Vector product - Defination

Chapter 10 Class 12 Vector Algebra
Concept wise

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Example 23 Find a unit vector perpendicular to each of the vectors π β + π β and π β β π β where π β = π Μ + π Μ + π Μ, b = π Μ + 2 π Μ + 3π Μ . Finding (π β + π β) and (π β β π β) (π β + π β) = (1 + 1) π Μ + (1 + 2) π Μ + (1 + 3) π Μ = 2π Μ + 3π Μ + 4π Μ (π β β π β) = (1 β 1) π Μ + (1 β 2) π Μ + (1 β 3) π Μ = 0π Μ β 1π Μ β 2π 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) π β = |β 8(π Μ&π Μ&π Μ@2&3&4@0&β1&β2)| = π Μ [(3Γβ2)β(β1Γ4)] βπ Μ [(2Γβ2)β(0Γ4)] + π Μ [(2Γβ1)β(0Γ3)] = π Μ [β6β(β4)] βπ Μ [β4β0] + π Μ [β2β0] = π Μ (β6 + 4) βπ Μ (β4) + π Μ(β2) = β2π Μ + 4π Μ β 2π Μ Since we need to find unit vector perpendicular Unit vector of π β = π/(π΄ππππππππ πππ β ) Γ π β = 1/β((β2)2 + (4)^2 + (β2)2) Γ (β2π Μ + 4π Μ β 2π Μ) = 1/β(4 + 16 + 4) Γ (β2π Μ + 4π Μ β 2π Μ) = 1/(2β6) Γ (β2π Μ + 4π Μ β 2π Μ) = (βπ)/βπ π Μ + π/βπ π Μ β π/βπ π Μ Note: There are always two perpendicular vectors So, another vector would be = β((β1)/β6 π Μ" + " 2/β6 π Μ" β " 1/β6 π Μ ) = π/βπ π Μ" β" π/βπ π Μ" + " π/βπ π Μ Hence, Perpendicular vectors are (β1)/β6 π Μ + 2/β6 π Μ β 1/β6 π Μ & 1/β6 π Μ" β" 2/β6 π Μ" + " 1/β6 π Μ