
Ex 11.2
Ex 11.2, 2
Ex 11.2, 3 Important
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Ex 11.2, 5 Important
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Ex 11.2, 7 Important You are here
Ex 11.2, 8
Ex 11.2, 9 Important
Ex 11.2, 10 (i) Important Deleted for CBSE Board 2022 Exams
Ex 11.2, 10 (ii)
Ex 11.2, 11 (i) Important Deleted for CBSE Board 2022 Exams
Ex 11.2, 11 (ii)
Ex 11.2, 12 Important
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Ex 11.2, 14 Important
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Last updated at Feb. 1, 2020 by Teachoo
Ex 11.2, 7 The Cartesian equation of a line is (𝑥 − 5)/3 = (𝑦 + 4)/7 = (𝑧 − 6)/2. Write its vector form.Cartesian equation : (𝑥 − 5)/3 = (𝑦 + 4)/7 = (𝑧 − 6)/2 (𝑥 − 5)/3 = (𝑦 − (−4))/7 = (𝑧 − 6)/2 Equation of a line in Cartesian form is given by (𝑥 − 𝑥1)/𝑎 = (𝑦 − 𝑦1)/𝑏 = (𝑧 − 𝑧1)/𝑐 Comparing (1) and (2), 𝑥1 = 5, 𝑦1 = −4, 𝑧1 = 6 𝑎 = 3, b = 7, c = 2 Cartesian equation : (𝑥 − 5)/3 = (𝑦 + 4)/7 = (𝑧 − 6)/2 (𝑥 − 5)/3 = (𝑦 − (−4))/7 = (𝑧 − 6)/2 Equation of a line in Cartesian form is given by (𝑥 − 𝑥1)/𝑎 = (𝑦 − 𝑦1)/𝑏 = (𝑧 − 𝑧1)/𝑐 Comparing (1) and (2), 𝑥1 = 5, 𝑦1 = −4, 𝑧1 = 6 𝑎 = 3, b = 7, c = 2 Equation of line in vector form is 𝑟 ⃗ = 𝑎 ⃗ + 𝜆𝑏 ⃗ Where 𝑎 ⃗ = 𝑥1𝑖 ̂ + y1𝑗 ̂ + z1𝑘 ̂ = 5𝑖 ̂ − 4𝑗 ̂ + 6𝑘 ̂ & 𝑏 ⃗ = 𝑎𝑖 ̂ + b𝑗 ̂ + c𝑘 ̂ = 3𝑖 ̂ + 7𝑗 ̂ + 2𝑘 ̂ Now, 𝑟 ⃗ = (5𝒊 ̂ − 4𝒋 ̂ + 6𝒌 ̂) + 𝜆 (3𝒊 ̂ + 7𝒋 ̂ + 2𝒌 ̂)