$x_{1}+x_{2}+x_{3}=0$ and $y_{1}+y_{2}+y_{3}=0$ and $x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}=0$
then value of $\frac{x^2_{1}}{x^2_{1}+x^2_{2}+x^2_{3}}+\frac{y^2_{1}}{y^2_{1}+y^2_{2}+y^2_{3}}$
answer is $2/3$
$n_{1}=x_{1}\hat{i}+x_{2}\hat{j}+x_{3}\hat{k}$ and $n_{2}=y_{1}\hat{i}+y_{2}\hat{j}+y_{3}\hat{k}$ how can i solve after this
Let $\mathbf{x}=x_1 \mathbf{i}+x_2 \mathbf{j} +x_3 \mathbf{k}$, $\mathbf{y}=y_1 \mathbf{i}+y_2 \mathbf{j} +y_3 \mathbf{k}$, and let $\mathbf{p}=\mathbf{i}+ \mathbf{j} + \mathbf{k}$.
For the result to make sense we also need the hypotheses $\mathbf{x}\not=\mathbf{0}$ and $\mathbf{y}\not=\mathbf{0}$, so assume that this is the case.
The other hypotheses are $\mathbf{x}\cdot\mathbf{p}=0$, $\mathbf{y}\cdot\mathbf{p}=0$, and $\mathbf{x}\cdot\mathbf{y}=0$; that is, these three vectors are mutually perpendicular.
We can therefore form a new orthonormal basis $\{\mathbf{a}, \mathbf{b}, \mathbf{c}\}$ by taking $$ \mathbf{a}:=\mathbf{x}/|\mathbf{x}|, \ \ \mathbf{b}:=\mathbf{y}/|\mathbf{y}|, \ \ \mathbf{c}:=\mathbf{p}/|\mathbf{p}|. $$ (To get a right-handed system we may have to take $-\mathbf{c}$, it's easy to check later that the sign makes mo difference.)
Now express $\mathbf{i}$ in terms of this basis, $$ \mathbf{i}= (\mathbf{i}\cdot\mathbf{a})\mathbf{a}+ (\mathbf{i}\cdot\mathbf{b})\mathbf{b}+ (\mathbf{i}\cdot\mathbf{c})\mathbf{c}= \frac{x_1\mathbf{a}}{\sqrt{x_1^2+x_2^2 +x_3^2}}+ \frac{y_1\mathbf{b}}{\sqrt{y_1^2+y_2^2 +y_3^2}}+ \frac{1}{\sqrt{3}}\mathbf{c}. $$ Hence $$ 1= \mathbf{i}\cdot\mathbf{i}= \frac{x_1^2}{x_1^2+x_2^2+x_3^2}+ \frac{y_1^2}{y_1^2+y_2^2+y_3^2}+ \frac{1}{3}, $$ which simplifies to give the result.
[I feel there should be a neat geometric argument that lets one just write down the answer. After all, the hypotheses are symmetric in the suffices $\{1,2,3\}$ so one expects to get the same value for the other two expressions we get by permuting the suffices; and the sum of the three of them is clearly $1+1=2$. But I can't make that rigorous.]