The length of the perpendicular from the orig | vedclass

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Question

(A) The intercept form of a plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b, c$ are the intercepts on the $X, Y, Z$ axes r

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$\frac{1}{5 \sqrt{2}}$

Vedclass · Answer

(A) The intercept form of a plane is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,where $a, b, c$ are the intercepts on the $X, Y, Z$ axes respectively. Given intercepts are $a = \frac{1}{3}, b = \frac{1}{4}, c = \frac{1}{5}$. Substituting these values,the equation of the plane is $\frac{x}{1/3} + \frac{y}{1/4} + \frac{z}{1/5} = 1$,which simplifies to $3x + 4y + 5z = 1$ or $3x + 4y + 5z - 1 = 0$. The length of the perpendicular from the origin $(0, 0, 0)$ to the plane $Ax + By + Cz + D = 0$ is given by $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$. Here,$A = 3, B = 4, C = 5$ and $D = -1$. Thus,$d = \frac{|-1|}{\sqrt{3^2 + 4^2 + 5^2}} = \frac{1}{\sqrt{9 + 16 + 25}} = \frac{1}{\sqrt{50}} = \frac{1}{5 \sqrt{2}}$.

The length of the perpendicular from the origin to the plane which makes intercepts $\frac{1}{3}, \frac{1}{4}$ and $\frac{1}{5}$ respectively on the coordinate axes is

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