Vector Algebra Question 51
Question: If $ \vec{a}=\hat{i}+\hat{j}+\hat{k},\vec{b}=4\hat{i}+3\hat{j}+4\hat{k} $ and $ \vec{c}=\hat{i}+\alpha \hat{j}+\beta \hat{k} $ are coplanar and $ | {\vec{c}} |=\sqrt{3} $ , then
Options:
A) $ \alpha =\sqrt{2},\beta =1 $
B) $ \alpha =1,\beta =\pm 1 $
C) $ \alpha =\pm 1,\beta =1 $
D) $ \alpha =\pm 1,\beta =-1 $
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Answer:
Correct Answer: C
Solution:
- [c] Since $ \vec{a},,\vec{b} $ and $ \vec{c} $ are coplanar therefore $ \begin{vmatrix} 1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta \\ \end{vmatrix} =0\Rightarrow \beta =1;| {\vec{c}} |=\sqrt{1+{{\alpha }^{2}}+{{\beta }^{2}}}=\sqrt{3} $
$ \Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}=2\Rightarrow {{\alpha }^{2}}=1\therefore \alpha =\pm 1 $
BETA
