The equation of a plane parallel to the x-axi | vedclass

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

Question

(C) The general equation of a plane is given by $ax + by + cz + d = 0$,where $(a, b, c)$ are the direction ratios of the normal to the plane.If a plan

Vedclass · Accepted Answer

$by + cz + d = 0$

Vedclass · Answer

(C) The general equation of a plane is given by $ax + by + cz + d = 0$,where $(a, b, c)$ are the direction ratios of the normal to the plane.If a plane is parallel to the $x$-axis,its normal must be perpendicular to the $x$-axis.The direction ratios of the $x$-axis are $(1, 0, 0)$.Since the normal $(a, b, c)$ is perpendicular to the $x$-axis $(1, 0, 0)$,their dot product must be zero:$a(1) + b(0) + c(0) = 0 \implies a = 0$.Substituting $a = 0$ into the general equation,we get $0x + by + cz + d = 0$,which simplifies to $by + cz + d = 0$.Therefore,the correct option is $C$.

The equation of a plane parallel to the $x$-axis is:

Similar Questions

The distance between the parallel planes $2x - 2y + z + 3 = 0$ and $4x - 4y + 2z + 5 = 0$ is:

The equation of the plane passing through $3 \hat{i}+2 \hat{j}+6 \hat{k}$ and parallel to the vectors $2 \hat{i}+\hat{j}+\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$ is

If the equation of the plane passing through the point $A(-2, 1, 3)$ and perpendicular to the vector $3 \hat{i} + \hat{j} + 5 \hat{k}$ is $ax + by + cz + d = 0$,then $\frac{a + b}{c + d} = $

Find the equation of a plane,given that the foot of the perpendicular drawn to the plane from the origin is $(2, 1, 2)$.

If the vector equation of the plane $\bar{r}=(2 \hat{i}+\hat{k})+\lambda \hat{i}+\mu(\hat{i}+2 \hat{j}-3 \hat{k})$ in scalar product form is given by $\bar{r} \cdot(3 \hat{j}+2 \hat{k})=\alpha$,then $\alpha=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module