Vector Algebra Question 95
Question: The value of ‘x’ for which the angle between the vectors $ \overset{\to }{\mathop{a}},=2x^{2}\hat{i}+4x\hat{j}+\hat{k} $ and $ \overset{\to }{\mathop{b}},=7\hat{i}-2\hat{j}+x\hat{k} $ is obtuse are
Options:
A) $ x < 0 $
B) $ x>\frac{1}{2} $
C) $ 0<x<\frac{1}{2} $
D) $ x\in R $
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Answer:
Correct Answer: C
Solution:
- [c] $ 0<x<\frac{1}{2} $ $ \overset{\to }{\mathop{a}},.\overset{\to }{\mathop{b}},=(2x^{2}\hat{i}+4x\hat{j}+\hat{k}).(7\hat{i}-2\hat{j}+x\hat{k}) $ $ =2x^{2}(7)+(4x)(-2)+(1)(x)=14x^{2}-7x $ $ \hat{i}\cdot \hat{i}=1s=\hat{j}\cdot \hat{j}=1 $ The angle between vectors $ \overset{\to }{\mathop{a}}, $ and $ \overset{\to }{\mathop{b}}, $ is obtuse
$ \Rightarrow ,\vec{a}.\vec{b}<0\Rightarrow 14x^{2}-7x<0 $
$ \Rightarrow 7x(2x-1)<0\Rightarrow 14x( x-\frac{1}{2} )<0 $
$ \Rightarrow x $ lies between 0 and $ \frac{1}{2} $ (By the Method of Intervals) i.e., $ 0<x<\frac{1}{2}. $ Hence, the angle between the given vectors is obtuse if $ x\in ( 0,\frac{1}{2} ) $ .
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