Question

The vector $\widehat{i}+x\widehat{j}+3\widehat{k}$ is rotated through an angle $\theta$ and is doubled in magnitude. It now becomes $4\widehat{i}+\left( 4x-2 \right)\widehat{j}+2\widehat{k}$ . The values of x are
(a) 1
(b) $-\dfrac{2}{3}$
(c) 2
(d) $\dfrac{4}{3}$

Answer 1

We have to denote $\overrightarrow{{{v}_{1}}}=\widehat{i}+x\widehat{j}+3\widehat{k}$ and $\overrightarrow{{{v}_{2}}}=4\widehat{i}+\left( 4x-2 \right)\widehat{j}+2\widehat{k}$ . Then, according to the given condition, we can write $\left| \overrightarrow{{{v}_{2}}} \right|=2\left| \overrightarrow{{{v}_{1}}} \right|$ . Then, we have to substitute the magnitudes of $\overrightarrow{{{v}_{1}}}$ and $\overrightarrow{{{v}_{2}}}$ in this equation and solve for x.

Complete step by step answer:
We are given that the vector $\widehat{i}+x\widehat{j}+3\widehat{k}$ is rotated through an angle $\theta$. Let us consider $\overrightarrow{{{v}_{1}}}=\widehat{i}+x\widehat{j}+3\widehat{k}$
After the rotation, we are given that the vector becomes $4\widehat{i}+\left( 4x-2 \right)\widehat{j}+2\widehat{k}$ . So let $\overrightarrow{{{v}_{2}}}=4\widehat{i}+\left( 4x-2 \right)\widehat{j}+2\widehat{k}$
We are also given that after the rotation, the magnitude is doubled. We can denote this mathematically as
$\left| \overrightarrow{{{v}_{2}}} \right|=2\left| \overrightarrow{{{v}_{1}}} \right|...\left( i \right)$
We know that for a vector $\overrightarrow{v}=a\widehat{i}+b\widehat{j}+c\widehat{k}$ , the magnitude is given by
$\left| \overrightarrow{v} \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}$
Therefore, we can write equation (i) as
$\begin{aligned} & \Rightarrow \sqrt{{{4}^{2}}+{{\left( 4x-2 \right)}^{2}}+{{2}^{2}}}=2\sqrt{{{1}^{2}}+{{x}^{2}}+{{3}^{2}}} \\\ & \Rightarrow \sqrt{16+{{\left( 4x-2 \right)}^{2}}+4}=2\sqrt{1+{{x}^{2}}+9} \\\ \end{aligned}$
Let us square both the sides of the above equation.
$\Rightarrow 16+{{\left( 4x-2 \right)}^{2}}+4=4\left( 1+{{x}^{2}}+9 \right)$
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ .
$\begin{aligned} & \Rightarrow 16+16{{x}^{2}}-\left( 2\times 4x\times 2 \right)+4+4=4\left( 1+{{x}^{2}}+9 \right) \\\ & \Rightarrow 16+16{{x}^{2}}-16x+4+4=4\left( {{x}^{2}}+10 \right) \\\ & \Rightarrow 24+16x-16{{x}^{2}}=4\left( {{x}^{2}}+10 \right) \\\ \end{aligned}$
Let us take common factor of 4 outside from the LHS.
$\Rightarrow 4\left( 6-4x+4{{x}^{2}} \right)=4\left( {{x}^{2}}+10 \right)$
We can now cancel 4 from both the sides.
$\Rightarrow 6-4x+4{{x}^{2}}={{x}^{2}}+10$
Let us take the terms in x and ${{x}^{2}}$ to the LHS and constant on the RHS.

