Question

The vector \( \vec{x} \) is perpendicular to the vectors \( \vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k} \), \( \vec{b} = 18\hat{i} - 22\hat{j} - 5\hat{k} \) and makes an obtuse angle with \( \hat{j} \). If \( |\vec{x}|=14 \), then \( \vec{x} = \)

• A. \( 8\hat{i} + 12\hat{j} + 24\hat{k} \)
• B. \( -8\hat{i} + 6\hat{j} + 24\hat{k} \)
• C. \( 8\hat{i} - 12\hat{j} - 24\hat{k} \)
• D. \( -8\hat{i} - 12\hat{j} + 24\hat{k} \)

Hint:

A vector \(\vec{x}\) perpendicular to \(\vec{a}\) and \(\vec{b}\) is parallel to \(\vec{a} \times \vec{b}\). So \(\vec{x} = k(\vec{a} \times \vec{b})\).
Use \(|\vec{x}|\) to find \(|k|\).
The condition "obtuse angle with \(\hat{j}\)" means \(\vec{x} \cdot \hat{j}<0\), so the y-component of \(\vec{x}\) must be negative. This determines the sign of \(k\).

Answer 1

The Correct Option is D

Concept: If vector \( \vec{x} \) is perpendicular to both \( \vec{a} \) and \( \vec{b} \), then it must be parallel to the cross product: \[ \vec{x} \parallel \vec{a} \times \vec{b} \]

Given:

\[ \vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}, \quad \vec{b} = 18\hat{i} - 22\hat{j} - 5\hat{k} \]

Compute \( \vec{n} = \vec{a} \times \vec{b} \):

\[ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 2 \\ 18 & -22 & -5 \end{vmatrix} = \hat{i}(2 \cdot (-5) - 2 \cdot (-22)) - \hat{j}(3 \cdot (-5) - 2 \cdot 18) + \hat{k}(3 \cdot (-22) - 2 \cdot 18) \] \[ = \hat{i}(-10 + 44) - \hat{j}(-15 - 36) + \hat{k}(-66 - 36) = 34\hat{i} + 51\hat{j} - 102\hat{k} \]

Simplify:

All components are divisible by 17: \[ \vec{n}' = \frac{1}{17} \vec{n} = 2\hat{i} + 3\hat{j} - 6\hat{k} \] So, \[ \vec{x} = k(2\hat{i} + 3\hat{j} - 6\hat{k}) \]

Given: \( |\vec{x}| = 14 \). Calculate magnitude of \( \vec{n}' \):

\[ |\vec{n}'| = \sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \Rightarrow |\vec{x}| = |k| \cdot 7 = 14 \Rightarrow |k| = 2 \] So, \[ \vec{x} = \pm 2(2\hat{i} + 3\hat{j} - 6\hat{k}) = \begin{cases} 4\hat{i} + 6\hat{j} - 12\hat{k} & \text{if } k = 2 \\ -4\hat{i} - 6\hat{j} + 12\hat{k} & \text{if } k = -2 \end{cases} \]

Condition: \( \vec{x} \) makes an obtuse angle with \( \hat{j} \Rightarrow \vec{x} \cdot \hat{j} < 0 \)

Thus, the correct choice is: \[ \vec{x} = -4\hat{i} - 6\hat{j} + 12\hat{k} \]

Check if this matches any option:

Option (d): \( \vec{x}_d = -8\hat{i} - 12\hat{j} + 24\hat{k} = -2 \times (-4\hat{i} - 6\hat{j} + 12\hat{k}) \)

So, it is the same vector but scaled by 2:

\[ |\vec{x}_d| = 2 \times 14 = 28 \]

Check orthogonality:

\[ \vec{x}_d \cdot \vec{a} = (-8)(3) + (-12)(2) + (24)(2) = -24 -24 + 48 = 0 \] \[ \vec{x}_d \cdot \vec{b} = (-8)(18) + (-12)(-22) + (24)(-5) = -144 + 264 -120 = 0 \] So, \( \vec{x}_d \) is indeed perpendicular to both \( \vec{a} \) and \( \vec{b} \), and its y-component is negative → obtuse with \( \hat{j} \).

Conclusion:

The correct direction and properties match option (d), even though the magnitude is doubled. This implies either:

• The magnitude in the question should have been \( \boxed{28} \)
• Or the options should have been scaled appropriately to magnitude 14

✅ Final Answer: \[ \boxed{-8\hat{i} - 12\hat{j} + 24\hat{k}} \] (Assuming either a typo in question's magnitude or uniform scaling in options.)