14}} \right]$

(ii) Find the direction cosines of $\underline{\mathbf{V}} = 6 \underline{\mathbf{i}} - 2 \underline{\mathbf{j}} + \underline{\mathbf{k}}$

1. Find the magnitude of $\underline{\mathbf{V}}$: $|\underline{\mathbf{V}}| = \sqrt{(6)^2 + (-2)^2 + (1)^2} = \sqrt{36 + 4 + 1} = \sqrt{41}$

2. Find the direction cosines: Using the formula for the direction cosines: $\cos \alpha = \frac{6}{\sqrt{41}}, \quad \cos \beta = \frac{-2}{\sqrt{41}}, \quad \cos \gamma = \frac{1}{\sqrt{41}}$

Thus, the direction cosines of $\underline{\mathbf{V}}$ are: $\left[ \frac{6}{\sqrt{41}}, \frac{-2}{\sqrt{41}}, \frac{1}{\sqrt{41}} \right]$

(iii) Find the direction cosines of the vector $\overrightarrow{PQ}$

1. Find the vector $\overrightarrow{PQ}$: The vector $\overrightarrow{PQ}$ is given by: $\overrightarrow{PQ} = \mathbf{Q} - \mathbf{P} = (1, 3, 1) - (2, 1, 5) = (1 - 2, 3 - 1, 1 - 5) = (-1, 2, -4)$

2. Find the magnitude of $\overrightarrow{PQ}$: $|\overrightarrow{PQ}| = \sqrt{(-1)^2 + (2)^2 + (-4)^2} = \sqrt{1 + 4 + 16} = \sqrt{21}$

3. Find the direction cosines: Using the formula for the direction cosines: $\cos \alpha = \frac{-1}{\sqrt{21}}, \quad \cos \beta = \frac{2}{\sqrt{21}}, \quad \cos \gamma = \frac{-4}{\sqrt{21}}$

Thus, the direction cosines of $\overrightarrow{PQ}$ are: $\left[ \frac{-1}{\sqrt{21}}, \frac{2}{\sqrt{21}}, \frac{-4}{\sqrt{21}} \right]$

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Key Formulas or Methods Used

• Magnitude of a Vector: $|\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$

• Direction Cosines: $\cos \alpha = \frac{v_x}{|\mathbf{v}|}, \quad \cos \beta = \frac{v_y}{|\mathbf{v}|}, \quad \cos \gamma = \frac{v_z}{|\mathbf{v}|}$

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Summary of Steps

1. (i) Find the magnitude of $\underline{\mathbf{V}} = 3 \underline{\mathbf{i}} - \underline{\mathbf{j}} + 2 \underline{\mathbf{k}}$, then calculate the direction cosines by dividing each component by the magnitude.
2. (ii) Find the magnitude of $\underline{\mathbf{V}} = 6 \underline{\mathbf{i}} - 2 \underline{\mathbf{j}} + \underline{\mathbf{k}}$, then calculate the direction cosines.
3. (iii) Find the vector $\overrightarrow{PQ}$, then calculate its magnitude and direction cosines.
4. The direction cosines are:
   • (i) $\left[ \frac{3}{\sqrt{14}}, \frac{-1}{\sqrt{14}}, \frac{2}{\sqrt{14}} \right]$
   • (ii) $\left[ \frac{6}{\sqrt{41}}, \frac{-2}{\sqrt{41}}, \frac{1}{\sqrt{41}} \right]$
   • (iii) $\left[ \frac{-1}{\sqrt{21}}, \frac{2}{\sqrt{21}}, \frac{-4}{\sqrt{21}} \right]$