The planes: 2x - y + 4z = 5 and 5x - 2.5y + 1 | vedclass

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

Question

(D) The equations of the planes are:$2x - y + 4z = 5$  $(1)$$5x - 2.5y + 10z = 6$  $(2)$For two planes $a_1x + b_1y + c_1z = d_1$ and $a_2x + b_2y + c

Vedclass · Accepted Answer

Parallel

Vedclass · Answer

(D) The equations of the planes are:$2x - y + 4z = 5$  $(1)$$5x - 2.5y + 10z = 6$  $(2)$For two planes $a_1x + b_1y + c_1z = d_1$ and $a_2x + b_2y + c_2z = d_2$,they are parallel if $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.Calculating the ratios of the coefficients of $x, y,$ and $z$:$\frac{a_1}{a_2} = \frac{2}{5} = 0.4$$\frac{b_1}{b_2} = \frac{-1}{-2.5} = \frac{1}{2.5} = 0.4$$\frac{c_1}{c_2} = \frac{4}{10} = 0.4$Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = 0.4$,the normal vectors of the planes are proportional.Also,check if the planes are coincident: $\frac{d_1}{d_2} = \frac{5}{6} 
eq 0.4$. Since the ratios of the coefficients are equal but not equal to the ratio of the constants,the planes are parallel and distinct.Therefore,the correct answer is $D$.

The planes: $2x - y + 4z = 5$ and $5x - 2.5y + 10z = 6$ are

Similar Questions

Find the equation of the plane passing through the points $(2, 2, 1)$ and $(9, 3, 6)$ and perpendicular to the plane $2x + 6y + 6z - 1 = 0$.

The Cartesian equation of the plane,passing through the points $(3,1,1)$,$(1,2,3)$ and $(-1,4,2)$,is

If $\alpha$ and $\beta$ are scalars and $\vec{r} = (2+\alpha-3\beta) \hat{i} + (\beta-3) \hat{j} + (2\alpha-5\beta-1) \hat{k}$ is the equation of a plane, then its equation in Cartesian form is:

The angle between the planes $x+y+2z=6$ and $2x-y+z=9$ is:

If the plane passing through the points $(1, 2, 3)$, $(2, 3, 1)$, and $(3, 1, 2)$ is $a x + b y + c z = 1$, then $a + 2 b + 3 c = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module