If $a+b+c=x+y+z=ax+by+cz=0$, then find value of $\frac{a^2}{a^2+b^2+c^2}+\frac{x^2}{x^2+y^2+z^2}$

Question

From the $a+b+c=0$ equality I tried taking systems of equations into consideration: $$\begin{cases} c=-(a+b)\\ z=-(x+y)\\ ax+by+(a+b)(x+y)=0 \end{cases} $$ which can be later on turned to : $$\begin{cases} c=-(a+b)\\ z=-(x+y)\\ (2a+b)x+(a+2b)y=0 \end{cases} $$ However this is did not put me anywhere near the solution. Any help is appreciated.

For whole numbers the solution is $x = y = z = a = b = c = 0$. Kinda like Euler's $e^{i\pi} + 1 = 0$. Nice! — Hudjefa, Mar 23 '24 at 13:53
@AgentSmith How is the first one a solution? $1+1+1$ is $3$ not $0$ — Noble Mushtak, Mar 23 '24 at 13:55
@NobleMushtak, error on my part, didn't see the $0$ at the end — Hudjefa, Mar 23 '24 at 13:56
$(a + b + c)(x + y + z) = ax + by + cz + \text{Lena}$ i.e $(a + b + c)^2 = ax + by + cz + \text{Lena} = (x + y + z)^2$ — Hudjefa, Mar 23 '24 at 13:59
$\begin {bmatrix}a & b & c \end {bmatrix} \times \begin {bmatrix} x \ y \ z \end {bmatrix}$ Can we use matrices to solve this SYSTEM? — Hudjefa, Mar 23 '24 at 14:03
@AgentSmith See my answer, I didn't use matrices, but I did use vectors and the fact that $ax+by+cz$ represents the dot-product of two 3D vectors. — Noble Mushtak, Mar 23 '24 at 14:22

score 4 · Answer 1 · answered Mar 23 '24 at 14:37

(i) If $a + 2b \ne 0$, then $$y = - \frac{(2a + b)x}{a + 2b}.$$

We have $$x^2 + xy + y^2 = x^2 - x \cdot \frac{(2a + b)x}{a + 2b} + \left(\frac{(2a + b)x}{a + 2b}\right)^2 = \frac{3x^2(a^2 + ab + b^2)}{(a + 2b)^2}.$$

We have \begin{align*} \frac{a^2}{a^2+b^2+c^2}+\frac{x^2}{x^2+y^2+z^2} &= \frac{a^2}{a^2+b^2+(a + b)^2}+\frac{x^2}{x^2+y^2+(x + y)^2}\\ &= \frac{a^2}{2a^2 + 2ab + 2b^2} + \frac{x^2}{2x^2 + 2xy + 2y^2}\\ &= \frac{a^2}{2a^2 + 2ab + 2b^2} + \frac{x^2}{\frac{6x^2(a^2 + ab + b^2)}{(a + 2b)^2}}\\ &= \frac{a^2}{2a^2 + 2ab + 2b^2} + \frac{(a+2b)^2}{6(a^2 + ab + b^2)}\\ &= \frac23. \end{align*}

(ii) If $a + 2b = 0$, then $a = -2b$ and $-3bx = 0$. Thus, $x = 0$ (Note. If $b = 0$, then $a = b = c = 0$). Thus, $$\frac{a^2}{a^2+b^2+c^2}+\frac{x^2}{x^2+y^2+z^2} = \frac23.$$

Noble Mushtak · Answer 2 · 2024-03-23T14:32:29.637

I'm going to rename the variables, because I think it will make my explanation less confusing: If $x_1+y_1+z_1=x_2+y_2+z_2=x_1x_2+y_1y_2+z_1z_2=0$, then find the value of: $$ \frac{x_1^2}{x_1^2+y_1^2+z_1^2}+\frac{x_2^2}{x_2^2+y_2^2+c_2^2} $$

Let's define some vectors: Let $\vec{v}_1=(x_1,y_1,z_1)$ and $\vec{v}_2=(x_2,y_2,c_1)$. Clearly, both $\vec{v}_1$ and $\vec{v}_2$ are on the plane $x+y+z=0$ since $x_1+y_1+z_1=x_2+y_2+z_2=0$. Moreover, $\vec{v}_1$ and $\vec{v}_2$ are perpendicular to each other, i.e. $\vec{v}_1\cdot \vec{v}_2=0$ (where $\cdot$ represents the dot product) because $x_1x_2+y_1y_2+z_1z_2=0$.

