If the plane 2x + 3y + 5z = 1 intersects the | vedclass

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

Question

(C) The given equation of the plane is $2x + 3y + 5z = 1$. To find the intercepts on the coordinate axes,we rewrite the equation in the intercept form

Vedclass · Accepted Answer

$\left(\frac{1}{6}, \frac{1}{9}, \frac{1}{15}\right)$

Vedclass · Answer

(C) The given equation of the plane is $2x + 3y + 5z = 1$. To find the intercepts on the coordinate axes,we rewrite the equation in the intercept form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. This gives $\frac{x}{(1/2)} + \frac{y}{(1/3)} + \frac{z}{(1/5)} = 1$. Thus,the intercepts on the $X, Y, Z$ axes are $a = 1/2$,$b = 1/3$,and $c = 1/5$ respectively. The coordinates of the points $A, B, C$ are $A = (1/2, 0, 0)$,$B = (0, 1/3, 0)$,and $C = (0, 0, 1/5)$. The centroid of $	riangle ABC$ is given by the formula $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}ight)$. Substituting the coordinates,we get $\left(\frac{1/2 + 0 + 0}{3}, \frac{0 + 1/3 + 0}{3}, \frac{0 + 0 + 1/5}{3}ight) = \left(\frac{1}{6}, \frac{1}{9}, \frac{1}{15}ight)$. Therefore,the correct option is $C$.

If the plane $2x + 3y + 5z = 1$ intersects the coordinate axes at the points $A, B, C$,then the centroid of $\triangle ABC$ is

Similar Questions

$A$ plane $\pi$ passing through the point $(1,1,1)$ is perpendicular to the line joining the points $(6,3,2)$ and $(1,-4,-9)$. If $ax+by+cz-23=0$ is the equation of the plane $\pi$,then $a+b-c=$

$A$ plane passes through the point $A(2, 1, -3)$. If the distance of this plane from the origin is maximum,then its equation is

The equation of the plane in normal form which passes through the points $(-2,1,3), (1,1,1)$ and $(2,3,4)$ is

Find the vector equation of the plane passing through the points $R(2, 5, -3)$,$S(-2, -3, 5)$,and $T(5, 3, -3)$.

Let $\pi$ be the plane that passes through the point $(-2, 1, -1)$ and is parallel to the plane $2x - y + 2z = 0$. Then the foot of the perpendicular drawn from the point $(1, 2, 1)$ to the plane $\pi$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module