& \Rightarrow -4x+4{{x}^{2}}-{{x}^{2}}=10-6 \\\ & \Rightarrow 3{{x}^{2}}-4x=4 \\\ \end{aligned}$$ We have to take 4 to the LHS from the RHS. $$\Rightarrow 3{{x}^{2}}-4x-4=0...\left( ii \right)$$ We can see that the above equation is of the form $a{{x}^{2}}+bx+c=0$ . We can write the solution of this standard form as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ From equation (ii), we can see that $a=3,b=-4,c=-4$ . $$\begin{aligned} & \Rightarrow x=\dfrac{-\left( -4 \right)\pm \sqrt{{{\left( -4 \right)}^{2}}-4\times 3\times -4}}{2\times 3} \\\ & \Rightarrow x=\dfrac{4\pm \sqrt{16+48}}{6} \\\ & \Rightarrow x=\dfrac{4\pm \sqrt{64}}{6} \\\ & \Rightarrow x=\dfrac{4\pm 8}{6} \\\ \end{aligned}$$ Therefore, we can write the value of x as $\begin{aligned} & \Rightarrow x=\dfrac{4+8}{6},\dfrac{4-8}{6} \\\ & \Rightarrow x=\dfrac{12}{6},-\dfrac{4}{6} \\\ & \Rightarrow x=2,-\dfrac{2}{3} \\\ \end{aligned}$ **So, the correct answer is “Option b and c”.** **Note:** Students have a chance of making mistake by writing the given condition as $$\left| \overrightarrow{{{v}_{1}}} \right|=2~\left| \overrightarrow{{{v}_{2}}} \right|$$ . They must know to take the magnitudes of a vector. Students can verify the solution by doing the following procedure. Let us consider the case when $x=2$ .Then we can write the two vectors as $\overrightarrow{{{v}_{1}}}=\widehat{i}+2\widehat{j}+3\widehat{k}$ $\begin{aligned} & \overrightarrow{{{v}_{2}}}=4\widehat{i}+\left( 4\times 2-2 \right)\widehat{j}+2\widehat{k} \\\ & \Rightarrow \overrightarrow{{{v}_{2}}}=4\widehat{i}+6\widehat{j}+2\widehat{k} \\\ \end{aligned}$ Let us take the magnitude of $\overrightarrow{{{v}_{1}}}$ . $\begin{aligned} & \Rightarrow \left| \overrightarrow{{{v}_{1}}} \right|=\sqrt{1+{{2}^{2}}+{{3}^{2}}} \\\ & \Rightarrow \left| \overrightarrow{{{v}_{1}}} \right|=\sqrt{1+4+9} \\\ & \Rightarrow \left| \overrightarrow{{{v}_{1}}} \right|=\sqrt{14}...\left( iii \right) \\\ \end{aligned}$ Let us take the magnitude of $\overrightarrow{{{v}_{2}}}$ . $\begin{aligned} & \Rightarrow \left| \overrightarrow{{{v}_{2}}} \right|=\sqrt{16+36+4} \\\ & \Rightarrow \left| \overrightarrow{{{v}_{2}}} \right|=\sqrt{56} \\\ & \Rightarrow \left| \overrightarrow{{{v}_{2}}} \right|=2\sqrt{14} \\\ \end{aligned}$ From (iii), we can write $\Rightarrow \left| \overrightarrow{{{v}_{2}}} \right|=2\left| \overrightarrow{{{v}_{1}}} \right|$ Therefore, the given condition is satisfied. Now, let us consider $x=-\dfrac{2}{3}$ . $\overrightarrow{{{v}_{1}}}=\widehat{i}-\dfrac{2}{3}\widehat{j}+3\widehat{k}$ $\begin{aligned} & \overrightarrow{{{v}_{2}}}=4\widehat{i}+\left( 4\times -\dfrac{2}{3}-2 \right)\widehat{j}+2\widehat{k} \\\ & \Rightarrow \overrightarrow{{{v}_{2}}}=4\widehat{i}-\dfrac{14}{3}\widehat{j}+2\widehat{k} \\\ \end{aligned}$ Let us take the magnitude of $\overrightarrow{{{v}_{1}}}$ . $\begin{aligned} & \Rightarrow \left| \overrightarrow{{{v}_{1}}} \right|=\sqrt{1+{{\left( -\dfrac{2}{3} \right)}^{2}}+{{3}^{2}}} \\\ & \Rightarrow \left| \overrightarrow{{{v}_{1}}} \right|=\sqrt{1+\dfrac{4}{9}+9} \\\ & \Rightarrow \left| \overrightarrow{{{v}_{1}}} \right|=\sqrt{\dfrac{94}{9}} \\\ \end{aligned}$ Let us take the magnitude of $\overrightarrow{{{v}_{2}}}$ . $\begin{aligned} & \Rightarrow \left| \overrightarrow{{{v}_{2}}} \right|=\sqrt{16+\dfrac{196}{9}+4} \\\ & \Rightarrow \left| \overrightarrow{{{v}_{2}}} \right|=\sqrt{\dfrac{376}{9}} \\\ & \Rightarrow \left| \overrightarrow{{{v}_{2}}} \right|=2\sqrt{\dfrac{94}{9}} \\\ \end{aligned}$ From (iii), we can write $\Rightarrow \left| \overrightarrow{{{v}_{2}}} \right|=2\left| \overrightarrow{{{v}_{1}}} \right|$ Therefore, the given condition is satisfied.