1 votes 1 votes If the vectors $e_1=(1,0,2)$, $e_2=(0,1,0)$ and $e_3=(-2,0,1)$ form an orthogonal basis of the three-dimensional real space $\mathbb{R}^3$, then the vector $\textbf{u}=(4,3,-3)\in \mathbb{R}^3 $ can be expressed as $\textbf{u}=-$$\large\frac{2}{5}$$e_1-3e_2-$$\large\frac{11}{5}$$e_3\\$ $\textbf{u}=-$$\large\frac{2}{5}$$e_1-3e_2+$$\large\frac{11}{5}$$e_3 \\$ $\textbf{u}=-$$\large\frac{2}{5}$$e_1+3e_2+$$\large\frac{11}{5}$$e_3 \\$ $\textbf{u}=-$$\large\frac{2}{5}$$e_1+3e_2-$$\large\frac{11}{5}$$e_3$ Vector Analysis gate2016-ec-3 vector-analysis + – Milicevic3306 asked Mar 27, 2018 • retagged Mar 1, 2021 by Lakshman Bhaiya Milicevic3306 16.0k points 288 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes $u= e_1* x_1+e_2*x_2+e_3*x_3$ $(4,3,-3)=(1,0,2)* x_1+(0,1,0)*x_2+(-2,0,1)*x_3$ $x_1-2 x_3=4$ $x_2=3$ $2x_1+x_3=-3$ $x_1=\frac{-2}{5}$ $x_3=\frac{-11}{5}$ $u= -\frac{2}{5}e_1+3e_2-\frac{11}{5}e_3$ $Ans= D$ Divyanshu Shukla answered Jun 20, 2021 Divyanshu Shukla 160 points comment Share ask related question Follow See all 0 reply Please log in or register to add a comment.