question_answer

The value of 'x' for which the angle between the vectors \[\overset{\to }{\mathop{a}}\,=2{{x}^{2}}\hat{i}+4x\hat{j}+\hat{k}\] and \[\overset{\to }{\mathop{b}}\,=7\hat{i}-2\hat{j}+x\hat{k}\] is obtuse are

A) \[\operatorname{x} < 0\]

B) \[x>\frac{1}{2}\]

C) \[0<x<\frac{1}{2}\]

D) \[x\in R\]

Correct Answer: C

Solution :

[c]             \[0<x<\frac{1}{2}\] \[\overset{\to }{\mathop{a}}\,.\overset{\to }{\mathop{b}}\,=(2{{x}^{2}}\hat{i}+4x\hat{j}+\hat{k}).(7\hat{i}-2\hat{j}+x\hat{k})\] \[=2{{x}^{2}}(7)+(4x)(-2)+(1)(x)=14{{x}^{2}}-7x\] \[\hat{i}\cdot \hat{i}=1s=\hat{j}\cdot \hat{j}=1\] The angle between vectors \[\overset{\to }{\mathop{a}}\,\] and \[\overset{\to }{\mathop{b}}\,\] is obtuse \[\Rightarrow \,\vec{a}.\vec{b}<0\Rightarrow 14{{x}^{2}}-7x<0\] \[\Rightarrow 7x(2x-1)<0\Rightarrow 14x\left( x-\frac{1}{2} \right)<0\] \[\Rightarrow x\] 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 \[x\in \left( 0,\frac{1}{2} \right)\].