Answer

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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.

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