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 π β = π Μ + π Μ + π Μ and π β = π Μ β 2π Μ + π Μ, find a vector of magnitude 5β3 units perpendicular to the vector π β and coplanar with vectors π β and π β Let π β be the required vector π β = aπ Μ + bπ Μ + cπ Μ Now, given that π β & π β are perpendicular Therefore, π β.π β = 0 (aπ Μ + bπ Μ + cπ Μ) . (π Μ β 2π Μ + π Μ) = 0 a Γ 1 + b Γ (β2) + c Γ 1 = 0 a β 2b + c = 0 Also, given that π β is coplanar with vectors π β and π β Therefore, [π β" " π β" " π β ] = 0 Finding [π β" " π β" " π β ] [π β" " π β" " π β ] = |β 8(1&1&1@1&β2&1@π&π&π)| = 1[β2πβπ ] β 1[πβπ] + 1[π+2π] = β2πβπβπ+π+π+2π = β3π+3π Since [π β" " π β" " π β ] = 0 β3π+3π=0 3π=3π π=π Now, from (1) a β 2b + c = 0 Putting c = a a β 2b + a = 0 2a β 2b = 0 2a = 2b a = b Therefore, a = b = c So, our vector π β becomes π β = aπ Μ + bπ Μ + cπ Μ = aπ Μ + aπ Μ + aπ Μ Also, given that magnitude of π β is 5β3 units |π β | = 5β3 β(π^2+π^2+π^2 ) = 5β3 β(γ3πγ^2 ) = 5β3 β3 π = 5β3 π = 5 So, our vector π β becomes π β = aπ Μ + aπ Μ + aπ Μ = 5π Μ + 5π Μ + 5π Μ
CBSE Class 12 Sample Paper for 2019 Boards
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CBSE Class 12 Sample Paper for 2019 Boards
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