Now, without loss of generality, we can assume $\vec{v}_1$ and $\vec{v}_2$ are unit vectors. We can assume this without losing generality is because, if $\vec{v}_1$ or $\vec{v}_2$ is not a unit vector, then we can just normalize that vector and all the conditions we need will still hold: Normalizing a vector just changes the length of that vector, so normalization will not change whether the vector is on the plane $x+y+z=0$ and will not change whether the vector is perpendicular to the other vector.

Because we assume $\vec{v}_1, \vec{v}_2$ are unit vectors, we have $$ x_1^2+y_1^2+z_1^2=x_2^2+y_2^2+z_2^2=1 $$ Therefore, the value we want to find becomes just $x_1^2+x_2^2$.

Next, notice that both $\vec{v}_1$ and $\vec{v}_2$ are perpendicular to the vector $(1,1,1)$, because $x_1+y_1+z_1=x_2+y_2+z_2=0$. Then, if we normalize $(1,1,1)$, we find that both $\vec{v}_1$ and $\vec{v}_2$ are perpendicular to $(1/\sqrt{3},1/\sqrt{3},1/\sqrt{3})$.

Thus, $\vec{v}_2$ is a unit vector which is perpendicular to both $\vec{v}_1$ and $(1/\sqrt{3},1/\sqrt{3},1/\sqrt{3})$. It is a common fact that, for any unit vector $\vec{a}\in\mathbb{R}^3$ which is perpendicular to two other unit vectors $\vec{b},\vec{c}\in\mathbb{R}^3$, either $\vec{a}=\vec{b}\times\vec{c}$ or $\vec{a}=-(\vec{b}\times\vec{c})$, where $\times$ here represents the 3D cross-product. Thus, one of the below equations must be true: $$\vec{v}_2=\vec{v}_1\times (1/\sqrt{3},1/\sqrt{3},1/\sqrt{3})\ \ \ \ \ \ \ \text{ or }\ \ \ \ \ \ \ \vec{v}_2=-(\vec{v}_1\times (1/\sqrt{3},1/\sqrt{3},1/\sqrt{3}))$$ By taking the cross product of $\vec{v}_1=(x_1,y_1,z_1)$ with $(1/\sqrt{3},1/\sqrt{3},1/\sqrt{3})$, we see that in the first case, the $x$-component of $\vec{v}_2$, which is $x_2$, is $y_1/\sqrt{3}-z_1/\sqrt{3}$. In the second case, the $x$-component of $\vec{v}_2$ is just the negative of this expression, i.e. $x_2=-(y_1/\sqrt{3}-z_1/\sqrt{3})$. However, in either case, when we square this expression, we find that: $$ x_2^2=\left(\frac{y_1}{\sqrt{3}}-\frac{z_1}{\sqrt{3}}\right)^2=\frac{y_1^2}{3}+\frac{z_1^2}{3}-\frac{2y_1z_1}{3} $$ Ergo, we have $$ x_1^2+x_2^2=x_1^2+\frac{y_1^2}{3}+\frac{z_1^2}{3}-\frac{2y_1z_1}{3} $$ Now, I'm going to split $x_1^2$ into $2x_1^2/3+x_1^2/3$, $y_1^2/3$ into $2y_1^2/3-y_1^2/3$, and $z_1^2/3$ into $2z_1^2/3-z_1^2/3$. By rearranging terms a little, the expression then becomes: $$ x_1^2+x_2^2=\frac{2}{3}x_1^2+\frac{2}{3}y_1^2+\frac{2}{3}z_1^2+\frac{1}{3}x_1^2-\frac{1}{3}y_1^2-\frac{1}{3}z_1^2-\frac{2}{3}y_1z_1 $$ Factor out $2/3$ from the first three terms and the $-$ sign from the last three terms: $$ x_1^2+x_2^2=\frac{2}{3}(x_1^2+y_1^2+z_1^2)+\frac{1}{3}x_1^2-\left(\frac{1}{3}y_1^2+\frac{1}{3}z_1^2+\frac{2}{3}y_1z_1\right) $$ We know $x_1^2+y_1^2+z_1^2=1$: $$ x_1^2+x_2^2=\frac{2}{3}+\frac{1}{3}x_1^2-\left(\frac{1}{3}y_1^2+\frac{1}{3}z_1^2+\frac{2}{3}y_1z_1\right) $$ Clearly, $(1/3)x_1^2=(x_1/\sqrt{3})^2$. Moreover, with the last three terms, by completing the square, we can see that those last three terms are actually $(y_1/\sqrt{3}+z_1/\sqrt{3})^2$: $$ x_1^2+x_2^2=\frac{2}{3}+\left(\frac{x_1}{\sqrt{3}}\right)^2-\left(\frac{y_1}{\sqrt{3}}+\frac{z_1}{\sqrt{3}}\right)^2 $$ Finally, I'll leave it as an exercise to you to show that $(x_1/\sqrt{3})^2-(y_1/\sqrt{3}+z_1/\sqrt{3})^2=0$. Hint: Use the fact that $x_1+y_1+z_1=0$.

