Question:
let a= 2i-j k , b=i 2j-k. and c=i j-2k be three vectors . prove that a vector in the plane of b and c whose projection on a is of length root2/root3 is 2i+3j-3k
Answer:
A vector in the plane of b and c is given by
d = αb + βc
=> d = α(i + 2j - k) + β(i + j - 2k)
=> d = i(α + β) + j(2α + β) + k(-α - 2β) ...................1
Given length of the projection of d on a = √(2/3)
=> (d . a)/|a| = √(2/3)
=> [{i(α + β) + j(2α + β) + k(-α - 2β)}.(2i - j + k)]/√{22 + (-1)2 + 12 } = √(2/3)
=> {2(α + β) - (2α + β) + (-α - 2β)}/√{4 + 1 + 1} = √(2/3)
=> {2α + 2β - 2α - β -α - 2β}/{√2 * √3} = √(2/3)
=> {2α + 2β - 2α - β -α - 2β}/√2 = √2
=> - α - β = √2 * √2
=> - α - β = 2
=> α + β = -2
This satisfies when α = -1 and β = -1
From eqaution 1, we get,
d = i(-1 - 1) + j(-2 - 1) + k(1 + 2)
=> d = -2i - 3j + 3k
=> d = -(2i + 3j - 3k)
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