MCQ
If the vectors $\hat{i}+2 \hat{k}, \hat{j}+\hat{k}$ and $\lambda \hat{i}+\mu \hat{j}$ are collinear, then
- A$\lambda=2, \mu=1$
- B$\lambda=2, \mu=-1$
- ✓$\lambda=-1, \mu=2$
- D$\lambda=-1, \mu=-2$
✓
Answer
Correct option: C.
$\lambda=-1, \mu=2$
(C) Let $\bar{a}=\hat{i}+2 \hat{k}, \bar{b}=\hat{j}+\hat{k}$ and $\bar{c}=\lambda \hat{i}+\mu \hat{j}$
$\therefore \quad \overline{ AB }= m \cdot \overline{ BC }$
$\Rightarrow-\hat{ i }+\hat{ j }-\hat{ k }= m [(\lambda \hat{ i }+(\mu-1) \hat{ j }-\hat{ k })]$
On comparing, we get
$\begin{array}{l}-1=-m \Rightarrow m=1 \\ -1=\lambda \cdot m \Rightarrow \lambda=-1\end{array}$
and $1=m(\mu-1) \Rightarrow \mu=2$
$\therefore \quad \overline{ AB }= m \cdot \overline{ BC }$
$\Rightarrow-\hat{ i }+\hat{ j }-\hat{ k }= m [(\lambda \hat{ i }+(\mu-1) \hat{ j }-\hat{ k })]$
On comparing, we get
$\begin{array}{l}-1=-m \Rightarrow m=1 \\ -1=\lambda \cdot m \Rightarrow \lambda=-1\end{array}$
and $1=m(\mu-1) \Rightarrow \mu=2$
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