Question 50

Mathematics Scalar and Vector Products Hard

If \( \vec{a}, \vec{b}, \vec{c} \) and \( \vec{d} \) are the unit vectors such that \( (\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})=1 \) and \( (\vec{a} \cdot \vec{c})=\frac{1}{2} \) , then

(A) only \( \vec{a}, \vec{b}, \vec{c} \) are non-coplanar
(B) only \( \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{d}} \) are non-coplanar
(C) Both \( \vec{a}, \vec{b}, \vec{c} \) and \( \vec{a}, \vec{b}, \vec{d} \) are non-coplanar
(D) Both \( \vec{a}, \vec{b}, \vec{c} \) and \( \vec{a}, \vec{b}, \vec{d} \) are coplanar
✓ Correct Answer: Option D Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

Step 1: Analyze the Given Conditions 1. Cross Product Condition: \[ (\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1 \] This implies that the vectors \(\vec{a} \times \vec{b}\) and \(\vec{c} \times \vec{d}\) are parallel and their dot product is 1. 2. Dot Product Condition: \[ \vec{a} \cdot \vec{c} = \frac{1}{2} \] This indicates that the angle between \(\vec{a}\) and \(\vec{c}\) is \(60^\circ\) since \(\cos^{-1}\left(\frac{1}{2}\right) = 60^\circ\). \[Step 2:\] Determine Coplanarity - Coplanar Vectors: Vectors are coplanar if their scalar triple product is zero, i.e., \(\vec{a} \cdot (\vec{b} \times \vec{c}) = 0\). Given the conditions, we need to determine if \(\vec{a}, \vec{b}, \vec{c}\) and \(\vec{a}, \vec{b}, \vec{d}\) are coplanar or non-coplanar. \[Step 3:\] Evaluate the Options - Option A: Only \(\vec{a}, \vec{b}, \vec{c}\) are non-coplanar. - Option B: Only \(\vec{a}, \vec{b}, \vec{d}\) are non-coplanar. - Option C: Both \(\vec{a}, \vec{b}, \vec{c}\) and \(\vec{a}, \vec{b}, \vec{d}\) are non-coplanar. - Option D: Both \(\vec{a}, \vec{b}, \vec{c}\) and \(\vec{a}, \vec{b}, \vec{d}\) are coplanar. Given the cross product condition \((\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1\), it suggests that \(\vec{a} \times \vec{b}\) and \(\vec{c} \times \vec{d}\) are parallel, which implies that \(\vec{a}, \vec{b}, \vec{c}, \vec{d}\) are coplanar. \[Step 4:\] Conclusion Based on the given conditions, the most accurate conclusion is that both \(\vec{a}, \vec{b}, \vec{c}\) and \(\vec{a}, \vec{b}, \vec{d}\) are coplanar. Correct Answer: \[ \boxed{D} \]

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