Question

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.​

Answer 1

Step-by-step explanation:

EXPLANATION.

Points A B C are the position vectors,

$\sf \implies (A) = 3 \hat {i} - y \hat {j} + 2 \hat{k}.$

$\sf \implies (B) = 5 \hat {i} - \hat {j} + \hat {k}.$

$\sf \implies ( C) =3x \hat {i} + 3 \hat{j} - \hat {k}.$

$\sf \implies \vec {AB} = (5 \hat {i} - \hat {j} + \hat {k} ) - [ 3 \hat {i} - y \hat {j} + 2 \hat {k}] .$

$\sf \implies \vec {AB} = 5 \hat {i} - \hat {j} + \hat {k} - 3 \hat {i} + y \hat {j} - 2 \hat {k}$

$\sf \implies \vec {AB} = 2 \hat {i} - (1 - y ) \hat {j} - 2 \hat {k} .$

$\sf \implies \vec {BC} = 3X \hat {i} + 3 \hat {j} - \hat {k} - [ 5 \hat {i} - \hat {j} + \hat {k} ] .$

$\sf \implies \vec {BC} = 3x \hat {i} + 3 \hat {j} - \hat {j} - 5 \hat {i} + \hat {j} - \hat {k} .$

$\sf \implies \vec {BC} = (3x - 5) \hat {i} + 4 \hat {j} - 2 \hat {k} .$

$\sf \implies as \ we\ know \ that = \vec {r} = \vec {a} + \lambda \vec {b} .$

$\sf \implies 2 \hat {i} - (1 - y) \hat {j} - 2 \hat {k} =\lambda [(3x - 5) \hat{i} + 4 \hat {j} - 2 \hat {k} ] .$

$\sf \implies 2 = \lambda(3x - 5). =(1).$

$\sf \implies -(1 - y) = 4 \lambda =(2).$

$\sf \implies -2 = -2 \lambda = (3)$

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).

$\sf \implies \vec {OA} = 3 \hat {i} - y \hat {j} + 2\hat {k}$

$\sf \implies \vec {OB} = 5 \hat {i} - \hat {j} +\hat {k}$

$\sf \implies \vec {OC} = 3x \hat {i} + 3 \hat {j} - \hat {k}$

$\sf \implies \vec {OB} = \dfrac{k \vec {(OC)} + 1\vec {(OA) }}{k + 1}$

$\sf \implies 5 \hat {i} - \hat {j} + \hat {k} = \dfrac{k[8 \hat {i} + 3 \hat {j} - \hat {k] + 1[3 \hat {i} - 5 \hat {j} + 2 \hat {k]}}}{k + 1}$

$\sf \implies 5 \hat {i} - \hat {j} + \hat {k} = \dfrac{(8k + 3) \hat {i} + (3k - 5) \hat {j} - (k - 2)\hat {k}}{k + 1}$

$\sf \implies 5 \hat {i} - \hat {j} + \hat {k} = \dfrac{(8k + 3) \hat {i}}{k + 1} = \dfrac{(3k - 5 ) \hat {j}}{k + 1} = \dfrac{-(k - 2) \hat {k}}{k + 1}$

$\sf \implies \dfrac{8k + 3}{k + 1} = 5$

⇒ 8k + 3 = 5 ( k + 1 ).

⇒ 8k + 3 = 5k + 5.

⇒ 8k - 5k = 5 - 3.

⇒ 3k = 2.

⇒ k = 2/3.

⇒ 3k - 5/k + 1 = -1.

⇒ 3k - 5 = - 1(k + 1).

⇒ 3k - 5 = -k - 1.

⇒ 3k + k = -1 + 5.

⇒ 4k = 4.

⇒ k = 1.

⇒ -(k - 2)/k + 1 = 1.

⇒ -k + 2 = k + 1.

⇒ 0.

Thus, B divides AC into ratio = 2/3 or 1