Vectors 5 Question 1
1. Let $\alpha=(\lambda-2) \mathbf{a}+\mathbf{b}$ and $\beta=(4 \lambda-2) \mathbf{a}+3 \mathbf{b}$ be two given vectors where vectors $\mathbf{a}$ and $\mathbf{b}$ are non-collinear. The value of $\lambda$ for which vectors $\alpha$ and $\beta$ are collinear, is
(2019 Main, 10 Jan II)
(a) 4
(b) -3
(c) 3
(d) -4
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Answer:
Correct Answer: 1. (d)
Solution:
- Two vectors $\mathbf{c}$ and $\mathbf{d}$ are said to be collinear if we can write $\mathbf{c}=\lambda \mathbf{b}$ for some non-zero scalar $\lambda$.
Let the vectors $\alpha=(\lambda-2) \mathbf{a}+\mathbf{b}$
and
$$ \beta=(4 \lambda-2) \mathbf{a}+3 \mathbf{b} \text { are } $$
collinear, where $\mathbf{a}$ and $\mathbf{b}$ are non-collinear.
$\therefore$ We can write
$$ \begin{array}{cc} & \alpha=k \beta, \text { for some } k \in R-{0} \\ \Rightarrow & (\lambda-2) \mathbf{a}+\mathbf{b}=k[(4 \lambda-2) \mathbf{a}+3 \mathbf{b}] \\ \Rightarrow & {[(\lambda-2)-k(4 \lambda-2)] \mathbf{a}+(1-3 k) \mathbf{b}=0} \end{array} $$
Now, as $\mathbf{a}$ and $\mathbf{b}$ are non-collinear, therefore they are linearly independent and hence $(\lambda-2)-k(4 \lambda-2)=0$ and $1-3 k=0$
$\Rightarrow \lambda-2=k(4 \lambda-2)$ and $3 k=1$
$\Rightarrow \lambda-2=\frac{1}{3}(4 \lambda-2) \quad\left[\because 3 k=1 \Rightarrow k=\frac{1}{3}\right]$
$\Rightarrow 3 \lambda-6=4 \lambda-2$
$\Rightarrow \quad \lambda=-4$
