for any vector a , prove that |a X i|2+|a X j|2+|a X k | = 2|a|2 Share with your friends Share 5 Manbar Singh answered this Let a→ = a1i^ + a2j^ + a3k^Now, a→ × i^ = a1i^ + a2j^ + a3k^×i^⇒a→ × i^ = a1 i^ × i^ + a2 j^ × i^ + a3 k^ × i^ = 0 - a2k^ + a3j^⇒a→ × i^ 2 = a22 + a32 a→ × j^ = a1i^ + a2j^ + a3k^×j^⇒a→ × j^ = a1 i^ × j^ + a2 j^ × j^ + a3 k^ ×j^ = a1k^ - a3i^⇒a→ × j^2 = a12 + a32 a→ × k^ = a1i^ + a2j^ + a3k^× k^⇒a→ × k^ = a1 i^ × k^ + a2 j^ × k^ + a3 k^ × k^ = -a1j^ + a2i^⇒a→ × k^2 = a12 + a22Now,a→ × i^ 2 + a→ × j^2 + a→ × k^2=a22 + a32 + a12 + a32 + a12 + a22=2 a12 + a22 + a32 =2a→2 16 View Full Answer Ansh Jain answered this I HAVE USED CURLY BRACKETS FOR MAGNITUDElet a=xi +yj +zkaXi=(xi+yj+zk)Xi= 0-yk+jzaXj=(xi+yj+zk)Xj= xk+0-izaXk=(xi+yj+zk)Xk= -xj+yi+0LHS=magnitude of (aXi)^2+ (aXj)^2+(aXk)^2=2(y^2 +z^2+x^2)=RHS 4 Ansh Jain answered this I HAVE USED CURLY BRACKETS FOR MAGNITUDElet a=xi +yj +zkaXi=(xi+yj+zk)Xi= 0-yk+jzaXj=(xi+yj+zk)Xj= xk+0-izaXk=(xi+yj+zk)Xk= -xj+yi+0LHS=magnitude of (aXi)^2+ (aXj)^2+(aXk)^2=2(y^2 +z^2+x^2)=RHS 8
