Get live Maths 1-on-1 Classs - Class 6 to 12
Supplementary examples and questions from CBSE
Supplementary Example 2 Important Deleted for CBSE Board 2023 Exams
Supplementary Example 3 Deleted for CBSE Board 2023 Exams
Supplementary Example 4 Deleted for CBSE Board 2023 Exams
Supplementary Example 5 Important Deleted for CBSE Board 2023 Exams
Supplementary Example 6 Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q1 Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q2 Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q3 Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q4 Important Deleted for CBSE Board 2023 Exams You are here
Supplementary Exercise Q5 Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q6 Important Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q7 Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q8 Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q9 Important Deleted for CBSE Board 2023 Exams
Supplementary Exercise Q10 Deleted for CBSE Board 2023 Exams
Supplementary examples and questions from CBSE
Last updated at March 30, 2023 by Teachoo
Supplementary Exercise Q4 Show that (i) the vectors 𝑎 ⃗ = 2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂, 𝑏 ⃗ = 𝑖 ̂ + 2𝑗 ̂ − 3𝑘 ̂, and 𝑐 ⃗ = 3𝑖 ̂ − 4𝑗 ̂ + 5𝑘 ̂ are coplanar. Three vectors 𝑎 ⃗, 𝑏 ⃗, 𝑐 ⃗ are coplanar if [ 𝒂 ⃗, 𝒃 ⃗, 𝒄 ⃗ ] = 0 Given, 𝑎 ⃗ = 2𝑖 ̂ − 𝑗 ̂ + 𝑘 ̂ 𝑏 ⃗ = 𝑖 ̂ + 2𝑗 ̂ − 3𝑘 ̂ 𝑐 ⃗ = 3𝑖 ̂ − 4𝑗 ̂ + 5𝑘 ̂ [𝑎 ⃗,𝑏,𝑐 ⃗ ] = |■8(2&−1&[email protected]&2&−[email protected]&−4&5)| = 2[(2×5)−(−4×−3)] − (−1) [(1×5)−(3×−3) ] + 1[(1×−4)−(3×2) ] = 2 [10−12]+[5+9] + [−4−6] = –4 + 14 – 10 = 0 ∴ [𝒂 ⃗,𝒃,𝒄 ⃗ ] = 0 Therefore, 𝑎 ⃗,𝑏 and 𝑐 ⃗ are coplanar. Supplementary Exercise Q4 Show that (ii) the vectors 𝑎 ⃗ = 𝑖 ̂ − 2𝑗 ̂ + 3𝑘 ̂, 𝑏 ⃗ = −2𝑖 ̂ + 3𝑗 ̂ − 4𝑘 ̂, and 𝑐 ⃗ = 𝑖 ̂ − 3𝑗 ̂ + 5𝑘 ̂ are coplanar Three vectors 𝑎 ⃗, 𝑏 ⃗, 𝑐 ⃗ are coplanar if [ 𝒂 ⃗, 𝒃 ⃗, 𝒄 ⃗ ] = 0 Given, 𝑎 ⃗ = 𝑖 ̂ − 2𝑗 ̂ + 3𝑘 ̂ 𝑏 ⃗ = −2𝑖 ̂ + 3𝑗 ̂ − 4𝑘 ̂ 𝑐 ⃗ = 𝑖 ̂ − 3𝑗 ̂ + 5𝑘 ̂ [𝑎 ⃗,𝑏,𝑐 ⃗ ] = |■8(1&−2&[email protected]−2&3&−[email protected]&−3&5)| = 1[(3×5)−(−3×4) ] − (−2) [(−2×5)−(1×−4) ] + 3[(−2×−3)−(1×3) ] = 1 [15−12]+[2(−10−(−4)] + 3 [6−3] = 1(3) + 2 (−6) + 3 (3) = 3 − 12 + 9 = 12 − 12 = 0 ∴ [𝒂 ⃗,𝒃,𝒄 ⃗ ] = 0 Therefore, 𝑎 ⃗,𝑏 and 𝑐 ⃗ are coplanar.