Question

Find the unit vector perpendicular to each of the vectors \( (\vec{a} + \vec{b}) \) and \( (\vec{a} - \vec{b}) \) where \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k} \).

Hint:

Remember that the cross product \( \vec{u} \times \vec{v} \) gives a vector perpendicular to the plane containing \( \vec{u} \) and \( \vec{v} \). To find a unit vector, always divide the resulting vector by its own magnitude. Be careful with the signs when calculating the determinant for the cross product.

Answer 1

Step 1: Understanding the Concept:
A vector perpendicular to two given vectors can be found using their cross product. A unit vector is a vector with a magnitude of 1. The process involves first finding the two vectors \( (\vec{a} + \vec{b}) \) and \( (\vec{a} - \vec{b}) \), then finding their cross product, and finally normalizing the resulting vector to get the unit vector.
Step 2: Key Formula or Approach:
Let \( \vec{u} = \vec{a} + \vec{b} \) and \( \vec{v} = \vec{a} - \vec{b} \).
The vector perpendicular to both \( \vec{u} \) and \( \vec{v} \) is given by their cross product, \( \vec{c} = \vec{u} \times \vec{v} \).
The unit vector in the direction of \( \vec{c} \) is \( \hat{c} = \frac{\vec{c}}{|\vec{c}|} \).
Step 3: Detailed Explanation or Calculation:
First, calculate \( \vec{a} + \vec{b} \) and \( \vec{a} - \vec{b} \).
Given: \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k} \).
\[ \vec{a} + \vec{b} = (\hat{i} + \hat{j} + \hat{k}) + (\hat{i} + 2\hat{j} + 3\hat{k}) = (1+1)\hat{i} + (1+2)\hat{j} + (1+3)\hat{k} = 2\hat{i} + 3\hat{j} + 4\hat{k} \] \[ \vec{a} - \vec{b} = (\hat{i} + \hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = (1-1)\hat{i} + (1-2)\hat{j} + (1-3)\hat{k} = 0\hat{i} - \hat{j} - 2\hat{k} \] Let \( \vec{u} = 2\hat{i} + 3\hat{j} + 4\hat{k} \) and \( \vec{v} = -\hat{j} - 2\hat{k} \).
Now, find the cross product \( \vec{c} = \vec{u} \times \vec{v} \).
\[ \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 3 & 4
0 & -1 & -2 \end{vmatrix} \] \[ \vec{c} = \hat{i}((3)(-2) - (4)(-1)) - \hat{j}((2)(-2) - (4)(0)) + \hat{k}((2)(-1) - (3)(0)) \] \[ \vec{c} = \hat{i}(-6 - (-4)) - \hat{j}(-4 - 0) + \hat{k}(-2 - 0) \] \[ \vec{c} = \hat{i}(-2) - \hat{j}(-4) + \hat{k}(-2) = -2\hat{i} + 4\hat{j} - 2\hat{k} \] Next, find the magnitude of \( \vec{c} \).
\[ |\vec{c}| = \sqrt{(-2)^2 + (4)^2 + (-2)^2} = \sqrt{4 + 16 + 4} = \sqrt{24} = 2\sqrt{6} \] Finally, find the unit vector \( \hat{c} \).
\[ \hat{c} = \frac{\vec{c}}{|\vec{c}|} = \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{2\sqrt{6}} = \frac{-1}{\sqrt{6}}\hat{i} + \frac{2}{\sqrt{6}}\hat{j} - \frac{1}{\sqrt{6}}\hat{k} \] Step 4: Final Answer:
The unit vector perpendicular to both \( (\vec{a} + \vec{b}) \) and \( (\vec{a} - \vec{b}) \) is \( \frac{-1}{\sqrt{6}}\hat{i} + \frac{2}{\sqrt{6}}\hat{j} - \frac{1}{\sqrt{6}}\hat{k} \). Note that the opposite vector \( \frac{1}{\sqrt{6}}\hat{i} - \frac{2}{\sqrt{6}}\hat{j} + \frac{1}{\sqrt{6}}\hat{k} \) is also a valid answer.