Question

Find the direction-cosines of the sum of the vectors \( \vec{a} = 3\hat{i} + 4\hat{j} - 3\hat{k} \) and \( \vec{b} = -2\hat{i} - 3\hat{j} + \hat{k} \).

Hint:

Finding the direction cosines is equivalent to finding the components of the unit vector along the given vector. Both processes involve dividing the vector's components by its magnitude.

Answer 1

Step 1: Understanding the Concept:
Direction cosines of a vector are the cosines of the angles the vector makes with the positive x, y, and z axes. For a vector \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \), the direction cosines are \( \frac{x}{|\vec{r}|} \), \( \frac{y}{|\vec{r}|} \), and \( \frac{z}{|\vec{r}|} \). We first need to find the sum of the given vectors.
Step 2: Key Formula or Approach:
1. Calculate the resultant vector \( \vec{s} = \vec{a} + \vec{b} \).
2. Find the magnitude of the resultant vector, \( |\vec{s}| \).
3. The direction cosines (l, m, n) are the components of the unit vector in the direction of \( \vec{s} \).
Step 3: Detailed Explanation:
Given vectors are: \[ \vec{a} = 3\hat{i} + 4\hat{j} - 3\hat{k} \] \[ \vec{b} = -2\hat{i} - 3\hat{j} + \hat{k} \] First, find the sum \( \vec{s} = \vec{a} + \vec{b} \): \[ \vec{s} = (3 - 2)\hat{i} + (4 - 3)\hat{j} + (-3 + 1)\hat{k} \] \[ \vec{s} = 1\hat{i} + 1\hat{j} - 2\hat{k} \] Next, find the magnitude of \( \vec{s} \): \[ |\vec{s}| = \sqrt{(1)^2 + (1)^2 + (-2)^2} \] \[ |\vec{s}| = \sqrt{1 + 1 + 4} = \sqrt{6} \] Now, calculate the direction cosines: \[ l = \frac{x}{|\vec{s}|} = \frac{1}{\sqrt{6}} \] \[ m = \frac{y}{|\vec{s}|} = \frac{1}{\sqrt{6}} \] \[ n = \frac{z}{|\vec{s}|} = \frac{-2}{\sqrt{6}} \] Step 4: Final Answer:
The direction-cosines of the sum of the vectors are \( \left(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}}\right) \).