Question 22
If vector p = i + j + k and q = i – 2j + k, find a vector of magnitude 5√3 units perpendicular to the vector q and coplanar with vectors p and q
Last updated at Oct. 1, 2019 by Teachoo
Question 22
If vector p = i + j + k and q = i – 2j + k, find a vector of magnitude 5√3 units perpendicular to the vector q and coplanar with vectors p and q
Transcript
Question 22 If ๐ โ = ๐ ฬ + ๐ ฬ + ๐ ฬ and ๐ โ = ๐ ฬ โ 2๐ ฬ + ๐ ฬ, find a vector of magnitude 5โ3 units perpendicular to the vector ๐ โ and coplanar with vectors ๐ โ and ๐ โ Let ๐ โ be the required vector ๐ โ = a๐ ฬ + b๐ ฬ + c๐ ฬ Now, given that ๐ โ & ๐ โ are perpendicular Therefore, ๐ โ.๐ โ = 0 (a๐ ฬ + b๐ ฬ + c๐ ฬ) . (๐ ฬ โ 2๐ ฬ + ๐ ฬ) = 0 a ร 1 + b ร (โ2) + c ร 1 = 0 a โ 2b + c = 0 Also, given that ๐ โ is coplanar with vectors ๐ โ and ๐ โ Therefore, [๐ โ" " ๐ โ" " ๐ โ ] = 0 Finding [๐ โ" " ๐ โ" " ๐ โ ] [๐ โ" " ๐ โ" " ๐ โ ] = |โ 8(1&1&1@1&โ2&1@๐&๐&๐)| = 1[โ2๐โ๐ ] โ 1[๐โ๐] + 1[๐+2๐] = โ2๐โ๐โ๐+๐+๐+2๐ = โ3๐+3๐ Since [๐ โ" " ๐ โ" " ๐ โ ] = 0 โ3๐+3๐=0 3๐=3๐ ๐=๐ Now, from (1) a โ 2b + c = 0 Putting c = a a โ 2b + a = 0 2a โ 2b = 0 2a = 2b a = b Therefore, a = b = c So, our vector ๐ โ becomes ๐ โ = a๐ ฬ + b๐ ฬ + c๐ ฬ = a๐ ฬ + a๐ ฬ + a๐ ฬ Also, given that magnitude of ๐ โ is 5โ3 units |๐ โ | = 5โ3 โ(๐^2+๐^2+๐^2 ) = 5โ3 โ(ใ3๐ใ^2 ) = 5โ3 โ3 ๐ = 5โ3 ๐ = 5 So, our vector ๐ โ becomes ๐ โ = a๐ ฬ + a๐ ฬ + a๐ ฬ = 5๐ ฬ + 5๐ ฬ + 5๐ ฬ
CBSE Class 12 Sample Paper for 2019 Boards
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CBSE Class 12 Sample Paper for 2019 Boards
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