The points A,B and C with position vectors 3i-yj+2k,5i-j+k and 3xi+3j-k are collinear. Find the values of x and y and also the ratio in which the point B divides AC.
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The points A,B and C with position vectors 3i-yj+2k,5i-j+k and 3xi+3j-k are collinear. Find the values of x and y and also the ratio in which the point B divides AC.
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EXPLANATION.
Points A B C are the position vectors,
\sf \implies (A) = 3 \hat {i} - y \hat {j} + 2 \hat{k}.⟹(A)=3i^−yj^+2k^.
\sf \implies (B) = 5 \hat {i} - \hat {j} + \hat {k}.⟹(B)=5i^−j^+k^.
\sf \implies ( C) =3x \hat {i} + 3 \hat{j} - \hat {k}.⟹(C)=3xi^+3j^−k^.
\sf \implies \vec {AB} = (5 \hat {i} - \hat {j} + \hat {k} ) - [ 3 \hat {i} - y \hat {j} + 2 \hat {k}] .⟹AB=(5i^−j^+k^)−[3i^−yj^+2k^].
\sf \implies \vec {AB} = 5 \hat {i} - \hat {j} + \hat {k} - 3 \hat {i} + y \hat {j} - 2 \hat {k}⟹AB=5i^−j^+k^−3i^+yj^−2k^
\sf \implies \vec {AB} = 2 \hat {i} - (1 - y ) \hat {j} - 2 \hat {k} .⟹AB=2i^−(1−y)j^−2k^.
\sf \implies \vec {BC} = 3X \hat {i} + 3 \hat {j} - \hat {k} - [ 5 \hat {i} - \hat {j} + \hat {k} ] .⟹BC=3Xi^+3j^−k^−[5i^−j^+k^].
\sf \implies \vec {BC} = 3x \hat {i} + 3 \hat {j} - \hat {j} - 5 \hat {i} + \hat {j} - \hat {k} .⟹BC=3xi^+3j^−j^−5i^+j^−k^.
\sf \implies \vec {BC} = (3x - 5) \hat {i} + 4 \hat {j} - 2 \hat {k} .⟹BC=(3x−5)i^+4j^−2k^.
\sf \implies as \ we\ know \ that = \vec {r} = \vec {a} + \lambda \vec {b} .⟹as we know that=r=a+λb.
\sf \implies 2 \hat {i} - (1 - y) \hat {j} - 2 \hat {k} =\lambda [(3x - 5) \hat{i} + 4 \hat {j} - 2 \hat {k} ] .⟹2i^−(1−y)j^−2k^=λ[(3x−5)i^+4j^−2k^].
\sf \implies 2 = \lambda(3x - 5). =(1).⟹2=λ(3x−5).=(1).
\sf \implies -(1 - y) = 4 \lambda =(2).⟹−(1−y)=4λ=(2).
\sf \implies -2 = -2 \lambda = (3)⟹−2=−2λ=(3)
From equation, (3) we get.
⇒ 2 = 2λ.
⇒ λ = 1.
Put the value of λ in equation, (1) and (2) we get.
⇒ 2 = 1(3x - 5).
⇒ 2 = 3x - 5.
⇒ 8 = 3x.
⇒ x = 8/3.
⇒ -(1 - y) = 4.
⇒ - 1 + y = 4.
⇒ y = 5.
Value of x = 8/3 y = 5 λ = 1
As the point divides in the ratio of k or 1, we get,
Co-ordinates are,
A = (3,-1,2).
B = (5,-1,1).
C = (8,3,-1).
\sf \implies \vec {OA} = 3 \hat {i} - y \hat {j} + 2\hat {k}
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