Question

Let \( \vec{a} = 2\vec{i} + \vec{j} - 2\vec{k} \) and \( \vec{b} = \vec{i} + \vec{j} \) be two vectors. If \( \vec{c} \) is a vector such that \( \vec{a} \cdot \vec{c} = |\vec{c}| \), \( |\vec{c} - \vec{a}| = 2\sqrt{2} \) and the angle between \( \vec{a} \times \vec{b} \) and \( \vec{c} \) is \( 30^\circ \), then \( |(\vec{a} \times \vec{b}) \times \vec{c}| = \)

• A. \( \frac{2}{3} \)
• B. \( \frac{3}{2} \)
• C. \( 2 \)
• D. \( 3 \) Correct Answer

Hint:

Use vector properties: \( |\vec{X}-\vec{Y}|^2 = (\vec{X}-\vec{Y})\cdot(\vec{X}-\vec{Y}) = |\vec{X}|^2 - 2\vec{X}\cdot\vec{Y} + |\vec{Y}|^2 \). Magnitude of cross product: \( |\vec{U} \times \vec{V}| = |\vec{U}| |\vec{V}| \sin\phi \), where \( \phi \) is the angle between \( \vec{U} \) and \( \vec{V} \). Calculate magnitudes and cross products carefully.

Answer 1

The Correct Option is B

Step 1: Calculate \( |\vec{a}| \).
\( \vec{a} = 2\vec{i} + \vec{j} - 2\vec{k} \) \( |\vec{a}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4+1+4} = \sqrt{9} = 3 \).

Step 2: Use the condition \( |\vec{c} - \vec{a}| = 2\sqrt{2} \).
Square both sides: \( |\vec{c} - \vec{a}|^2 = (2\sqrt{2})^2 = 8 \).
\( (\vec{c} - \vec{a}) \cdot (\vec{c} - \vec{a}) = 8 \) \( |\vec{c}|^2 - 2(\vec{a} \cdot \vec{c}) + |\vec{a}|^2 = 8 \).
Substitute \( \vec{a} \cdot \vec{c} = |\vec{c}| \) and \( |\vec{a}| = 3 \): \( |\vec{c}|^2 - 2|\vec{c}| + 3^2 = 8 \) \( |\vec{c}|^2 - 2|\vec{c}| + 9 = 8 \) \( |\vec{c}|^2 - 2|\vec{c}| + 1 = 0 \) This is \( (|\vec{c}| - 1)^2 = 0 \).
So, \( |\vec{c}| - 1 = 0 \implies |\vec{c}| = 1 \).

Step 3: Calculate \( \vec{a} \times \vec{b} \).
\( \vec{a} = (2, 1, -2) \), \( \vec{b} = (1, 1, 0) \) \[ \vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = \vec{i}((1)(0) - (-2)(1)) - \vec{j}((2)(0) - (-2)(1)) + \vec{k}((2)(1) - (1)(1)) \] \[ = \vec{i}(0+2) - \vec{j}(0+2) + \vec{k}(2-1) = 2\vec{i} - 2\vec{j} + \vec{k} \]
Step 4: Calculate \( |\vec{a} \times \vec{b}| \).
\( |\vec{a} \times \vec{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4+4+1} = \sqrt{9} = 3 \).

Step 5: Calculate \( |(\vec{a} \times \vec{b}) \times \vec{c}| \).
Let \( \vec{v} = \vec{a} \times \vec{b} \).
We need to find \( |\vec{v} \times \vec{c}| \).
Using the definition of the magnitude of a cross product: \( |\vec{v} \times \vec{c}| = |\vec{v}| |\vec{c}| \sin \theta \), where \( \theta \) is the angle between \( \vec{v} \) and \( \vec{c} \).
We are given that the angle between \( \vec{a} \times \vec{b} \) (which is \( \vec{v} \)) and \( \vec{c} \) is \( 30^\circ \).
So \( \theta = 30^\circ \).
We have \( |\vec{v}| = |\vec{a} \times \vec{b}| = 3 \) and \( |\vec{c}| = 1 \).
\( \sin 30^\circ = \frac{1}{2} \).
\[ |(\vec{a} \times \vec{b}) \times \vec{c}| = (3)(1)\left(\frac{1}{2}\right) = \frac{3}{2} \] This matches option (2).