E(1,0,0), F(0,2,0), G(0,0,3) are respectively | vedclass

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(A) Let the vertices of $	riangle ABC$ be $A(x_1, y_1, z_1), B(x_2, y_2, z_2)$ and $C(x_3, y_3, z_3)$.Given that $E, F, G$ are mid-points of $AB, BC,

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$\frac{26}{41}$

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(A) Let the vertices of $	riangle ABC$ be $A(x_1, y_1, z_1), B(x_2, y_2, z_2)$ and $C(x_3, y_3, z_3)$.Given that $E, F, G$ are mid-points of $AB, BC, CA$ respectively:$\frac{x_1+x_2}{2}=1, \frac{y_1+y_2}{2}=0, \frac{z_1+z_2}{2}=0$ $(1)$$\frac{x_2+x_3}{2}=0, \frac{y_2+y_3}{2}=2, \frac{z_2+z_3}{2}=0$ $(2)$$\frac{x_3+x_1}{2}=0, \frac{y_3+y_1}{2}=0, \frac{z_3+z_1}{2}=3$ $(3)$Solving these, we get $A(1, -2, 3), B(1, 2, -3), C(-1, 2, 3)$.Direction ratios of $AF$ (where $F$ is $(0, 2, 0)$): $a_1 = 0-1 = -1, b_1 = 2-(-2) = 4, c_1 = 0-3 = -3$. Thus $a_1^2+b_1^2+c_1^2 = (-1)^2+4^2+(-3)^2 = 1+16+9 = 26$.Direction ratios of $BG$ (where $G$ is $(0, 0, 3)$): $a_2 = 0-1 = -1, b_2 = 0-2 = -2, c_2 = 3-(-3) = 6$. Thus $a_2^2+b_2^2+c_2^2 = (-1)^2+(-2)^2+6^2 = 1+4+36 = 41$.Therefore, $\frac{a_1^2+b_1^2+c_1^2}{a_2^2+b_2^2+c_2^2} = \frac{26}{41}$.

$E(1,0,0), F(0,2,0), G(0,0,3)$ are respectively the mid-points of the sides $AB, BC, CA$ of $\triangle ABC$. If $a_1, b_1, c_1$ and $a_2, b_2, c_2$ are respectively the direction ratios of $AF$ and $BG$, then $\frac{a_1^2+b_1^2+c_1^2}{a_2^2+b_2^2+c_2^2}=$

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