The x-intercept of a plane pi passing through | vedclass

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Question

(A) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Given $a = \frac{5}{2}$,the equation becomes $\frac{2x}{5} + \frac

Vedclass · Accepted Answer

$-\frac{5}{3}$

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(A) Let the equation of the plane be $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Given $a = \frac{5}{2}$,the equation becomes $\frac{2x}{5} + \frac{y}{b} + \frac{z}{c} = 1$. Since the plane passes through $(1,1,1)$,we have $\frac{2}{5} + \frac{1}{b} + \frac{1}{c} = 1$,which implies $\frac{1}{b} + \frac{1}{c} = \frac{3}{5}$. The perpendicular distance from the origin $(0,0,0)$ to the plane $\frac{2x}{5} + \frac{y}{b} + \frac{z}{c} - 1 = 0$ is given by $\frac{|-1|}{\sqrt{(\frac{2}{5})^2 + (\frac{1}{b})^2 + (\frac{1}{c})^2}} = \frac{5}{7}$. Squaring both sides,we get $\frac{4}{25} + \frac{1}{b^2} + \frac{1}{c^2} = (\frac{7}{5})^2 = \frac{49}{25}$. Using $(\frac{1}{b} + \frac{1}{c})^2 = \frac{1}{b^2} + \frac{1}{c^2} + \frac{2}{bc}$,we substitute $\frac{1}{b^2} + \frac{1}{c^2} = \frac{49}{25} - \frac{4}{25} = \frac{45}{25} = \frac{9}{5}$. Thus,$(\frac{3}{5})^2 = \frac{9}{5} + \frac{2}{bc} \Rightarrow \frac{9}{25} - \frac{45}{25} = \frac{2}{bc} \Rightarrow \frac{2}{bc} = -\frac{36}{25} \Rightarrow bc = -\frac{50}{36} = -\frac{25}{18}$. Now,$\frac{b+c}{bc} = \frac{3}{5} \Rightarrow b+c = \frac{3}{5} 	imes (-\frac{25}{18}) = -\frac{5}{6}$. We have $b+c = -\frac{5}{6}$ and $bc = -\frac{25}{18}$. These are roots of $t^2 - (b+c)t + bc = 0$,i.e.,$t^2 + \frac{5}{6}t - \frac{25}{18} = 0$. Multiplying by $18$,$18t^2 + 15t - 25 = 0$. Solving for $t$,$(6t-5)(3t+5) = 0$,so $t = \frac{5}{6}$ or $t = -\frac{5}{3}$. Since the $y$-intercept $b$ is negative,$b = -\frac{5}{3}$.

The $x$-intercept of a plane $\pi$ passing through the point $(1,1,1)$ is $\frac{5}{2}$ and the perpendicular distance from the origin to the plane $\pi$ is $\frac{5}{7}$. If the $y$-intercept of the plane $\pi$ is negative and the $z$-intercept is positive,then its $y$-intercept is

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