Question

If $\vec{a}=xi+\left( x-1 \right)j+k$ and $\vec{b}=\left( x+1 \right)i+j+ak$ always make an acute angle with each other for every value of $x\in \mathbb{R}$, then
(a) $a\in \left( -\infty ,2 \right)$
(b) $a\in \left( 2,\infty \right)$
(c) $a\in \left( -\infty ,1 \right)$
(d) $a\in \left( 1,\infty \right)$

Answer 1

Hint: We will apply the formula of two vectors which forms an acute angle between them. The formula is given by $\cos \left( \theta \right)=\dfrac{\left| \left. \overrightarrow{a}\centerdot \overrightarrow{b} \right| \right.}{\left| \left. \overrightarrow{a} \right|\centerdot \right.\left| \left. \overrightarrow{b} \right| \right.}$ where $\overrightarrow{a}={{a}_{1}}i+{{b}_{1}}j+{{c}_{1}}k$ and $\overrightarrow{b}={{a}_{2}}i+{{b}_{2}}j+{{c}_{2}}k$.

Complete step-by-step solution -
Now, we will consider the two vectors given as $\vec{a}=xi+\left( x-1 \right)j+k$ and $\vec{b}=\left( x+1 \right)i+j+ak$. After this we will find out the dot product between the two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$. For finding the dot product of the two vectors by the formula given by,
$\left| \left. \overrightarrow{a}\centerdot \overrightarrow{b} \right| \right.=\left| \left. \left( xi+\left( x-1 \right)j+k \right)\centerdot \left( \left( x+1 \right)i+j+ak \right) \right| \right.$
As we know that all the dot products except for $i\centerdot i=1\,,\,j\centerdot j=1\,,\,k\centerdot k=1$ are zero. Therefore, we have
$\begin{align}
  & \left| \left. \overrightarrow{a}\centerdot \overrightarrow{b} \right| \right.=\left| \left. \left( xi+\left( x-1 \right)j+k \right)\centerdot \left( \left( x+1 \right)i+j+ak \right) \right| \right. \\
 & \Rightarrow \left| \left. \overrightarrow{a}\centerdot \overrightarrow{b} \right| \right.=\left| \left. x\times \left( x+1 \right)+\left( x-1 \right)\times 1+1\times a \right| \right. \\
 & \Rightarrow \left| \left. \overrightarrow{a}\centerdot \overrightarrow{b} \right| \right.=\left| \left. {{x}^{2}}+x+x-1+a \right| \right. \\
\end{align}$
Now, we will find the value of the angle. For this we will apply the formula which is used for finding the acute angel between them. The formula is given by $\cos \left( \theta \right)=\dfrac{\left| \left. \overrightarrow{a}\centerdot \overrightarrow{b} \right| \right.}{\left| \left. \overrightarrow{a} \right|\centerdot \right.\left| \left. \overrightarrow{b} \right| \right.}$ where $\overrightarrow{a}={{a}_{1}}i+{{b}_{1}}j+{{c}_{1}}k$ and $\overrightarrow{b}={{a}_{2}}i+{{b}_{2}}j+{{c}_{2}}k$.
Before substituting the values of the vectors we will first see that since the angle is not ${{90}^{\circ }}$ that means the angle must be less than ${{90}^{\circ }}$. Now, if we were told the angle equal to ${{90}^{\circ }}$ then we must have manipulated the formula $\cos \left( \theta \right)=\dfrac{\left| \left. \overrightarrow{a}\centerdot \overrightarrow{b} \right| \right.}{\left| \left. \overrightarrow{a} \right|\centerdot \right.\left| \left. \overrightarrow{b} \right| \right.}$ as
$\begin{align}
  & \cos \left( \theta \right)=\dfrac{\left| \left. \overrightarrow{a}\centerdot \overrightarrow{b} \right| \right.}{\left| \left. \overrightarrow{a} \right|\centerdot \right.\left| \left. \overrightarrow{b} \right| \right.} \\
 & \Rightarrow \cos \left( {{90}^{\circ }} \right)=\dfrac{\left| \left. \overrightarrow{a}\centerdot \overrightarrow{b} \right| \right.}{\left| \left. \overrightarrow{a} \right|\centerdot \right.\left| \left. \overrightarrow{b} \right| \right.} \\
