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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 23 Find a unit vector perpendicular to each of the vectors 𝑎 ⃗ + 𝑏 ⃗ and 𝑎 ⃗ − 𝑏 ⃗ where 𝑎 ⃗ = 𝑖 ̂ + 𝑗 ̂ + 𝑘 ̂, b = 𝑖 ̂ + 2 𝑗 ̂ + 3𝑘 ̂ . Finding (𝒂 ⃗ + 𝒃 ⃗) and (𝒂 ⃗ − 𝒃 ⃗) (𝒂 ⃗ + 𝒃 ⃗) = (1 + 1) 𝑖 ̂ + (1 + 2) 𝑗 ̂ + (1 + 3) 𝑘 ̂ = 2𝒊 ̂ + 3𝒋 ̂ + 4𝒌 ̂ (𝒂 ⃗ − 𝒃 ⃗) = (1 − 1) 𝑖 ̂ + (1 − 2) 𝑗 ̂ + (1 − 3) 𝑘 ̂ = 0𝒊 ̂ − 1𝒋 ̂ − 2𝒌 ̂ Now, we need to find a vector perpendicular to both 𝑎 ⃗ + 𝑏 ⃗ and 𝑎 ⃗ − 𝑏 ⃗, We know that (𝑎 ⃗ × 𝑏 ⃗) is perpendicular to 𝑎 ⃗ and 𝑏 ⃗ Replacing 𝑎 ⃗ by (𝑎 ⃗ + 𝑏 ⃗) & 𝑏 ⃗ by (𝑎 ⃗ − 𝑏 ⃗) (𝒂 ⃗ + 𝒃 ⃗) × (𝒂 ⃗ − 𝒃 ⃗) will be perpendicular to (𝑎 ⃗ + 𝑏 ⃗) and (𝑎 ⃗ − 𝑏 ⃗) Let 𝑐 ⃗ = (𝑎 ⃗ + 𝑏 ⃗) × (𝑎 ⃗ − 𝑏 ⃗) 𝒄 ⃗ = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@2&3&4@0&−1&−2)| = 𝑖 ̂ [(3×−2)−(−1×4)] −𝑗 ̂ [(2×−2)−(0×4)] + 𝑘 ̂ [(2×−1)−(0×3)] = 𝑖 ̂ [−6−(−4)] −𝑗 ̂ [−4−0] + 𝑘 ̂ [−2−0] = 𝑖 ̂ (−6 + 4) −𝑗 ̂ (−4) + 𝑘 ̂(−2) = −2𝒊 ̂ + 4𝒋 ̂ − 2𝒌 ̂ Since we need to find unit vector perpendicular Unit vector of 𝑐 ⃗ = 𝟏/(𝑴𝒂𝒈𝒏𝒊𝒕𝒖𝒅𝒆 𝒐𝒇𝒄 ⃗ ) × 𝒄 ⃗ = 1/√((−2)2 + (4)^2 + (−2)2) × (−2𝑖 ̂ + 4𝑗 ̂ − 2𝑘 ̂) = 1/√(4 + 16 + 4) × (−2𝑖 ̂ + 4𝑗 ̂ − 2𝑘 ̂) = 1/(2√6) × (−2𝑖 ̂ + 4𝑗 ̂ − 2𝑘 ̂) = (−𝟏)/√𝟔 𝒊 ̂ + 𝟐/√𝟔 𝒋 ̂ − 𝟏/√𝟔 𝒌 ̂ Note: There are always two perpendicular vectors So, another vector would be = −((−1)/√6 𝑖 ̂" + " 2/√6 𝑗 ̂" − " 1/√6 𝑘 ̂ ) = 𝟏/√𝟔 𝒊 ̂" −" 𝟐/√𝟔 𝒋 ̂" + " 𝟏/√𝟔 𝒌 ̂ Hence, Perpendicular vectors are (−1)/√6 𝑖 ̂ + 2/√6 𝑗 ̂ − 1/√6 𝑘 ̂ & 1/√6 𝑖 ̂" −" 2/√6 𝑗 ̂" + " 1/√6 𝑘 ̂