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Scalar Or Dot Product — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsScalar Or Dot Product1 Mark
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 $\frac{\pi}{6}.$
$\therefore0<\frac{\text{x}}{\sqrt{53+\text{x}^2}}<\cos\frac{\pi}{6}$
$\Rightarrow 0<\text{x}<\frac{\sqrt{3}}{2}\sqrt{53+\text{x}^2}\dots(2)$
From $(1)$ and $(2)$ we get
$0<\text{x}<\frac{1}{2}$

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