\end{align}$
As we know that the value of $\cos \left( {{90}^{\circ }} \right)=0$. Therefore, we have
$\begin{align}
  & \cos \left( {{90}^{\circ }} \right)=\dfrac{\left| \left. \overrightarrow{a}\centerdot \overrightarrow{b} \right| \right.}{\left| \left. \overrightarrow{a} \right|\centerdot \right.\left| \left. \overrightarrow{b} \right| \right.} \\
 & \Rightarrow 0=\dfrac{\left| \left. \overrightarrow{a}\centerdot \overrightarrow{b} \right| \right.}{\left| \left. \overrightarrow{a} \right|\centerdot \right.\left| \left. \overrightarrow{b} \right| \right.} \\
 & \Rightarrow \left| \left. \overrightarrow{a}\centerdot \overrightarrow{b} \right| \right.=0 \\
\end{align}$
Now, since the angle given to us is not ${{90}^{\circ }}$ thus we have that $\cos \left( \theta \right)>\cos \left( {{90}^{\circ }} \right)$ where $\theta $ is any acute angle. This results into the inequality,
$\begin{align}
  & \cos \left( \theta \right)>\cos \left( {{90}^{\circ }} \right) \\
 & \Rightarrow \dfrac{\left| \left. \overrightarrow{a}\centerdot \overrightarrow{b} \right| \right.}{\left| \left. \overrightarrow{a} \right|\centerdot \right.\left| \left. \overrightarrow{b} \right| \right.}>0 \\
 & \Rightarrow \left| \left. \overrightarrow{a}\centerdot \overrightarrow{b} \right| \right.>0 \\
\end{align}$
Now, we will substitute the value of the dot product which is given by $\left| \left. \overrightarrow{a}\centerdot \overrightarrow{b} \right| \right.=\left| \left. {{x}^{2}}+x+x-1+a \right| \right.$. Therefore, we have
$\begin{align}
  & \left| \left. \overrightarrow{a}\centerdot \overrightarrow{b} \right| \right.>0 \\
 & \Rightarrow \left| \left. {{x}^{2}}+x+x-1+a \right| \right.>0 \\
\end{align}$
As the modulus is positive here and the value is also positive. So, we can have that
$\begin{align}
  & \Rightarrow \left| \left. {{x}^{2}}+x+x-1+a \right| \right.>0 \\
 & \Rightarrow {{x}^{2}}+x+x-1+a>0 \\
 & \Rightarrow {{x}^{2}}+2x-1+a>0 \\
\end{align}$
Now, we apply the square root formula which is given by
$\begin{align}
  & x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\
 & \Rightarrow x=\dfrac{-2\pm \sqrt{{{\left( 2 \right)}^{2}}-4\left( 1 \right)\left( a-1 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{-2\pm \sqrt{4-4\left( a-1 \right)}}{2} \\
 & \Rightarrow x=\dfrac{-2\pm \sqrt{4-4a-4}}{2} \\
 & \Rightarrow x=\dfrac{-2\pm \sqrt{-4a}}{2} \\
\end{align}$
Clearly, we can see that in the equation $x=\dfrac{-2\pm \sqrt{-4a}}{2}$ we have that $-4a$ is less than 0. So we will consider its value as $4-4\left( a-1 \right)$ since these two equations are equal if we see in the solution of the square root formula above. Therefore, we have that $4-4\left( a-1 \right)>0$ only if
$\begin{align}
  & 4>4\left( a-1 \right) \\
 & \Rightarrow 1>a-1 \\
 & \Rightarrow 1+1>a \\
 & \Rightarrow 2>a \\
\end{align}$
Clearly, we have that the value of a is greater than 2.
Hence the correct option is (b).

Note: We could have solved the expression ${{x}^{2}}+2x-1+a$ by hit and trial method. But here as we did not know that value of a so this method is not successful here. That is why we have solved the expression by square root formula. We also need to take care of the fact that since we got the negative sign inside the root then it does not mean that the answer is wrong. We need to find the value of a and not x so, we will consider the expression ${{b}^{2}}-4ac$ greater than 0.