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# The set of values for which the angle between the vectors $cx\widehat i - 6\widehat j + 3\widehat k$ and $x\widehat i - 2\widehat j + 2cx\widehat k$ is acute for every $x \in R$ isA) $(0,\dfrac{4}{3})$B) $(0,\dfrac{4}{3})$C) $\left( {\dfrac{{11}}{9},\dfrac{4}{3}} \right)$D) $(0,\dfrac{4}{3})$

Last updated date: 17th Sep 2024
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Hint: We will use the formula of the angle between any two vectors to find the required set of values. We know that the angle between any two vectors is $\cos \theta = \dfrac{{\overrightarrow u .\overrightarrow v }}{{\left\| {\overrightarrow u } \right\|.\left\| {\overrightarrow v } \right\|}}$, $\left\| {\overrightarrow u } \right\|$ means the length of the vector $\overrightarrow u$.

Complete step by step solution:
We know that the angle between any two vectors is $\cos \theta = \dfrac{{\overrightarrow u .\overrightarrow v }}{{\left\| {\overrightarrow u } \right\|.\left\| {\overrightarrow v } \right\|}}$.
Thus, we assume $\overrightarrow u$ to be $cx\widehat i - 6\widehat j + 3\widehat k$ and $\overrightarrow v$ to be $x\widehat i - 2\widehat j + 2cx\widehat k$.
Using this formula, we multiply the vectors using scalar multiplication.
$\cos \theta = \dfrac{{\overrightarrow u .\overrightarrow v }}{{\left\| {\overrightarrow u } \right\|.\left\| {\overrightarrow v } \right\|}} \\ \Rightarrow \cos \theta = \dfrac{{(cx)x + ( - 6)( - 2) + (3)(2cx)}}{{\sqrt {{{(cx)}^2} + {{( - 6)}^2} + {3^2}} .\sqrt {{x^2} + {{( - 2)}^2} + {{(2cx)}^2}} }} \\ \Rightarrow \cos \theta = \dfrac{{c{x^2} + 12 + 6cx}}{{\sqrt {{c^2}{x^2} + 36 + 9} .\sqrt {{x^2} + 4 + 4{c^2}{x^2}} }} \\ \Rightarrow \cos \theta = \dfrac{{c{x^2} + 12 + 6cx}}{{\sqrt {{c^2}{x^2} + 45} .\sqrt {(4{c^2} + 1){x^2} + 4} }} \\ \Rightarrow \cos \theta = \dfrac{{c{x^2} + 12 + 6cx}}{{\sqrt {({c^2}{x^2} + 45)((4{c^2} + 1){x^2} + 4)} }} \\$
We are told that we need an acute angle and we know that $\cos \theta$ is positive between $0^\circ\;\text{and}\;90^\circ$.
So, $0^\circ < \theta < 90^\circ$
$\cos \theta$ is positive which means that the numerator is also positive. So, $c{x^2} + 6cx + 12 > 0$
Given the options we take c to be greater than zero, $c > 0$
We know that if a quadratic equation is greater than 0 then its discriminant is less than 0.
Therefore, discriminant for $c{x^2} + 6cx + 12 < 0$
Discriminant$= {b^2} - 4ac$ where $a = c,b = 6c,c = 12$
$= {(6c)^2} - 4 \times c \times 12 \\ = 36{c^2} - 48c \\$
Since, $D < 0$
$\Rightarrow 36{c^2} - 48c < 0$
Dividing by $12$ on both sides, we get

$\Rightarrow {3\text{c}^{2}-4\text{c}}<0 \\ \Rightarrow \text{c}\left(3\text{c}-{4}\right)<0 \\ \Rightarrow \text{c}<0,\;\text{and}\;\text{c}<\dfrac{4}{3} \\ \Rightarrow 0>\text{c}<\dfrac{4}{3}$

The set of values of c will be $\left( {0,\dfrac{4}{3}} \right)$ and not $[0,\dfrac{4}{3}]$ or $[0,\dfrac{4}{3})$ according to the options given because the range of values of c lies between $0$ and $\dfrac{4}{3}$which implies that c will not be equal to those values, so we use open brackets or first brackets to represent it. $[0,\dfrac{4}{3}]$ implies that the range of values of c also includes $0$and $\dfrac{4}{3}$. $[0,\dfrac{4}{3})$ implies that the range of values of c includes$0$ but not $\dfrac{4}{3}$.
Therefore, the set of values for which the angle between the vectors $cx\widehat i - 6\widehat j + 3\widehat k$ and $x\widehat i - 2\widehat j + 2cx\widehat k$ is acute for every $x \in R$ is $\left( {0,\dfrac{4}{3}} \right)$.
Thus, the answer is option A.

Note: This is a three dimensional vector as there are three coordinates given. We know that if a quadratic equation is greater than 0 then its discriminant is less than 0 but then it becomes a complex number, so we use the discriminant to find the roots of the equations.