Question

The values of x for which the angle between $\overrightarrow{a}=2x^2\hat{i}+4x\hat{j}+\hat{k},\overrightarrow{b}=7\hat{i}-2\hat{j}+x\hat{k}$ is obtuse and the angle between $\overrightarrow{b}$and the z-axis is acute and less than $\frac{\mathrm{\pi }}{6}$ are
(a) $x>\frac{1}{2}\mathrm{or}x<0$

(b) $0<x<\frac{1}{2}$

(c) $\frac{1}{2}<x<15$

(d) ϕ

Answer 1

(b) $0<x<\frac{1}{2}$

$\overrightarrow{a}=2x^2\hat{i}+4x\hat{j}+\hat{k},\overrightarrow{b}=7\hat{i}-2\hat{j}+x\hat{k}$

Let the angle between vector a and vector b be A.

$$\begin{gathered} \therefore \cos A=\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{\left(2x^2\hat{i}+4x\hat{j}+\hat{k}\right).\left(7\hat{i}-2\hat{j}+x\hat{k}\right)}{\left|2x^2\hat{i}+4x\hat{j}+\hat{k}\right|\left|7\hat{i}-2\hat{j}+x\hat{k}\right|} \\ =\frac{14x^2-8x+x}{\sqrt{4x^4+16x^2+1}\sqrt{49+4+x^2}} \\ =\frac{14x^2-7x}{\sqrt{4x^4+16x^2+1}\sqrt{53+x^2}} \\ \mathrm{Now},\angle A\mathrm{is}\text{an}\mathrm{obtuse}\mathrm{angle}. \\ \therefore \cos A<0 \\ \Rightarrow \frac{14x^2-7x}{\sqrt{4x^4+16x^2+1}\sqrt{53+x^2}}<0 \\ \Rightarrow 14x^2-7x<0 \\ \Rightarrow 2x^2-x<0 \\ \Rightarrow x\left(2x-1\right)<0 \\ \Rightarrow x<0\&2x-1>0\mathrm{or}x>0\&2x-1<0 \\ \Rightarrow x<0\&x>\frac{1}{2}\mathrm{or}x>0\&x<\frac{1}{2} \\ \Rightarrow x>0\&x<\frac{1}{2}\left(\mathrm{As}\mathrm{there}\mathrm{cannot}\mathrm{be}\mathrm{any}\mathrm{number}\mathrm{less}\mathrm{than}\mathrm{zero}\mathrm{and}\mathrm{greater}\mathrm{than}1/2\right) \\ \Rightarrow x\in \left(0,\frac{1}{2}\right)...\left(\mathrm{i}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{equation}\mathrm{of}\text{the}z-\mathrm{axis}\text{be}z\hat{k}. \\ \mathrm{And}\mathrm{let}\mathrm{the}\mathrm{angle}\mathrm{between}\overrightarrow{b}\mathrm{and}z-\text{axis}\text{be}B. \\ \therefore \cos B=\frac{\left(7\hat{i}-2\hat{j}+x\hat{k}\right).\left(z\hat{k}\right)}{\left|7\hat{i}-2\hat{j}+x\hat{k}\right|\left|z\hat{k}\right|} \\ =\frac{xz}{z\sqrt{49+4+x^2}} \\ =\frac{x}{\sqrt{53+x^2}} \\ \mathrm{Now},\mathrm{angle}B\mathrm{is}\mathrm{acute}\mathrm{and}\mathrm{less}\mathrm{than}\mathrm{\pi }/6. \\ \therefore 0<\frac{\mathrm{x}}{\sqrt{53+{\mathrm{x}}^2}}<\cos \frac{\mathrm{\pi }}{6} \\ \Rightarrow 0<\mathrm{x}<\frac{\sqrt{3}}{2}\sqrt{53+{\mathrm{x}}^2}...\left(\mathrm{ii}\right) \\ \mathrm{From}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\mathrm{we}\mathrm{get} \\ 0<x<\frac{1}{2} \end{gathered}$$