1.

Two vector a and b have magnitudes 6 &8 .find |a-b|.if the angle between vectors is pi/2

Answer»

Answer:

  • The VALUE of | a - b | is 10 units.

Given:

  1. Vector (a) = 6 units.
  2. Vector (b) = 8 units.
  3. Angle (θ) = π / 2

Explanation:

\rule{300}{1.5}

From the formula we know,

⇒ | a - b | = √ a² + b² - 2 a b cosθ

Where,

  • a & b Are vectors.
  • θ Denotes the angle between them.

Now,

⇒ | a - b | = √ a² + b² - 2 a b cosθ

Substituting the values,

⇒ | a - b | = √ (6)² + (8)² - 2 × 6 × 8 × cos(π/2)

In USUAL we take,

⇒ π = 180°

∴ π / 2 = 90°

⇒ | a - b | = √ (6)² + (8)² - 2 × 6 × 8 × cos 90°

⇒ | a - b | = √ (6)² + (8)² - 2 × 6 × 8 × 0

∵ [ cos 90° = 0 ]

⇒ | a - b | = √ (6)² + (8)² - 0

⇒ | a - b | = √ (6)² + (8)²

⇒ | a - b | = √ 36 + 64

⇒ | a - b | = √ 100

⇒ | a - b | = 10

⇒ | a - b | = 10

∴ The value of | a - b | is 10 units.

\rule{300}{1.5}


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