The distance between the planes x + 2y + 3z + | vedclass

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(A) The given equations of the planes are $x + 2y + 3z + 7 = 0$ and $2x + 4y + 6z + 7 = 0$. First,we rewrite the second equation to have the same coef

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$\frac{\sqrt{7}}{2\sqrt{2}}$

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(A) The given equations of the planes are $x + 2y + 3z + 7 = 0$ and $2x + 4y + 6z + 7 = 0$. First,we rewrite the second equation to have the same coefficients for $x, y, z$ as the first plane by dividing it by $2$: $x + 2y + 3z + \frac{7}{2} = 0$. The distance $d$ between two parallel planes $Ax + By + Cz + D_1 = 0$ and $Ax + By + Cz + D_2 = 0$ is given by the formula $d = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$. Here,$A = 1, B = 2, C = 3, D_1 = 7$,and $D_2 = \frac{7}{2}$. Substituting these values into the formula: $d = \frac{|7 - \frac{7}{2}|}{\sqrt{1^2 + 2^2 + 3^2}} = \frac{|\frac{14 - 7}{2}|}{\sqrt{1 + 4 + 9}} = \frac{\frac{7}{2}}{\sqrt{14}}$. Simplifying the expression: $d = \frac{7}{2\sqrt{14}} = \frac{7}{2\sqrt{2} 	imes \sqrt{7}} = \frac{\sqrt{7}}{2\sqrt{2}}$. Thus,the correct option is $A$.

The distance between the planes $x + 2y + 3z + 7 = 0$ and $2x + 4y + 6z + 7 = 0$ is

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