Vector product - Solving

Chapter 10 Class 12 Vector Algebra
Concept wise

### Transcript

Misc 19 If ΞΈ is the angle between any two vectors π β and π β, then |π β" β" π β | = |π β" Γ " π β | when ΞΈ is eq |π β" β" π β | π β" β" π β = |π β ||π β | cosβ‘"ΞΈ" |π β" β" π β | = ||π β ||π β |cosβ‘"ΞΈ" | = |π β ||π β ||πππβ‘"ΞΈ" | |π β" " Γ" " π β | π β Γ π β = |π β ||π β | π ππβ‘"ΞΈ" π Μ |π β Γ π β | = ||π β ||π β |β‘γsin" ΞΈ " π Μ γ | = |π β ||π β ||sin" ΞΈ|" |π Μ | = |π β ||π β ||π¬π’π§" ΞΈ|" As π Μ is unit vector, |π Μ | = 1 Given, |π β" β" π β | = |π βΓπ β | Putting values |π β ||π β | |πππ β‘"ΞΈ" | = |π β ||π β | |π ππβ‘"ΞΈ" | |πππβ‘"ΞΈ" | = |πππβ‘"ΞΈ" | This is only possible for ΞΈ = π/π Therefore, the angle between the vectors π β and π β is π/4,= So, option (B) is correct. ual to (A) 0 (B) π/4 (C) π/2 (D) Ο