If the plane x - 3y + 5z = d passes through t | vedclass

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(A) Given the plane equation is $x - 3y + 5z = d$. Since the plane passes through the point $(1, 2, 4)$,we substitute these coordinates into the equat

Vedclass · Accepted Answer

$15, -5, 3$

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(A) Given the plane equation is $x - 3y + 5z = d$. Since the plane passes through the point $(1, 2, 4)$,we substitute these coordinates into the equation: $1 - 3(2) + 5(4) = d$ $1 - 6 + 20 = d$ $d = 15$ Thus,the equation of the plane is $x - 3y + 5z = 15$. To find the intercepts,we divide the entire equation by $15$: $\frac{x}{15} + \frac{-3y}{15} + \frac{5z}{15} = \frac{15}{15}$ $\frac{x}{15} + \frac{y}{-5} + \frac{z}{3} = 1$ Comparing this with the intercept form of a plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$,the intercepts on the $x, y, z$ axes are $15, -5, 3$ respectively.

If the plane $x - 3y + 5z = d$ passes through the point $(1, 2, 4)$,then the lengths of intercepts cut by it on the axes of $x, y, z$ are respectively

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