| Statement 1: Vectors \[\vec{a}=2\hat{i}+\hat{k},\,\vec{b}=3\hat{j}+4\hat{k}\] and \[\vec{c}=8\hat{i}-3\hat{j}\] are co-planar then \[\vec{c}=4\vec{a}-\vec{b}\]. |
| Statement 2: A set of vectors \[\overrightarrow{{{a}_{1}}},\,\overrightarrow{{{a}_{2}}},\,\overrightarrow{{{a}_{3}}}...\overrightarrow{{{a}_{n}}}\] is said to be linearly independent if every relation of the form \[{{\ell }_{1}}\overrightarrow{{{a}_{1}}}+{{\ell }_{2}}\,\overrightarrow{{{a}_{2}}}+{{\ell }_{3}}\,\overrightarrow{{{a}_{3}}}+.....+{{\ell }_{n}}\overrightarrow{{{a}_{n}}}=\vec{0}\] implies that \[{{\ell }_{1}}={{\ell }_{2}}={{\ell }_{3}}=....={{\ell }_{n}}=0\] |
A) Both statements are true, and Statement-2 explains Statement-1.
B) Both statements are true, but Statement-2 does not explain Statement-1.
C) Statement-1 is True, Statement-2 is False.
D) Statement-1 is False, Statement-2 is true.
Correct Answer: B
Solution :
\[\because \,\,\vec{a},\vec{b},\vec{c}\] are co-planar, \[\Rightarrow \,\,\vec{c}=x\vec{a}+y\vec{b}\] \[\Rightarrow \,8\vec{i}-3\vec{j}=x(2\hat{c}+\hat{x})+y(3\hat{j}+4\hat{k})\] Solving we get \[x=4,\,y=-1\].
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