Question

If $x_1+x_2+x_3=0,y_1+y_2+y_3=0$ and $x_1{y}_1+x_2{y}_2+x_3{y}_3=0,$ then the value of $\frac{x_1^2}{x_1^2+x_2^2+x_3^2}+\frac{y_1^2}{y_1^2+y_2^2+y_3^2}$ is
(correct answer + 2, wrong answer - 0.50)

• A. $\frac{2}{5}$
• B. $\frac{2}{3}$
• C. $\frac{5}{2}$
• D. $\frac{3}{2}$

Answer 1

The correct option is B $\frac{2}{3}$

Assuming three vectors as
${\overrightarrow{n}}_1=(x_1,x_2,x_3){\overrightarrow{n}}_2=(y_1,y_2,y_3){\overrightarrow{n}}_3=(1,1,1)$

Now,
$x_1+x_2+x_3=0\Rightarrow {\overrightarrow{n}}_1\cdot {\overrightarrow{n}}_3=0y_1+y_2+y_3=0\Rightarrow {\overrightarrow{n}}_2\cdot {\overrightarrow{n}}_3=0x_1{y}_1+x_2{y}_2+x_3{y}_3=0\Rightarrow {\overrightarrow{n}}_1\cdot {\overrightarrow{n}}_2=0$
$\therefore {\overrightarrow{n}}_1,{\overrightarrow{n}}_2$ and ${\overrightarrow{n}}_3$ are mutually perpendicular vectors.

Now, let $\overrightarrow{n_4}=(1,0,0)$
$\frac{x_1^2}{x_1^2+x_2^2+x_3^2}={\left(\frac{\overrightarrow{n_1}\cdot \overrightarrow{n_4}}{|\overrightarrow{n_1}||\overrightarrow{n_4}|}\right)}^2\frac{y_1^2}{y_1^2+y_2^2+y_3^2}={\left(\frac{\overrightarrow{n_2}\cdot \overrightarrow{n_4}}{|\overrightarrow{n_2}||\overrightarrow{n_4}|}\right)}^2\frac{1}{3}={\left(\frac{\overrightarrow{n_3}\cdot \overrightarrow{n_4}}{|\overrightarrow{n_3}||\overrightarrow{n_4}|}\right)}^2$
They are squares of the projections of vector $\overrightarrow{n_4}=(1,0,0)$ on ${\overrightarrow{n}}_1,{\overrightarrow{n}}_2,{\overrightarrow{n}}_3$ respectively and hence,
$\frac{x_1^2}{x_1^2+x_2^2+x_3^2}+\frac{y_1^2}{y_1^2+y_2^2+y_3^2}+\frac{1}{3}=1\therefore \frac{x_1^2}{x_1^2+x_2^2+x_3^2}+\frac{y_1^2}{y_1^2+y_2^2+y_3^2}=\frac{2}{3}$