Question

The vector $\hat{i}+x\hat{j}+3\hat{k}$ is rotated about its initial point through an angle of ${\cos }^{-1}\left(\frac{11}{14}\right)$ and its magnitude is doubled. If the vector in the new position is given by $4\hat{i}+(4x-2)\hat{j}+2\hat{k}.$ Then which of the following can't be the value of $x$?

• A. $\frac{-2}{3}$
• B. $\frac{2}{3}$
• C. $\frac{-20}{17}$
• D. $2$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\frac{-2}{3}$
B $\frac{2}{3}$
C $\frac{-20}{17}$


$\cos \theta =\frac{11}{14}\overrightarrow{a}=\hat{i}+x\hat{j}+3\hat{k}\overrightarrow{b}=4\hat{i}+(4x-2)\hat{j}+2\hat{k}2|\overrightarrow{a}|=|\overrightarrow{b}|4(1+x^2+9)=16+(4x-2{)}^2+43x(x-2)+2(x-2)=0x=2,x=\frac{-2}{3}\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}=\frac{11}{14}\Rightarrow \frac{4+x(4x-2)+6}{2(10+x^2)}=\frac{11}{14}17x^2-14x-40=0x=2,x=-\frac{20}{17}$