Vectors Ques 45
- Let $\alpha \in R$ and the three vectors $\mathbf{a}=\alpha \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}, \mathbf{b}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\alpha \hat{\mathbf{k}}$ and $\mathbf{c}=\alpha \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$. Then, the set $ S = {\alpha: \mathbf{a}, \mathbf{b} \quad \text {and} \quad \mathbf{c}}$ are coplanar
(2019 Main, 12 April II)
(a) is singleton
(b) is empty
(c) contains exactly two positive numbers
(d) contains exactly two numbers only one of which is positive
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Answer:
Correct Answer: 45.(b)
Solution:
Formula:
- Given three vectors are
$$ \begin{aligned} & \mathbf{a}=\alpha \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & \mathbf{b}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\alpha \hat{\mathbf{k}} \\ & \text { and } \quad \mathbf{c}=\alpha \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & \text { Clearly, }[\mathbf{a} \mathbf{b} \mathbf{c}]=\left|\begin{array}{ccc} \alpha & 1 & 3 \\ 2 & 1 & -\alpha \\ \alpha & -2 & 3 \end{array}\right| \\ & =\alpha(3-2 \alpha)-1\left(6+\alpha^{2}\right)+3(-4-\alpha) \\ & =-3 \alpha^{2}-18=-3\left(\alpha^{2}+6\right) \end{aligned} $$
$\because$ There is no value of $\alpha$ for which $-3\left(\alpha^{2}+6\right)$ becomes zero, so $=\left|\begin{array}{ccc}\alpha & 1 & 3 \ 2 & 1 & -\alpha \ \alpha & -2 & 3\end{array}\right|[\mathbf{a} \mathbf{b} \mathbf{c}] \neq 0$
$\Rightarrow$ vectors $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ are not coplanar for any value $\alpha \in R$.
So, the $\operatorname{set} S={\alpha: \mathbf{a}, \mathbf{b} \quad \text {and} \quad \mathbf{c}}$ are coplanar is empty set.
BETA
