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: Write the given vectors in component form.
We are given \[ \vec{a}=\hat{i}+2\hat{j}-2\hat{k} \] and \[ \vec{b}=2\hat{i}+\hat{j}+2\hat{k} \] We need a unit vector perpendicular to both \[ \vec{a}+\vec{b} \quad \text{and} \quad \vec{a}-\vec{b} \] A vector perpendicular to both is obtained using the cross product: \[ (\vec{a}+\vec{b}) \times (\vec{a}-\vec{b}) \]

Step 2: Find \( \vec{a}+\vec{b} \) and \( \vec{a}-\vec{b} \).
First, \[ \vec{a}+\vec{b} = (\hat{i}+2\hat{j}-2\hat{k})+(2\hat{i}+\hat{j}+2\hat{k}) \] \[ =3\hat{i}+3\hat{j} \] So, \[ \vec{a}+\vec{b}=3\hat{i}+3\hat{j}+0\hat{k} \] Now, \[ \vec{a}-\vec{b} = (\hat{i}+2\hat{j}-2\hat{k})-(2\hat{i}+\hat{j}+2\hat{k}) \] \[ =-\hat{i}+\hat{j}-4\hat{k} \] Thus, \[ \vec{a}-\vec{b}=-\hat{i}+\hat{j}-4\hat{k} \]

Step 3: Compute the cross product.
Now, \[ (\vec{a}+\vec{b}) \times (\vec{a}-\vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} 3 & 3 & 0 -1 & 1 & -4 \end{vmatrix} \] Expanding along the first row, \[ = \hat{i} \begin{vmatrix} 3 & 0 1 & -4 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 0 -1 & -4 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 3 -1 & 1 \end{vmatrix} \] \[ = \hat{i}(3\cdot(-4)-0\cdot 1) -\hat{j}(3\cdot(-4)-0\cdot(-1)) +\hat{k}(3\cdot1-3\cdot(-1)) \] \[ = \hat{i}(-12)-\hat{j}(-12)+\hat{k}(3+3) \] \[ =-12\hat{i}+12\hat{j}+6\hat{k} \]

Step 4: Simplify the direction vector.
The vector \[ -12\hat{i}+12\hat{j}+6\hat{k} \] can be simplified by dividing by \( -6 \): \[ 2\hat{i}-2\hat{j}-\hat{k} \] So a vector perpendicular to both required vectors is \[ 2\hat{i}-2\hat{j}-\hat{k} \]

Step 5: Find its magnitude.
The magnitude of \[ 2\hat{i}-2\hat{j}-\hat{k} \] is \[ \sqrt{2^2+(-2)^2+(-1)^2} = \sqrt{4+4+1} = \sqrt{9} = 3 \]

Step 6: Write the unit vector.
Therefore, the required unit vector is \[ \frac{2\hat{i}-2\hat{j}-\hat{k}}{3} \] This has magnitude \( 1 \) and is perpendicular to both \( \vec{a}+\vec{b} \) and \( \vec{a}-\vec{b} \).

Step 7: Final conclusion.
Hence, the required unit vector is \[ \boxed{\frac{2\hat{i}-2\hat{j}-\hat{k}}{3}} \] Therefore, the correct option is \[ \boxed{(2)\ \dfrac{2\hat{i}-2\hat{j}-\hat{k}}{3}} \]