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?

Arun
25757 Points
6 years ago
Given |a|= 1, |b|= 1, |c|= 2

Now
a x( a x c) + b = 0
(a.c)a - (a.a) c = -b
(1.2 cos@) - c = - b
2 cos@ a = c -b             (1)
2 cos@ a.a = (c-b)a
2 cos @ = c.a - b.a
2 cos@ = 1.2 cos@ - a.b
Hence
a.b = 0
Again from (1)
2 cos@ a.b = (c-b).b = c.b - b.b
0 = c.b -1
b.c = 1                (2)
Now squaring (1)
4 cos²@ a² = c² + b² - 2b.c
4 cos²@ = 4 +1 - 2
Cos² @ = ¾
Cos @ = $\large \sqrt$3/2
Hence
@ = $\large \pi$/6
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

Given |a|= 1, |b|= 1, |c|= 2
Now
a x( a x c) + b = 0
(a.c)a - (a.a) c = -b
(1.2 cos@) - c = - b
2 cos@ a = c -b (1)
2 cos@ a.a = (c-b)a
2 cos @ = c.a - b.a
2 cos@ = 1.2 cos@ - a.b
Hence
a.b = 0
Again from (1)
2 cos@ a.b = (c-b).b = c.b - b.b
0 = c.b -1
b.c = 1 (2)
Now squaring (1)
4 cos²@ a² = c² + b² - 2b.c
4 cos²@ = 4 +1 - 2
Cos² @ = ¾
Cos @ = √3/2
Hence
@ = π/6

Thanks and Regards