The distance of the plane 6x - 3y + 2z - 14 = | vedclass

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

Question

(A) The distance of a plane $ax + by + cz + d = 0$ from the origin $(0, 0, 0)$ is given by the formula: Distance $= \frac{|d|}{\sqrt{a^2 + b^2 + c^2}}

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(A) The distance of a plane $ax + by + cz + d = 0$ from the origin $(0, 0, 0)$ is given by the formula: Distance $= \frac{|d|}{\sqrt{a^2 + b^2 + c^2}}$ Here,the equation of the plane is $6x - 3y + 2z - 14 = 0$. Comparing this with the standard form,we have $a = 6$,$b = -3$,$c = 2$,and $d = -14$. Substituting these values into the formula: Distance $= \frac{|-14|}{\sqrt{6^2 + (-3)^2 + 2^2}}$ $= \frac{14}{\sqrt{36 + 9 + 4}}$ $= \frac{14}{\sqrt{49}}$ $= \frac{14}{7} = 2$. Thus,the distance of the plane from the origin is $2$ units.

The distance of the plane $6x - 3y + 2z - 14 = 0$ from the origin is

Similar Questions

If planes $x - c y - b z = 0$,$c x - y + a z = 0$ and $b x + a y - z = 0$ pass through a straight line then $a^2 + b^2 + c^2 =$

The sum of intercepts of the plane $4x + 3y + 2z = 2$ on the coordinate axes is

The equation of the plane passing through the point $(-1, 3, 2)$ and perpendicular to each of the planes $x + 2y + 3z = 5$ and $3x + 3y + z = 0$ is:

Find the equation of the plane passing through the intersection of the planes $\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 5$ and $\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 3$,and passing through the point $(2, 1, -2)$.

Find the equation of the plane passing through the intersection of the planes $3x - y + 2z - 4 = 0$ and $x + y + z - 2 = 0$ and the point $(2, 2, 1)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module