Vector Algebra
Vector triple product
Grade None
Question:
<p>Given \(\frac{1}{6}\hat{i} - \frac{1}{3}\hat{j} + \frac{1}{k}\hat{k} = (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d})\) where \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are coplanar. Find \(\vec{c}\).</p>
<p>A) \(\dfrac{\hat{i} - 2\hat{j} + 2\hat{k}}{3}\)</p>
<p>B) \(-\dfrac{\hat{i} - 2\hat{j} + 2\hat{k}}{3}\)</p>
<p>C) \(\dfrac{\hat{i} + 2\hat{j} - 2\hat{k}}{3}\)</p>
<p>D) \(-\dfrac{\hat{i} + 2\hat{j} - 2\hat{k}}{3}\)</p>
Step-by-Step Solution
Key Concept: When vectors are coplanar, one can be expressed as a linear combination of the others. Use the vector triple product formula and the coplanarity condition to reduce the given cross product expression to a manageable form.
Step 1: Since vectors a , b , and c are coplanar, we have c = α a + β b for some scalars α, β. Step 2: This means a × b is perpendicular to c , so c × d is perpendicular to a × b . Step 3: Apply vector triple product: ( a × b ) × ( c × d ) = [( a × b ) · d ] c - [( a × b ) · c ] d Step 4: Since a , b , c are coplanar: ( a × b ) · c = 0 (scalar triple product is zero) Step 5: Therefore: ( a × b ) × ( c × d ) = [( a × b ) · d ] c Step 6: Given expression = (1/6) î - (1/3) ĵ + (1/k) k̂ must equal k· c for some scalar k. Thus c is parallel to the given vector. Step 7: Solving for k using magnitude or component matching with standard basis vectors, we find the specific form of c . ∴ Answer: A
Correct Answer: A