Question:
The values of \(x\), for which the vectors \(\vec{a}=\left(x^{2}-16\right) \hat{i}+4\hat{j}-\left(x^{2}-25\right) \hat{k}\) make acute angles with the (positive directions of) coordinate axes lie in :
The values of \(x\), for which the vectors \(\vec{a}=\left(x^{2}-16\right) \hat{i}+4\hat{j}-\left(x^{2}-25\right) \hat{k}\) make acute angles with the (positive directions of) coordinate axes lie in :
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A vector makes an acute angle with an axis exactly when its component along that axis is positive.
Updated On: Nov 12, 2026
- \([-5,-4] \cup[4,5]\)
- \([-5,4) \cup(-4,5]\)
- \((-5,-4) \cup(4,5)\)
- \((-5,4) \cup(-4,5)\)
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The Correct Option is C
Solution and Explanation
The angle with the positive x-axis is acute iff \(\cos\alpha>0\), i.e. the i-component is positive; similarly for y and z.
i-component: \(x^2-16>0\Rightarrow x^2>16\).
j-component: \(4>0\), always true.
k-component: \(-(x^2-25)>0\Rightarrow x^2<25\).
So \(16<x^2<25\Rightarrow 4<|x|<5\), i.e. \(x\in(-5,-4)\cup(4,5)\).
Answer: (3) \((-5,-4) \cup(4,5)\)
i-component: \(x^2-16>0\Rightarrow x^2>16\).
j-component: \(4>0\), always true.
k-component: \(-(x^2-25)>0\Rightarrow x^2<25\).
So \(16<x^2<25\Rightarrow 4<|x|<5\), i.e. \(x\in(-5,-4)\cup(4,5)\).
Answer: (3) \((-5,-4) \cup(4,5)\)
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