Example 29 - Chapter 10 Class 12 Vector Algebra

Transcript

Example 29 (Method 1) Three vectors 𝑎 ⃗, 𝑏 ⃗ and 𝑐 ⃗ satisfy the condition 𝑎 ⃗ + 𝑏 ⃗ + 𝑐 ⃗ = 0 ⃗ . Evaluate the quantity μ = 𝑎 ⃗ ⋅𝑏 ⃗ + 𝑏 ⃗ ⋅ 𝑐 ⃗ + 𝑐 ⃗ ⋅ 𝑎 ⃗, if |𝑎 ⃗|=1, |𝑏 ⃗|= 4 and |c ⃗|= 2.Given |𝑎 ⃗|=1, |𝑏 ⃗|= 4 and |c ⃗|= 2 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 ⃗ + 𝑎 ⃗ . 𝑐 ⃗ + 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 + 42 + 22 + 2(𝑎 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ + 𝑐 ⃗. 𝑎 ⃗) = 1 + 16 + 4 + 2(𝑎 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ + 𝑐 ⃗. 𝑎 ⃗) = 21 + 2 (𝒂 ⃗. 𝒃 ⃗ + 𝒃 ⃗. 𝒄 ⃗ + 𝒄 ⃗. 𝒂 ⃗) So, |𝑎 ⃗+𝑏 ⃗+𝑐 ⃗ |2 = 21 + 2 (𝑎 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ + 𝑐 ⃗. 𝑎 ⃗) Now, given that |𝒂 ⃗" + " 𝒃 ⃗" + " 𝒄 ⃗ | = 0 |𝑎 ⃗" + " 𝑏 ⃗" + " 𝑐 ⃗ |2 = 0 21 + 2 (𝒂 ⃗. 𝒃 ⃗ + 𝒃 ⃗. 𝒄 ⃗ + 𝒄 ⃗. 𝒂 ⃗) = 0 2(𝑎 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ + 𝑐 ⃗. 𝑎 ⃗) = −21 (𝑎 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ + 𝑐 ⃗. 𝑎 ⃗) = (−21)/2 Therefore, 𝝁 = 𝒂 ⃗. 𝒃 ⃗ + 𝒃 ⃗. 𝒄 ⃗ + 𝒄 ⃗ . 𝒂 ⃗ = (−𝟐𝟏)/𝟐 Example 29 (Method 2) Three vectors 𝑎 ⃗, 𝑏 ⃗ and 𝑐 ⃗ satisfy the condition 𝑎 ⃗ + 𝑏 ⃗ + 𝑐 ⃗ = 0 ⃗ . Evaluate the quantity μ = 𝑎 ⃗ ⋅𝑏 ⃗ + 𝑏 ⃗ ⋅ 𝑐 ⃗ + 𝑐 ⃗ ⋅ 𝑎 ⃗, if |𝑎 ⃗|=1, |𝑏 ⃗|= 4 and |c ⃗|= 2.Given |𝑎 ⃗| = 1, |𝑏 ⃗|= 4 and |c ⃗|= 2 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 ⃗ (Using prop : 𝑎 ⃗ . 𝑎 ⃗ = |𝑎 ⃗ |2 ) 0 ⃗ = |𝑎 ⃗ |2 + 𝑎 ⃗. 𝑏 ⃗ + 𝒄 ⃗. 𝒂 ⃗ 0 = 12 + 𝑎 ⃗. 𝑏 ⃗ + 𝑐 ⃗. 𝑎 ⃗ 0 = 1 + 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 ⃗ |2 + 𝑏 ⃗. 𝑐 ⃗ 0 = 𝑎 ⃗. 𝑏 ⃗ + 42 + 𝑏 ⃗ . 𝑐 ⃗ 0 = 𝑎 ⃗. 𝑏 ⃗ + 16 + 𝑏 ⃗ . 𝑐 ⃗ 𝑎 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ = −16 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 ⃗ |2 0 = 𝑐 ⃗. 𝑎 ⃗ + 𝑏 ⃗ . 𝑐 ⃗ + 22 0 = 𝑐 ⃗. 𝑎 ⃗ + 𝑏 ⃗ . 𝑐 ⃗ + 4 𝑐 ⃗. 𝑎 ⃗ + 𝑏 ⃗. 𝑐 ⃗ = −4 Adding (1), (2) and (3), (𝒂 ⃗. 𝒃 ⃗ + 𝒄 ⃗. 𝒂 ⃗) + (𝒂 ⃗. 𝒃 ⃗ + 𝒃 ⃗. 𝒄 ⃗) + (𝒄 ⃗. 𝒂 ⃗ + 𝒃 ⃗. 𝒄 ⃗) = −1 + (–16) + (–4) 2𝑎 ⃗. 𝑏 ⃗ + 2𝑐 ⃗. 𝑎 ⃗ + 2𝑏 ⃗. 𝑐 ⃗ = −21 2(𝑎 ⃗. 𝑏 ⃗ + 𝑏. 𝑐 ⃗ + 𝑐 ⃗. 𝑎 ⃗) = −21 𝑎 ⃗. 𝑏 ⃗ + 𝑏 ⃗. 𝑐 ⃗ + 𝑐 ⃗. 𝑎 ⃗ = (−21)/2 Therefore, 𝝁 = 𝒂 ⃗. 𝒃 ⃗ + 𝒃 ⃗. 𝒄 ⃗ + 𝒄 ⃗ . 𝒂 ⃗ = (−𝟐𝟏)/𝟐