If the vectors a-b-c-a-b-2c- -a-b-c- are linearly dependent then -

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Standard XII

Mathematics

Fundamental Theorem in 2D

If the vector...

Question

If the vectors $\to a + \to b + \to c, \to a + λ \to b + 2 \to c, - \to a + \to b + \to c$ are linearly dependent then $λ =$

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Solution

The correct option is B $2$
$a + b + c = k (a + d b + 2 c) + k_{1} (- a + b + c)$
(property of linearty dependent)
$k - k_{1} = 1$
$2 k + k_{1} = 1$
$k = \frac{2}{3}$ $k_{1} = \frac{- 1}{3}$
$1 = k d + k_{1}$
$1 = \frac{2}{3} d - \frac{1}{3}$
$d = 2$

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If the vectors →a+→b+→c,→a+λ→b+2→c,−→a+→b+→c are linearly dependent then λ=

The correct option is B 2a+b+c=k(a+db+2c)+k1(−a+b+c)(property of linearty dependent) k−k1=12k+k1=1k=23 k1=−131=kd+k11=23d−13d=2

If the vectors $\to a + \to b + \to c, \to a + λ \to b + 2 \to c, - \to a + \to b + \to c$ are linearly dependent then $λ =$

The correct option is B $2$
$a + b + c = k (a + d b + 2 c) + k_{1} (- a + b + c)$
(property of linearty dependent)
$k - k_{1} = 1$
$2 k + k_{1} = 1$
$k = \frac{2}{3}$ $k_{1} = \frac{- 1}{3}$
$1 = k d + k_{1}$
$1 = \frac{2}{3} d - \frac{1}{3}$
$d = 2$