Question

10. If lā| = 11,|b| = 23, la-b| = 30, then la+b) is : (A) 10 (B) 20 (C) 30(D) 40
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Answer 1

$option : (B) \: 40$

Explanation:

$Given, \: \: |a| = 11, \: |b| = 23 \: \: \: and \: \: \: |a - b| = 30 \\ we \: \: know \: \: \boxed{C ^{2} = {A}^{2} + {B}^{2} - 2AB \cos \theta} \\ Then, \: \: \: \: { |a - b| }^{2} = { |a| }^{2} + { |b| }^{2} + 2 |a| |b| \cos \theta \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {30}^{2} = {11}^{2} + {23}^{2} + 2 \times 11 \times 23 \times \cos \theta \\ 900 = 121 + 529 + 506 \cos \theta \\ 506 \cos \theta = 900 - 121 - 529 \\ 506 \cos \theta = 250 \\ \boxed{\cos \theta = \frac{ 250}{506} } \\ Now, \: \: \: { |a + b| }^{2} = { |a| }^{2} + { |b| }^{2} - 2 |a| |b| \cos \theta \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: { |a + b| }^{2} = {11}^{2} + {23}^{2} - 2 \times 11 \times 23 \times \frac{ -250}{506} \\{ |a + b| }^{2} = 121 + 529 - 250 \\ { |a + b| }^{2} = 400 \\ \: \: \: \: \: |a + b| = \sqrt{400} \\ \boxed{ |a + b| = 20}$