Question

Let a→=a1i^+a2j^+a3k^, b→=b1i^+b2j^+b3k^ be three non-zero vectors such that c→ is a vector perpendicular to both a→ and b→. If the angle between a→ and b→ is π6 then |a1b1c1a2b2c2a3b3c3|=?

• A. (A) 0
• B. (B) 1
• C. (C) 14(a12+a22+a32)(b12+b22+b32)
• D. (D) 34(a12+a22+a32)(b12+b22+b32)(c12+c22+c32)

Answer 1

Answer: (C)

(C) 14(a12+a22+a32)(b12+b22+b32)

Answer 2

Explanation:
Given:a→=a1i^+a2j^+a3k^, b→=b1i^+b2j^+b3k^ be three non-zero vectors such that c→ is a vector perpendicular to both a→ and b→.Angle between a→ and b→ = π6 We know that |a1b1c1a2b2c2a3b3c3|=[(a→×b→)⋅c→]2So,=[|a→×b→||c→|cos⁡0∘]2(∵a→×b→ is parellel to vector c→ as c→ is perpendicular to both a→ and b→)=(|a→||b→|sin⁡π6)2=|a→|2|b→|2(12)2=14|a→|2|b→|2We know that,|a→|2=(a12+a22+a32)|b→|2=(b12+b22+b32)So,14|a→|2|b→|2=14[(a12+a22+a32)(b12+b22+b32)]Hence, the correct option is (C).