Question

If \( \vec{a}=\hat{i}+2\hat{j}-2\hat{k} \) and \( \vec{b}=2\hat{i}+\hat{j}+2\hat{k} \), then a unit vector perpendicular to \( \vec{a}+\vec{b} \) and \( \vec{a}-\vec{b} \) is

• A. \( \dfrac{2\hat{i}-2\hat{j}+\hat{k}}{3} \)
• B. \( \dfrac{2\hat{i}-2\hat{j}-\hat{k}}{3} \)
• C. \( \dfrac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}} \)
• D. \( \dfrac{\hat{i}+2\hat{j}+2\hat{k}}{3} \)
• E. \( \dfrac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}} \)

Hint:

Whenever a vector is required to be perpendicular to two given vectors, first take their cross product. Then divide by its magnitude to get the unit vector.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A vector that is perpendicular to two given vectors can be found by taking their cross product. In this problem, we first need to find the vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\), and then compute their cross product. The result will be a vector perpendicular to both. Finally, we need to normalize this vector to make it a unit vector.
Step 2: Key Formula or Approach:
1. Calculate \(\vec{v}_1 = \vec{a} + \vec{b}\). 2. Calculate \(\vec{v}_2 = \vec{a} - \vec{b}\). 3. Calculate the perpendicular vector \(\vec{p} = \vec{v}_1 \times \vec{v}_2\). 4. The required unit vector is \(\hat{p} = \frac{\vec{p}}{|\vec{p}|}\).
Step 3: Detailed Explanation:
1. Calculate \(\vec{v}_1 = \vec{a} + \vec{b}\):
\[ \vec{v}_1 = (\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) + (2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) = 3\mathbf{i} + 3\mathbf{j} + 0\mathbf{k} \] 2. Calculate \(\vec{v}_2 = \vec{a} - \vec{b}\):
\[ \vec{v}_2 = (\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}) - (2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) = -\mathbf{i} + \mathbf{j} - 4\mathbf{k} \] 3. Calculate the cross product \(\vec{p} = \vec{v}_1 \times \vec{v}_2\):
\[ \vec{p} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}
3 & 3 & 0
-1 & 1 & -4 \end{vmatrix} \] \[ \vec{p} = \mathbf{i}(3(-4) - 0(1)) - \mathbf{j}(3(-4) - 0(-1)) + \mathbf{k}(3(1) - 3(-1)) \] \[ \vec{p} = \mathbf{i}(-12) - \mathbf{j}(-12) + \mathbf{k}(3+3) \] \[ \vec{p} = -12\mathbf{i} + 12\mathbf{j} + 6\mathbf{k} \] We can simplify this vector by factoring out a common scalar, say -6. A vector in the same or opposite direction is \(2\mathbf{i} - 2\mathbf{j} - \mathbf{k}\). Let's call this \(\vec{q}\). Any unit vector perpendicular to \(\vec{v}_1\) and \(\vec{v}_2\) will be a multiple of \(\hat{q}\).
4. Find the unit vector:
Let's find the magnitude of \(\vec{q} = 2\mathbf{i} - 2\mathbf{j} - \mathbf{k}\).
\[ |\vec{q}| = \sqrt{2^2 + (-2)^2 + (-1)^2} = \sqrt{4+4+1} = \sqrt{9} = 3 \] The unit vector is \(\hat{q} = \frac{\vec{q}}{|\vec{q}|}\).
\[ \hat{q} = \frac{2\mathbf{i} - 2\mathbf{j} - \mathbf{k}}{3} \] This matches option (B). Note that \(-\hat{q}\) would also be a correct answer, but it's not among the options.
Step 4: Final Answer:
The unit vector perpendicular to \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\) is \(\frac{2\mathbf{i}-2\mathbf{j}-\mathbf{k}}{3}\). This corresponds to option (B).