Question

Let $\vec{\alpha} = (\lambda - 2)a + b$ and $\vec{\beta} = (4\lambda - 2)a + 3b$ where a & b are non collinear. Find the value of $\lambda$ for which $\sf \alpha \ and \beta$ are collinear.
1) - 3
2) 4
3) 3
4) -4 ​

Answer 1

EXPLANATION.

• GIVEN

$\bold{\vec{a} = ( \lambda \: - 2)a \: + b}$

$\bold{\vec{b} = (4 \lambda \: - 2)a \: + 3b}$

a and b are non collinear

Find the value of lambda for which

a and b are collinear.

Let a, b are non collinear.

$\bold{let \: \: \vec{a} = \mu \vec{b}}$

$\bold{( \lambda \: - 2) \vec{a} + \vec{b} = \mu \: (4 \lambda \: - \: 2) \vec{a} + 3 \vec{b}}$

vector b coefficient

$\bold{= > \: \: 1 = 3\mu = \mu = \frac{1}{3} }$

vector a coefficient

$\bold{( \lambda \: - 2) \: = \mu \: (4 \lambda \: - 2)}$

put the value of mu in equation

we get,

$\bold{( \lambda \: - 2) = \frac{1}{3} ( 4\lambda \: - 2)}$

$\bold{3 \lambda \: - 6 \: = 4 \lambda \: - 2}$

$= > \: \: \lambda \: = - 4$

Therefore,

value = -4

option [D ] is correct

Answer 2

◇ANSWER ◇:- OPTION (4) -4.

☆CONCEPT: -

》Two vectors are said to be collinear if their directed line segments are parallel disregards to their direction. 

If a and b are collinear , then a = Kb  where K € R.

 

》Two vectors are said to be collinear if and only if there exists m such as that  a = mb

Where m is a Scalar.

SOLUTION: - REFER TO THE ATTACHMENT

HOPE IT HELPS :)