Find the distance of the point (2, 1, 0) from | vedclass

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

Question

(B) The distance $d$ of a point $(x_1, y_1, z_1)$ from a plane $ax + by + cz + d = 0$ is given by the formula: $d = \frac{|ax_1 + by_1 + cz_1 + d|}{\s

Vedclass · Accepted Answer

$10/3$

Vedclass · Answer

(B) The distance $d$ of a point $(x_1, y_1, z_1)$ from a plane $ax + by + cz + d = 0$ is given by the formula: $d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$ Given point is $(2, 1, 0)$ and the plane is $2x + y + 2z + 5 = 0$. Here,$a = 2, b = 1, c = 2, d = 5$ and $x_1 = 2, y_1 = 1, z_1 = 0$. Substituting these values into the formula: $d = \frac{|2(2) + 1(1) + 2(0) + 5|}{\sqrt{2^2 + 1^2 + 2^2}}$ $d = \frac{|4 + 1 + 0 + 5|}{\sqrt{4 + 1 + 4}}$ $d = \frac{|10|}{\sqrt{9}}$ $d = \frac{10}{3}$ Thus,the distance is $10/3$ units.

Find the distance of the point $(2, 1, 0)$ from the plane $2x + y + 2z + 5 = 0$.

Similar Questions

The distance of the point $(-1, 2, -2)$ from the line of intersection of the planes $2x + 3y + 2z = 0$ and $x - 2y + z = 0$ is:

The foot of the perpendicular from a point on the circle $x^{2} + y^{2} = 1, z = 0$ to the plane $2x + 3y + z = 6$ lies on which one of the following curves?

The length and foot of the perpendicular from the point $(7, 14, 5)$ to the plane $2x + 4y - z = 2$ are

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect,then $k$ is equal to:

The line $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is parallel to which of the following planes?

Vedclass Test Series

Exam Paper Generator

Online Exam Module