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Give a point P(1, 0, -1) and a plane H: -4x + y + z = 4. A) Complete the equation of the line...

Question:

Give a point {eq}P(1, 0, -1) {/eq} and a plane {eq}H: -4x + y + z = 4 {/eq}.

A) Complete the equation of the line passing through {eq}P {/eq} and perpendicular to the plane {eq}H {/eq}. What is the distance from the origin to this line?

B) Compute the distance between the point and the plane.

Line and Plane in 3D:

The above question concerns the topic of the 3-dimensional geometry. The equation of the line passing through the point {eq}(x_1,y_1,z_1) {/eq}, having direction ratio {eq}(a,b,c) {/eq} is given by

{eq}\displaystyle \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} {/eq}.

Generalizsed equation of the plane is given by {eq}ax+by+cz+d=0 {/eq}, where {eq}(a,b,c) {/eq} is the direction ratio of the normal to the plane.

The distance between the point {eq}(x_1,y_1,z_1) {/eq} and plane {eq}ax+by+cz+d=0 {/eq} is given by

{eq}\displaystyle D=\frac{a(x_1)+b(y_1)+c(z_1)+d}{\sqrt{a^2+b62+c^2}} {/eq}.

Answer and Explanation:

Given a point {eq}P(1, 0, -1) {/eq} and a plane {eq}H: -4x + y + z = 4 {/eq}.

A) We have to find the equation of the line passing through {eq}P {/eq} and perpendicular to the plane {eq}H {/eq}.

Since line is perpendicular to the plane so it is parallel to the normal of the plane. The direction ratio of the normal to the plane is given as {eq}a=-4, \ b=1, \ c=1 {/eq}.

Now equation of the line passing through the point {eq}(1,0,-1) {/eq} with direction ratio {eq}a=-4, \ b=1, \ c=1 {/eq} is given by

{eq}\displaystyle \frac{x-1}{-4}= \frac{y-0}{1}= \frac{z-(-1)}{1} \\ \displaystyle \frac{x-1}{-4}= \frac{y}{1}= \frac{z+1}{1} \\ {/eq}

Now the point {eq}P(1,0,-1) {/eq} lies on XZ plane so minimum distance between origin and line is distance between origin and that point lies on XZ plane. So distance is

{eq}D_A=\sqrt{(1-0)^2+(0-0)^2+(-1-0)^2}\\ D_A=\sqrt{2} {/eq}

B) Distance between the point and the plane is given by

{eq}\displaystyle D_B=\left | \frac{-4(1)+(0)+(-1)}{\sqrt{(-4)^2+(1)^2+(1)^2}} \right |\\ \displaystyle D_B=\left | \frac{5}{\sqrt{18}} \right | {/eq}


Learn more about this topic:

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How to Find the Distance Between a Point & a Line

from FTCE Mathematics 6-12 (026): Practice & Study Guide

Chapter 30 / Lesson 5
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