Chapter 10: Q 39. (page 846)

Find the distance from the point $(2, 0, - 3)$ to the plane $3 x - 4 y + 5 z = 1$

Short Answer

Expert verified

The distance from the point to the plane is $\sqrt{2}$

Step by step solution

Given information

The planes $3 x - 4 y + 5 z = 1$ and the point $P = (2, 0, - 3)$

Calculation

The goal is to calculate the distance between a specified point and the plane. The distance between a point and the plane is calculated as follows:

$|{comp}_{N} \vec{R P}| = \frac{| N \cdot \vec{R P} |}{‖ N ‖}$

The normal vector of the plane $3 x - 4 y + 5 z = 1$ is:

$N = ⟨ 3, - 4, 5 ⟩$

The point $R = (0, 0, \frac{1}{5})$ is on the plane $3 x - 4 y + 5 z = 1$

The vector $\vec{R P}$ is given by:

$\vec{R P} = <2 - 0, 0 - 0, - 3 - \frac{1}{5}> = <2, 0, - \frac{16}{5}>$

Calculation

The distance from $P$ to the plane is:

|{comp}_{N} \vec{R P}| = \frac{| N \cdot \vec{R P} |}{‖ N ‖} = \frac{|⟨ 3, - 4, 5 ⟩ \cdot <2, 0, - \frac{16}{5})|}{| | (3, - 4, 5) ∣} (Substitution) = \frac{| 6 + 0 - 16 |}{\sqrt{9 + 16 + 25}} (Dot product) = \frac{| - 10 |}{\sqrt{50}} = \frac{10}{5 \sqrt{2}} (Simplify) = \frac{2}{\sqrt{2}} = \sqrt{2}

Therefore, the distance from the point to the plane is $\sqrt{2}$

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Find the distance from the point $(2, 0, - 3)$ to the plane $3 x - 4 y + 5 z = 1$

Short Answer

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Given information

Calculation

Calculation

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Find the distance from the point (2,0,−3) to the plane 3x−4y+5z=1

Short Answer

Step by step solution

Given information

Calculation

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Applied Mathematics

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Logic and Functions

Study anywhere. Anytime. Across all devices.

Find the distance from the point $(2, 0, - 3)$ to the plane $3 x - 4 y + 5 z = 1$