The values of x for which the angle between a =2x^2 i +4x j + k , b =7 i -2 j +x k is obtuse and the angle between b and the z- axis is acute and less than 6 are :

The values of x for which the angle between a =2x^2 i +4x j + k ,…

MCQ

The values of $x$ for which the angle between $\vec{\text{a}}=2\text{x}^2\hat{\text{i}}+4\text{x}\hat{\text{j}}+\hat{\text{k}},\vec{\text{b}}=7\hat{\text{i}}-2\hat{\text{j}}+\text{x}\hat{\text{k}}$ is obtuse and the angle between $\vec{\text{b}}$ and the $z-$ axis is acute and less than $\frac{\pi}{6}$ are :

A
$\text{x} > \frac{1}{2}$ or $\text{x} < 0$
✓
$0 < \text{x} < \frac{1}{2}$
C
$\frac{1}{2} < \text{x} < 15$
D
$\phi$

✓

Answer

Correct option: B.

$0 < \text{x} < \frac{1}{2}$

$\vec{\text{a}}=2\text{x}^2\hat{\text{i}}+4\text{x}\hat{\text{j}}+\hat{\text{k}}, \vec{\text{b}}=7\hat{\text{i}}-2\hat{\text{j}}+\text{x}\hat{\text{k}}$
Let the angle between vector $a $ and vector $b$ be $A$.
$\therefore\cos\text{A}=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|}=\frac{\big(2\text{x}^2\hat{\text{i}}+4\text{x}\hat{\text{j}}+\hat{\text{k}}\big).\big(7\hat{\text{i}}-2\hat{\text{j}}+\text{x}\hat{\text{k}}\big)}{\big|2\text{x}^2\hat{\text{i}}+4\text{x}\hat{\text{j}}+\hat{\text{k}}\big|\big|7\hat{\text{i}}-2\hat{\text{j}}+\text{x}\hat{\text{k}}\big|}$
$=\frac{14\text{x}^2-8\text{x}+\text{x}}{\sqrt{4\text{x}^4+16\text{x}^2+1}\sqrt{49+4+\text{x}^2}}$
$=\frac{14\text{x}^2-8\text{x}+\text{x}}{\sqrt{4\text{x}^4+16\text{x}^2+1}\sqrt{53+\text{x}^2}}$
Now, $\angle\text{A}$ is an obtuse angle.
$\therefore\cos\text{A} < 0$
$\Rightarrow\frac{14\text{x}^2-7\text{x}}{\sqrt{4\text{x}^4+16\text{x}^2+1}\sqrt{53+\text{x}^2}}<0$
$\Rightarrow14\text{x}^2-7\text{x} < 0$
$\Rightarrow2\text{x}^2-\text{x} < 0$
$\Rightarrow\text{x}(2\text{x}-1) < 0$
$\Rightarrow\text{x} < 0\ \ \ 2\text{x}-1>0$ or $\text{x} > 0\ \ \ 2\text{x}-1<0$
$\Rightarrow\text{x} < 0\ \ \ \text{x} > \frac{1}{2}$ or $\text{x} > 0\ \ \ \text{x}<\frac{1}{2}$
$\Rightarrow\text{x} > 0\ \ \ \text{x} < \frac{1}{2}\ ($As there cannot be any number less than zero and greater than $\frac{1}{2})$
$\Rightarrow\text{x}\in(0,\frac{1}{2})\dots(1)$
Let the equation of the $z-$ axis be $\text{z}\hat{\text{k}}.$
And let the angle between $\vec{\text{b}}$ and $z-$ axis be $B.$
$\therefore\cos\text{B}=\frac{\big(7\hat{\text{i}}-2\hat{\text{j}}+\text{x}\hat{\text{k}}\big).\big(\text{z}\hat{\text{k}}\big)}{\big|7\hat{\text{i}}-2\hat{\text{j}}+\text{x}\hat{\text{k}}\big|\big|\text{z}\hat{\text{k}}\big|}$
$=\frac{\text{xz}}{\text{z}\sqrt{49+4+\text{x}^2}}$
$=\frac{\text{x}}{\sqrt{53+\text{x}^2}}$
Now, angle $B$ is acute and less than $\frac{\pi}{6}.$
$\therefore0 < \frac{\text{x}}{\sqrt{53+\text{x}^2}} < \cos\frac{\pi}{6}$
$\Rightarrow0 < \text{x} < \frac{\sqrt{3}}{2}\sqrt{53+\text{x}^2}\dots(2)$
From $(1)$ and $(2)$ we get
$0 < \text{x} < \frac{1}{2}$

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