The vector \( \left\{(a^{2}+a-2)\hat{\imath}+(a^{2}-4)\hat{\jmath}-(a^{2}-5a+6)\hat{k}\right\} \) makes acute angles with coordinate axes, then values of \( a \) lies in :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( (2,3) \)
The angle with an axis is acute when the component along that axis is positive, so all three components must be positive.
\( \displaystyle a^{2}+a-2\gt0 \)
\( \displaystyle \Rightarrow\ (a+2)(a-1)\gt0 \)
\( \displaystyle \Rightarrow\ a\lt-2\ \text{or}\ a\gt1 \)
\( \displaystyle a^{2}-4\gt0 \)
\( \displaystyle \Rightarrow\ (a-2)(a+2)\gt0 \)
\( \displaystyle \Rightarrow\ a\lt-2\ \text{or}\ a\gt2 \)
The k-component has a minus sign outside the bracket, so
\( \displaystyle -(a^{2}-5a+6)\gt0 \)
\( \displaystyle \Rightarrow\ (a-2)(a-3)\lt0 \)
\( \displaystyle \Rightarrow\ 2\lt a\lt3 \)
This interval already satisfies the first two conditions, so \( a\in(2,3) \).