Question

A vector of magnitude 6 and perpendicular to \( \vec{a} = 2i + 2j + k \) and \( \vec{b} = i - 2j + 2k \), is:

• A. \( \pm (2\vec i - \vec j - 2\vec k) \)
• B. \( \pm 2(2\vec i - \vec j + 2\vec k) \)
• C. \( \pm (2\vec i - \vec j + 2\vec k) \)
• D. \( \pm 2(2\vec i + \vec j - 2\vec k) \)
• E. \( \pm 2(2\vec i - \vec j - 2\vec k) \)

Hint:

To find a perpendicular vector, use the cross product and normalize it to the desired magnitude.

Answer 1

Answer: (E)

We are given two vectors \( \vec{a} = 2\vec{i} + 2\vec{j} + \vec{k} \) and \( \vec{b} = \vec{i} - 2\vec{j} + 2\vec{k} \). We need to find a vector that is perpendicular to both \( \vec{a} \) and \( \vec{b} \), and whose magnitude is 6. 
Step 1: Find the Cross Product of \( \vec{a} \) and \( \vec{b} \)
The vector that is perpendicular to both \( \vec{a} \) and \( \vec{b} \) is given by the cross product \( \vec{a} \times \vec{b} \). 
The formula for the cross product of two vectors \( \vec{a} = a_1\vec{i} + a_2\vec{j} + a_3\vec{k} \) and \( \vec{b} = b_1\vec{i} + b_2\vec{j} + b_3\vec{k} \) is: \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}. \] Substitute the components of \( \vec{a} \) and \( \vec{b} \): \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ 1 & -2 & 2 \end{vmatrix}. \] Now, compute the determinant: \[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} 2 & 1 \\ -2 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 2 \\ 1 & -2 \end{vmatrix}. \] Simplifying each of the 2x2 determinants: \[ = \hat{i} \left( (2)(2) - (1)(-2) \right) - \hat{j} \left( (2)(2) - (1)(1) \right) + \hat{k} \left( (2)(-2) - (2)(1) \right) \] \[ = \hat{i} (4 + 2) - \hat{j} (4 - 1) + \hat{k} (-4 - 2) \] \[ = 6\hat{i} - 3\hat{j} - 6\hat{k}. \] Thus, the cross product is: \[ \vec{a} \times \vec{b} = 6\vec{i} - 3\vec{j} - 6\vec{k}. \] 
Step 2: Find the Magnitude of the Cross Product
The magnitude of the cross product \( |\vec{a} \times \vec{b}| \) is: \[ |\vec{a} \times \vec{b}| = \sqrt{(6)^2 + (-3)^2 + (-6)^2} = \sqrt{36 + 9 + 36} = \sqrt{81} = 9. \] 
Step 3: Scale the Cross Product to Have Magnitude 6
We need a vector that is perpendicular to both \( \vec{a} \) and \( \vec{b} \) and has magnitude 6. The current cross product has magnitude 9, so we scale it by a factor of \( \frac{6}{9} = \frac{2}{3} \). Thus, the required vector is: \[ \frac{2}{3} \times (6\vec{i} - 3\vec{j} - 6\vec{k}) = 4\vec{i} - 2\vec{j} - 4\vec{k}. \] 
Step 4: Final Answer
The vector that is perpendicular to both \( \vec{a} \) and \( \vec{b} \) and has magnitude 6 is \( 2(2\vec{i} - \vec{j} - 2\vec{k}) \). 
Thus, the correct answer is \( \boxed{2(2\vec{i} - \vec{j} - 2\vec{k})} \), corresponding to option (E).