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Question 22
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If vector p = i + j + k and q = i β 2j + k, find a vector of magnitude 5β3 units perpendicular to the vector q and coplanar with vectors p and q

CBSE Class 12 Sample Paper for 2019 Boards

Paper Summary

Question 1 Important

Question 2

Question 3

Question 4 (Or 1st) Important

Question 4 (Or 2nd)

Question 5

Question 6

Question 7 Important

Question 8 (Or 1st) Important

Question 8 (Or 2nd)

Question 9

Question 10 (Or 1st) Important

Question 10 (Or 2nd)

Question 11 Important

Question 12 (Or 1st)

Question 12 (Or 2nd)

Question 13 (Or 1st) Important

Question 13 (Or 2nd)

Question 14 Important

Question 15

Question 16 (Or 1st)

Question 16 (Or 2nd) Important

Question 17

Question 18

Question 19 Important

Question 20 Important

Question 21 (Or 1st)

Question 21 (Or 2nd) Important

Question 22 You are here

Question 23 Important

Question 24 (Or 1st)

Question 24 (Or 2nd) Important

Question 25

Question 26 (Or 1st) Important

Question 26 (Or 2nd)

Question 27 (Or 1st) Important

Question 27 (Or 2nd) Important

Question 28

Question 29 Important

Class 12

Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Oct. 1, 2019 by Teachoo

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Question 22
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If vector p = i + j + k and q = i β 2j + k, find a vector of magnitude 5β3 units perpendicular to the vector q and coplanar with vectors p and q

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