Vector product - Defination

Chapter 10 Class 12 Vector Algebra
Concept wise

This video is only available for Teachoo black users

Introducing your new favourite teacher - Teachoo Black, at only βΉ83 per month

### Transcript

Ex 10.4, 2 Find a unit vector perpendicular to each of the vector π β + π β and π β β π β, where π β = 3π Μ + 2π Μ + 2π Μ and π β = π Μ + 2π Μ β 2π Μ .π β = 3π Μ + 2π Μ + 2π Μ π β = 1π Μ + 2π Μ β 2π Μ (π β + π β) = (3 + 1) π Μ + (2 + 2) π Μ + (2 β 2) π Μ = 4π Μ + 4π Μ + 0π Μ (π β β π β) = (3 β 1) π Μ + (2 β 2) π Μ + (2 β (β2)) π Μ = 2π Μ + 0π Μ + 4π Μ Now, we need to find a vector perpendicular to both π β + π β and π β β π β, We know that (π β Γ π β) is perpendicular to π β and π β Replacing π β by (π β + π β) & π β by (π β β π β) (π β + π β) Γ (π β β π β) will be perpendicular to (π β + π β) and (π β β π β) Let π β = (π β + π β) Γ (π β β π β) β΄ π β = |β 8(π Μ&π Μ&π Μ@β(4@2)&β(4@0)&β(0@4))| = π Μ [(4Γ4)β(0Γ0)] β π Μ [(4Γ4)β(2Γ0)] + π Μ [(4Γ0)β(2Γ4)] = π Μ (16 β 0) β π Μ (16 β 0) + π Μ (0 β 8) = 16 π Μ β 16π Μ β 8π Μ β΄ π β = 16 π Μ β 16π Μ β 8π Μ Now, Unit vector of π β = 1/(ππππππ‘π’ππ πππ β ) Γ π β Magnitude of π β = β(162+(β16)2+(β8)2) |π β | = β(256+256+64) = β576 = 24 Unit vector of π β = 1/|π β | Γ π β = 1/24 Γ ["16" π Μ" β 16" π Μ" β 8" π Μ ] = π/π π Μ β π/π π Μ β π/π π Μ . Therefore the required unit vector is 2/3 π Μ β 2/3 π Μ β 1/3 π Μ . Note: There are always two perpendicular vectors So, another vector would be = β(π/π " " π Μ" β " π/π " " π Μ" β " π/π " " π Μ ) = (βπ)/π π Μ + π/π π Μ + π/π π Μ Hence, the perpendicular vectors are 2/3 " " π Μ" β " 2/3 " " π Μ" β " 1/3 " " π Μ & (β2)/3 π Μ + 2/3 π Μ + 1/3 π Μ