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Let a, b and c be three vectors with magnitudes 1, 1 and 2 respectively. If ax(axc) + b = 0, then the acute angle between a and c is?
Given |a|= 1, |b|= 1, |c|= 2 Nowa x( a x c) + b = 0(a.c)a - (a.a) c = -b(1.2 cos@) - c = - b2 cos@ a = c -b (1)2 cos@ a.a = (c-b)a2 cos @ = c.a - b.a2 cos@ = 1.2 cos@ - a.b Hencea.b = 0Again from (1)2 cos@ a.b = (c-b).b = c.b - b.b0 = c.b -1b.c = 1 (2)Now squaring (1)4 cos²@ a² = c² + b² - 2b.c4 cos²@ = 4 +1 - 2Cos² @ = ¾Cos @ = 3/2Hence @ = /6
Dear Student,Please find below the solution to your problem.Given |a|= 1, |b|= 1, |c|= 2Nowa x( a x c) + b = 0(a.c)a - (a.a) c = -b(1.2 cos@) - c = - b2 cos@ a = c -b (1)2 cos@ a.a = (c-b)a2 cos @ = c.a - b.a2 cos@ = 1.2 cos@ - a.bHencea.b = 0Again from (1)2 cos@ a.b = (c-b).b = c.b - b.b0 = c.b -1b.c = 1 (2)Now squaring (1)4 cos²@ a² = c² + b² - 2b.c4 cos²@ = 4 +1 - 2Cos² @ = ¾Cos @ = √3/2Hence@ = π/6Thanks and Regards
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