Question

The vector $T+2\overline{y}+2k$ restated through an angle $\theta$ and doubled in magnitude then it becomes $2T+(2x+2)\}+(6x-2)k$ values of $x$ are

• A. $1,\frac{1}{3}$
• B. $-1,\frac{1}{3}=$
• C. $1,\frac{-1}{3}$
• D. $(0,3)$

Answer 1

Answer: (C)

$1,\frac{-1}{3}$

$\begin{array}{c}let,V=T+2\overline{y}+2k\\ \left|V\right|=\sqrt{1+4x^2+4}\\ And,U=\sqrt{2T+(2x+2)y+(6x-2)k}\\ \left|U\right|=\sqrt{4+4{(x+1)}^2+4{(3x-1)}^2}\\ Accordingtothecondition:2\left|V\right|=\left|U\right|\\ \Rightarrow 2\left(\sqrt{1+4x^2+4}\right)=\sqrt{4+4{(x+1)}^2+4{(3x-1)}^2}\\ \Rightarrow 2^2\left(1+4x^2+4\right)=4+4(x+1{)}^2+4(3x-1{)}^2\\ \Rightarrow 4\left(5+4x^2\right)=4\left[1+{(x+1)}^2+{(3x-1)}^2\right]\\ \Rightarrow \left(5+4x^2\right)=1+(x+1{)}^2+(3x-1{)}^2\\ \Rightarrow \left(5+4x^2\right)=1+x^2+1+2x+9x^2+1-6x\\ \Rightarrow 5+4x^2=3+10x^2-4x\\ \Rightarrow 6x^2-4x-2=0\\ \Rightarrow (x-1)(6x+2)=0\\ \therefore x=1,\frac{-2}{6}=\frac{-1}{3}\\ x=1,\frac{-1}{3}\\ sothatthecorrectoptionisC.\end{array}$