Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Scalar product - Solving
Last updated at May 29, 2023 by Teachoo
Ex 10.3, 13 (Method 1) If 𝑎 ⃗ ,𝑏 ⃗, 𝑐 ⃗ are unit vectors such that 𝑎 ⃗ + 𝑏 ⃗ + 𝑐 ⃗ = 0 ⃗, find the value of 𝑎 ⃗ .𝑏 ⃗+𝑏 ⃗ . 𝑐 ⃗ + 𝑐 ⃗. 𝑎 ⃗ . Given 𝑎 ⃗ ,𝑏 ⃗, 𝑐 ⃗ are unit vectors Magnitude of 𝑎 ⃗ ,𝑏 ⃗, 𝑐 ⃗ is 1 So, |𝒂 ⃗ | = |𝒃 ⃗ | = |𝒄 ⃗ | = 1 Also, 𝑎 ⃗ + 𝑏 ⃗ + 𝑐 ⃗ = 0 ⃗ So, |𝒂 ⃗" + " 𝒃 ⃗" + " 𝒄 ⃗ | = |𝟎 ⃗ | = 0 Now, |𝒂 ⃗+𝒃 ⃗+𝒄 ⃗ |2 = (𝒂 ⃗ + 𝒃 ⃗ + 𝒄 ⃗) . (𝒂 ⃗ + 𝒃 ⃗ + 𝒄 ⃗) = 𝑎 ⃗. 𝑎 ⃗ + 𝑎 ⃗ . 𝑏 ⃗ + 𝒂 ⃗ . 𝒄 ⃗ + 𝒃 ⃗ . 𝒂 ⃗ + 𝑏 ⃗ . 𝑏 ⃗ + 𝑏 ⃗ . 𝑐 ⃗ + 𝑐 ⃗ . 𝑎 ⃗ + 𝒄 ⃗ . 𝒃 ⃗ + 𝑐 ⃗ . 𝑐 ⃗ = 𝑎 ⃗. 𝑎 ⃗ + 𝑎 ⃗ . 𝑏 ⃗ + 𝒄 ⃗ . 𝒂 ⃗ + 𝒂 ⃗ . 𝒃 ⃗ + 𝑏 ⃗ . 𝑏 ⃗ + 𝑏 ⃗ . 𝑐 ⃗ + 𝑐 ⃗ . 𝑎 ⃗ + 𝒃 ⃗ . 𝒄 ⃗ + 𝑐 ⃗ . 𝑐 ⃗ = 𝑎 ⃗ . 𝑎 ⃗ + 𝑏 ⃗ . 𝑏 ⃗ + 𝑐 ⃗ . 𝑐 ⃗ + 2𝑎 ⃗. 𝑏 ⃗ + 2𝑏 ⃗. 𝑐 ⃗ + 2𝑐 ⃗. 𝑎 ⃗ = 𝒂 ⃗ . 𝒂 ⃗ + 𝒃 ⃗ . 𝒃 ⃗ + 𝒄 ⃗ . 𝒄 ⃗ + 2(𝑎 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ + 𝑐 ⃗. 𝑎 ⃗) = |𝒂 ⃗ |𝟐 + |𝒃 ⃗ |𝟐 + |𝒄 ⃗ |𝟐 + 2 (𝑎 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ + 𝑐 ⃗ . 𝑎 ⃗) = 12 + 12 + 12 + 2(𝑎 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ + 𝑐 ⃗. 𝑎 ⃗) = 1 + 1 + 1 + 2(𝑎 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ + 𝑐 ⃗. 𝑎 ⃗) = 3 + 2 (𝑎 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ + 𝑐 ⃗. 𝑎 ⃗) ∴ |𝑎 ⃗+𝑏 ⃗+𝑐 ⃗ |2 = 3 + 2 (𝑎 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ + 𝑐 ⃗. 𝑎 ⃗) Now, |𝑎 ⃗" + " 𝑏 ⃗" + " 𝑐 ⃗ | = 0 |𝑎 ⃗" + " 𝑏 ⃗" + " 𝑐 ⃗ |^2 = 0 3 + 2 (𝑎 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ + 𝑐 ⃗. 𝑎 ⃗) = 0 2(𝑎 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ + 𝑐 ⃗. 𝑎 ⃗) = −3 (𝒂 ⃗. 𝒃 ⃗ + 𝒃 ⃗. 𝒄 ⃗ + 𝒄 ⃗. 