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प्रश्न
Let \[\vec{a} = x^2 \hat{i} + 2 \hat{j} - 2 \hat{k} , \vec{b} = \hat{i} - \hat{j} + \hat{k} \text{ and } \vec{c} = x^2 \hat{i} + 5 \hat{j} - 4 \hat{k}\] be three vectors. Find the values of x for which the angle between \[\vec{a} \text{ and } \vec{b}\ \] is acute and the angle between \[\vec{b} \text{ and } \vec{c}\] is obtuse.
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उत्तर
\[\text{ We have }\]
\[ \vec{a} = = x^2 \stackrel\frown{i} + 2 \hat{j} - 2 \hat{k} , \vec{b} = \hat{i} - \hat{j} + \hat{k} \text{ and } \vec{c} = x^2 \hat{i} + 5 \hat{j} - 4 \hat{k} \]
\[ {\text{ Let } \theta}_1 \text{ be the angle between } \vec{a} \text{ and } \vec{b} \text{ and }\theta_2 \text{ be the angle between } \vec{b} \text{ and } \vec{c} .\]
\[ {\text{ Given that } \theta}_1 {\text { is acute and } \theta}_2 \text{ is obtuse }.\]
\[ \Rightarrow \cos \theta_1 > 0 \text{ and } \cos \theta_2 < 0\]
\[ \Rightarrow \frac{\vec{a} . \vec{b}}{\left| \vec{a} \right| . \left| \vec{b} \right|} > o\text{ and } \frac{\vec{b} . \vec{c}}{\left| \vec{b} \right| . \vec{\left| c \right|}} < 0\]
\[ \Rightarrow \frac{x^2 - 4}{\sqrt{x^4 + 4 + 4}\sqrt{1 + 1 + 1}} > 0 \text{ and } \frac{x^2 - 9}{\sqrt{1 + 1 + 1}\sqrt{x^4 + 25 + 16}} < 0\]
\[ \Rightarrow x^2 - 4 > 0 \text{ and } x^2 - 9 < 0\]
\[ \Rightarrow x \in \left( - \infty , - 2 \right) \cup \left( 2, \infty \right) \text{ and } x \in \left( - 3, 3 \right)\]
\[ \Rightarrow x \in \left( - 3, - 2 \right) \cup \left( 2, 3 \right)\]
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