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Example 26 (Supplementary NCERT) Deleted for CBSE Board 2025 Exams

Example 27 (Supplementary NCERT) Deleted for CBSE Board 2025 Exams

Example 28 (Supplementary NCERT) Important Deleted for CBSE Board 2025 Exams

Example 29 (Supplementary NCERT) Important Deleted for CBSE Board 2025 Exams

Example 30 (Supplementary NCERT) Important Deleted for CBSE Board 2025 Exams

Example 31 (Supplementary NCERT) Important Deleted for CBSE Board 2025 Exams

Last updated at April 16, 2024 by Teachoo

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