Vectors 5 Question 2
2. Let $\mathbf{a}=2 \hat{\mathbf{i}}+\lambda _1 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \quad \mathbf{b}=4 \hat{\mathbf{i}}+\left(3-\lambda _2\right) \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$ and $\mathbf{c}=3 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+\left(\lambda _3-1\right) \hat{\mathbf{k}}$ be three vectors such that $\mathbf{b}=2 \mathbf{a}$ and $\mathbf{a}$ is perpendicular to $\mathbf{c}$. Then a possible value of $\left(\lambda _1, \lambda _2, \lambda _3\right)$ is
(2019 Main, 10 Jan I)
(a) $(1,3,1)$
(b) $(1,5,1)$
(c) $\left(-\frac{1}{2}, 4,0\right)$
(d) $\left(\frac{1}{2}, 4,-2\right)$
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Answer:
Correct Answer: 2. (c)
Solution:
- We have, $\mathbf{a}=2 \hat{\mathbf{i}}+\lambda _1 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} ; \quad \mathbf{b}=4 \hat{\mathbf{i}}+\left(3-\lambda _2\right) \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$
and $\quad \mathbf{c}=3 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+\left(\lambda _3-1\right) \hat{\mathbf{k}}$,
such that $\quad \mathbf{b}=2 \mathbf{a}$
Now, $\quad \mathbf{b}=2 \mathbf{a}$
$\Rightarrow 4 \hat{\mathbf{i}}+\left(3-\lambda _2\right) \hat{\mathbf{j}}+6 \hat{\mathbf{k}}=2\left(2 \hat{\mathbf{i}}+\lambda _1 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\right)$
$\Rightarrow 4 \hat{\mathbf{i}}+\left(3-\lambda _2\right) \hat{\mathbf{j}}+6 \hat{\mathbf{k}}=4 \hat{\mathbf{i}}+2 \lambda _1 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$
$\Rightarrow \quad\left(3-2 \lambda _1-\lambda _2\right) \hat{\mathbf{j}}=\mathbf{0}$
$\Rightarrow \quad 3-2 \lambda _1-\lambda _2=0$
$$ \Rightarrow \quad 2 \lambda _1+\lambda _2=3 $$
Also, as a is perpendicular to $\mathbf{c}$, therefore $\mathbf{a} \cdot \mathbf{c}=0$
$\Rightarrow\left(2 \hat{\mathbf{i}}+\lambda _1 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\right) \cdot\left(3 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+\left(\lambda _3-1\right) \hat{\mathbf{k}}\right)=0$
$\Rightarrow 6+6 \lambda _1+3\left(\lambda _3-1\right)=0$
$\Rightarrow 6 \lambda _1+3 \lambda _3+3=0$
$\Rightarrow \quad 2 \lambda _1+\lambda _3=-1$
Now, from Eq. (i), $\lambda _2=3-2 \lambda _1$ and from Eq. (ii)
$\lambda _3=-2 \lambda _1-1$
$\therefore \quad\left(\lambda _1, \lambda _2, \lambda _3\right) \equiv\left(\lambda _1, 3-2 \lambda _1,-2 \lambda _1-1\right)$
If $\quad \lambda _1=-\frac{1}{2}$, then
$$ \lambda _2=4, \text { and } \lambda _3=0 $$
Thus, a possible value of $\left(\lambda _1, \lambda _2, \lambda _3\right)=\left(-\frac{1}{2}, 4,0\right)$
