Vector Algebra
Direction Cosines and Direction Ratios
Grade None
Question:
<p>The direction cosines of the vector $3\mathbf{i} - 4\mathbf{j} + 5\mathbf{k}$ are</p>
<p>(a) $\frac{3}{5}, \frac{-4}{5}, \frac{1}{5}$</p>
<p>(b) $\frac{3}{5\sqrt{2}}, \frac{-4}{5\sqrt{2}}, \frac{1}{\sqrt{2}}$</p>
<p>(c) $\frac{3}{\sqrt{2}}, \frac{-4}{\sqrt{2}}, \frac{1}{\sqrt{2}}$</p>
<p>(d) $\frac{3}{5\sqrt{2}}, \frac{4}{5\sqrt{2}}, \frac{1}{\sqrt{2}}$</p>
Step-by-Step Solution
Key Concept: Direction cosines are obtained by dividing each component of a vector by its magnitude. The magnitude must be calculated correctly using the formula $|\mathbf{r}| = \sqrt{a^2 + b^2 + c^2}$.
Step 1: Given vector $\mathbf{r} = 3\mathbf{i} - 4\mathbf{j} + 5\mathbf{k}$ Step 2: Find magnitude: $|\mathbf{r}| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$ Step 3: Direction cosines are found by dividing each component by the magnitude: Direction cosines = $\left(\frac{3}{5\sqrt{2}}, \frac{-4}{5\sqrt{2}}, \frac{5}{5\sqrt{2}}\right) = \left(\frac{3}{5\sqrt{2}}, \frac{-4}{5\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ ∴ Answer is (b).
Correct Answer: B