𝒂 ⃗) = (−𝟑)/𝟐 Ex 10.3, 13 (Method 2) If 𝑎 ⃗ ,𝑏 ⃗, 𝑐 ⃗ are unit vectors such that 𝑎 ⃗ + 𝑏 ⃗ + 𝑐 ⃗ = 0, find the value of 𝑎 ⃗ .𝑏 ⃗+𝑏 ⃗ . 𝑐 ⃗ + 𝑐 ⃗. 𝑎 ⃗ . Given 𝑎 ⃗ ,𝑏 ⃗, 𝑐 ⃗ are unit vectors So, |𝒂 ⃗ | = |𝒃 ⃗ | = |𝒄 ⃗ | = 1 Also, ( 𝑎 ⃗ + 𝑏 ⃗ + 𝑐 ⃗ ) = 0 ⃗ Now, 𝒂 ⃗ . (𝒂 ⃗ + 𝒃 ⃗ + 𝒄 ⃗) = 𝑎 ⃗ . 𝑎 ⃗ + 𝑎 ⃗. 𝑏 ⃗ + 𝑎 ⃗ . 𝑐 ⃗ 𝑎 ⃗ . 0 ⃗ = 𝑎 ⃗. 𝑎 ⃗ + 𝑎 ⃗. 𝑏 ⃗ + 𝑎 ⃗. 𝑐 ⃗ 0 = 𝒂 ⃗. 𝒂 ⃗ + 𝑎 ⃗. 𝑏 ⃗ + 𝑎 ⃗. 𝑐 ⃗ 0 ⃗ = |𝒂 ⃗ |𝟐 + 𝑎 ⃗. 𝑏 ⃗ + 𝑎 ⃗. 𝑐 ⃗ 0 ⃗ = |𝒂 ⃗ |𝟐 + 𝑎 ⃗. 𝑏 ⃗ + 𝑎 ⃗. 𝑐 ⃗ 0 ⃗ = |𝑎 ⃗ |2 + 𝑎 ⃗. 𝑏 ⃗ + 𝒄 ⃗. 𝒂 ⃗ 0 = 12 + 𝑎 ⃗. 𝑏 ⃗ + 𝑐 ⃗. 𝑎 ⃗ 𝑎 ⃗. 𝑏 ⃗ + 𝑐 ⃗. 𝑎 ⃗ = −1 Also, 𝒃 ⃗ . (𝒂 ⃗ + 𝒃 ⃗ + 𝒄 ⃗) = 𝑏 ⃗ . 𝑎 ⃗ + 𝑏 ⃗. 𝑏 ⃗ + 𝑏 ⃗ . 𝑐 ⃗ 𝑏 ⃗ . 0 ⃗ = 𝑏 ⃗. 𝑎 ⃗ + 𝑏 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ 0 = 𝒃 ⃗. 𝒂 ⃗ + 𝑏 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ 0 = 𝒂 ⃗. 𝒃 ⃗ + 𝑏 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ 0 = 𝑎 ⃗. 𝑏 ⃗ + 𝒃 ⃗. 𝒃 ⃗ + 𝑏 ⃗. 𝑐 ⃗ 0 = 𝑎 ⃗. 𝑏 ⃗ + |𝒃 ⃗ |2 + 𝑏 ⃗ . 𝑐 ⃗ 0 = 𝑎 ⃗. 𝑏 ⃗ + 12 + 𝑏 ⃗ . 𝑐 ⃗ 𝑎 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ = −1 Also 𝒄 ⃗ . (𝒂 ⃗ + 𝒃 ⃗ + 𝒄 ⃗) = 𝑐 ⃗ . 𝑎 ⃗ + 𝑐 ⃗ . 𝑏 ⃗ + 𝑐 ⃗ . 𝑐 ⃗ 𝑐 ⃗. 0 ⃗ = 𝑐 ⃗. 𝑎 ⃗ + 𝑐 ⃗. 𝑏 ⃗ + 𝑐 ⃗. 𝑐 ⃗ 0 = 𝑐 ⃗. 𝑎 ⃗ + 𝒄 ⃗. 𝒃 ⃗ + 𝑐 ⃗. 𝑐 ⃗ 0 = 𝑐 ⃗. 𝑎 ⃗ + 𝒃 ⃗. 𝒄 ⃗ + 𝑐 ⃗. 𝑐 ⃗ 0 = 𝑐 ⃗. 𝑎 ⃗ + 𝑏 ⃗. 𝑐 ⃗ + 𝒄 ⃗. 𝒄 ⃗ 0 = 𝑐 ⃗. 𝑎 ⃗ + 𝑏 ⃗. 𝑐 ⃗ + |𝑐 ⃗ |2 0 = 𝑐 ⃗. 𝑎 ⃗+ 𝑏 ⃗ . 𝑐 ⃗ + 12 𝑐 ⃗. 𝑎 ⃗ + 𝑏 ⃗. 𝑐 ⃗ = −1 Adding (1), (2) and (3) (𝑎 ⃗. 𝑏 ⃗ + 𝑐 ⃗. 𝑎 ⃗) + (𝑎 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗) + (𝑐 ⃗. 𝑎 ⃗ + 𝑏 ⃗. 𝑐 ⃗) = −1 + (–1) + (–1) 2𝑎 ⃗. 𝑏 ⃗ + 2𝑐 ⃗. 𝑎 ⃗ + 2𝑏 ⃗. 𝑐 ⃗ = −3 2(𝑎 ⃗. 𝑏 ⃗ + 𝑏. 𝑐 ⃗ + 𝑐 ⃗. 𝑎 ⃗) = −3 𝒂 ⃗. 𝒃 ⃗ + 𝒃 ⃗. 𝒄 ⃗ + 𝒄 ⃗. 𝒂 ⃗ = (−𝟑)/𝟐