If -a-1-b-2 and the angle between a and b is 120- then a-3 b -3 a-b2-

The correct option is D 300{ (a+3b)× (3a b) }^2={ 3a × a a × b+9b+a 3b × b }^2 ={ 10(b × a) }^2=10^2 |b|^2|a|^2 sin^2(a,b)=100(2)^2(1)^2 sin^2(120^∘)=400 ( √( ...

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Byju's Answer

Standard IX

Mathematics

Cross Product of Two Vectors

If |a|=1,|b|=...

Question

If |a|=1, |b|=2 and the angle between a and b is $120^{\circ}$ , then ${((a + 3 b) \times (3 a - b))}^{2}$ =

425

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375

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325

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300

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Solution

The correct option is D 300
${(a + 3 b) \times (3 a - b)}^{2} = {3 a \times a - a \times b + 9 b + a - 3 b \times b}^{2}$
$= {10 (b \times a)}^{2} = 10^{2} | b |^{2} | a |^{2} s i n^{2} (a, b) = 100 (2)^{2} (1)^{2} s i n^{2} (120^{\circ}) = 400 {(\frac{\sqrt{3}}{2})}^{2} = 300$

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If |a|=1, |b|=2 and the angle between a and b is 120∘, then ((a+3b)×(3a−b))2 =

The correct option is D 300{(a+3b)×(3a−b)}2={3a×a−a×b+9b+a−3b×b}2 ={10(b×a)}2=102|b|2|a|2sin2(a,b)=100(2)2(1)2sin2(120∘)=400(√32)2=300

If |a|=1, |b|=2 and the angle between a and b is $120^{\circ}$ , then ${((a + 3 b) \times (3 a - b))}^{2}$ =

The correct option is D 300
${(a + 3 b) \times (3 a - b)}^{2} = {3 a \times a - a \times b + 9 b + a - 3 b \times b}^{2}$
$= {10 (b \times a)}^{2} = 10^{2} | b |^{2} | a |^{2} s i n^{2} (a, b) = 100 (2)^{2} (1)^{2} s i n^{2} (120^{\circ}) = 400 {(\frac{\sqrt{3}}{2})}^{2} = 300$