Using this final hint, we find that: $$ x_1^2+x_2^2=\frac{2}{3} $$ so the final answer is $2/3$.

Excellent!!! You somewhere wrote "We know that ${x_1}^2+{y_1}^2+{z_1}^2=0"$...it should be $1$ ig — Blue Cat Blues, Mar 23 '24 at 14:31
$\frac{a^2}{a^2 + b^2 + c^2}$ and $\frac{x^2}{x^2 + y^2 + z^2}$ does resemble the 3D Pythagorean theorem. Did you, perchance, find the magnitude of these vectors? — Hudjefa, Mar 23 '24 at 15:12
@AgentSmith I don't need to find the magnitudes of the vectors to solve this problem. I just assume without loss of generality that the vectors are unit vectors, i.e. the magnitudes are 1. — Noble Mushtak, Mar 23 '24 at 18:35

score 1 · Answer 3 · answered Mar 23 '24 at 14:40

First note there is an implicit assumption that neither $\langle a,b,c \rangle$ nor $\langle x,y,z \rangle$ is the zero vector for the problem to make sense.

Next, since $a + b + c = 0$ we have $c = - a - b$, so the vector $\langle a,b,c \rangle$ can be rewritten as $\langle a,b, - a - b\rangle$.

The assumptions imply that $\langle x,y,z\rangle$ is perpendicular to both $\langle a,b,c\rangle$ and $\langle 1,1,1\rangle$. So it has to be a multiple of the cross product $\langle a,b,c\rangle \times \langle 1,1,1\rangle$ $= \langle c - b, a - c, b - a\rangle$ $= \langle -a-2b, 2a + b, b - a\rangle$. Since the expression you are looking at is invariant under multiplication of $\langle x,y,z\rangle$ by a nonzero scalar, without loss of generality we can assume $\langle x,y,z\rangle = \langle -a-2b, 2a + b, b - a\rangle$.

So we have $$\langle a,b,c \rangle = \langle a,b, - a - b\rangle$$ $$\langle x,y,z\rangle = \langle -a-2b, 2a + b, b - a\rangle$$ At this point we can just insert these vectors into ${\displaystyle \frac{a^2}{a^2+b^2+c^2}+\frac{x^2}{x^2+y^2+z^2}}$ and obtain $$\frac{a^2}{2a^2 + 2ab + 2b^2 } + \frac{a^2 + 4ab + 4b^2}{6a^2 + 6ab + 6b^2}$$ $$= \frac{4a^2 + 4ab + 4b^2}{6a^2 + 6ab + 6b^2 }$$ $$=\frac{2}{3}$$

If $a+b+c=x+y+z=ax+by+cz=0$, then find value of $\frac{a^2}{a^2+b^2+c^2}+\frac{x^2}{x^2+y^2+z^2}$

3 Answers3