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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$15, -5, 3$

Vedclass · Answer

(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

Similar Questions

The plane passing through the points $(1,1,1)$,$(1,-1,1)$ and $(-7,-3,-5)$ is

For which values of $a$ do the two points $(1, a, 1)$ and $(-3, 0, a)$ lie on opposite sides of the plane $3x + 4y - 12z + 13 = 0$?

$A$ bisector of the angle between the normals of the planes $4x + 3y = 5$ and $x + 2y + 2z = 4$ is along the vector

The distance between the two parallel planes $2x + y + 2z = 8$ and $4x + 2y + 4z + 5 = 0$ is: (in $/2$)

If a variable plane,at a distance of $3 \ units$ from the origin,intersects the coordinate axes at $A, B$,and $C$,then the locus of the centroid of $\Delta ABC$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module