Vectors 3 Question 11
11. If $\overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \overrightarrow{\mathbf{b}}=4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}+\alpha \hat{\mathbf{j}}+\beta \hat{\mathbf{k}}$ are linearly dependent vectors and $|\overrightarrow{\mathbf{c}}|=\sqrt{3}$, then
(a) $\alpha=1, \beta=-1$
(b) $\alpha=1, \beta= \pm 1$
(c) $\alpha=-1, \beta= \pm 1$
(d) $\alpha= \pm 1, \beta=1$
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Answer:
Correct Answer: 11. (d)
Solution:
- Since, $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}$ are linearly dependent vectors.
Applying $C _2 \rightarrow C _2-C _1, C _3 \rightarrow C _3-C _1$,
$$ \left|\begin{array}{ccc} 1 & 0 & 0 \\ 4 & -1 & 0 \\ 1 & \alpha-1 & \beta-1 \end{array}\right|=0 \Rightarrow-(\beta-1)=0 \Rightarrow \beta=1 $$
Also, $\quad|\overrightarrow{\mathbf{c}}|=\sqrt{3} \quad$ [given]
$\Rightarrow \quad 1+\alpha^{2}+\beta^{2}=3 \quad$ [given, $c=\hat{\mathbf{i}}+\alpha \hat{\mathbf{j}}+\beta \hat{\mathbf{k}}$ ]
$\Rightarrow \quad 1+\alpha^{2}+1=3 \Rightarrow \alpha^{2}=1 \Rightarrow \alpha= \pm 1$
