Question

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Answer 1

★ Question:

→ If a, b and c be 3 vectors such that a + b + c = 0 and |a| = 3, |b| = 5, |c| = 7 find the angle between a and b.

★ Solution:

It is given that:

a + b + c = 0

This can also be written as:

a + b = -c

Squaring on both sides,

(a + b)² = (-c)²

→ a² + b² + 2ab = (-1)² × c²

→ a² + b² + 2ab = c²

These a, b and c are vectors and so let's consider the angle between them as θ. Now the above equation becomes:

a² + b² + 2ab cos θ = c²

Substitute the given values.

→ 3² + 5² + 2(3)(5) cos θ = 7²

→ 9 + 25 + 30 cos θ = 49

→ 34 + 30 cos θ = 49

→ 30 cos θ = 49 - 34

→ 30 cos θ = 15

→ cos θ = 15 ÷ 30

→ cos θ = ½

→ cos θ = cos 60°

→ θ = 60°

∴ The angle between a vector and b vector is 60°

Short trick:

When the question is asked about the angle between a and b vector, then consider;

tan θ = b ÷ a

So you get:

tan θ = 5/3

⇒ tan θ = tan 60°

⇒ θ = 60°

Note that this method can be used only for verification purposes only.

Answer 2

$\pink{Solution}$

It is given that:

$\sf \: a + b + c = 0$

This can also be written as:

$\sf \: a + b = -c$

Squaring on both sides,

$\sf \: (a + b)² = (-c)²$

$\sf \: a² + b² + 2ab = (-1)² × c²$

$\sf \: a² + b² + 2ab = c²$

These a, b and c are vectors and so let's consider the angle between them as θ. Now the above equation becomes

$\sf \: a² + b² + 2ab cos θ = c²$

Substitute the given values.

$\sf \: 3² + 5² + 2(3)(5) cos θ = 7²$

$\sf \: 9 + 25 + 30 cos θ = 49$

$\sf \: 34 + 30 cos θ = 49$

$\sf \: 30 cos θ = 49 - 34$

$\sf \: 30 cos θ = 15$

$\sf \: cos θ = 15 ÷ 30$

$\sf \: cos θ = ½</p><p>cos θ = cos 60° \\ θ = 60°$

∴ Vector and b vector is 60°