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Vector of magnitude 5 units and in the direction opposite to

2i + 3j − 6k  is  ____________


Transcript

Question 15 (OR 2nd Question) Vector of magnitude 5 units and in the direction opposite to 2𝑖 ̂ + 3𝑗 ̂ − 6𝑘 ̂ is ____________ Let 𝑎 ⃗ = 2𝑖 ̂ + 3𝑗 ̂ – 6𝑘 ̂ Magnitude of 𝑎 ⃗ = √(22+32+(−6)2) |𝑎 ⃗ | = √(4+9+36) = √49 = 7 Unit vector opposite to direction of 𝑎 ⃗ = –1 × 1/|𝑎 ⃗ | . 𝑎 ⃗ = (−1)/7 (2𝑖 ̂ + 3𝑗 ̂ – 6𝑘 ̂) Thus, Vector with magnitude 1 opposite to 𝑎 ⃗ = (−1)/7 (2𝑖 ̂ + 3𝑗 ̂ – 6𝑘 ̂) Vector with magnitude 5 opposite to 𝑎 ⃗ = (−5)/7 (2𝑖 ̂ + 3𝑗 ̂ – 6𝑘 ̂) = 5/7 (–2𝑖 ̂ – 3𝑗 ̂ + 6𝑘 ̂) Hence, the required vector is 𝟓/𝟕 (–2𝒊 ̂ – 3𝒋 ̂ + 6𝒌 ̂)

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo