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Determine the volume of the parallelepiped that has a, b, and c as sides. Given the plane P: 3x -...

Question:

Determine the volume of the parallelepiped that has a, b, and c as sides.

Given the plane P: 3x - y + 4z = 3, the line {eq}L: \frac{x - 1}{-2} = \frac{y + 4}{2} =\frac{ z + 3}{2} {/eq}, and the point A(-3, 0, 5).

a) Determine whether P and L are parallel.

b) Determine whether A lies on L. If it doesn't, find the distance from A to L.

c) Determine the parametric equations for the line M that passes through A and is perpendicular to P.

d) Find the point of intersection of M and P.

Condition To Check If A Line Is Parallel To A Plane:


Given a plane {eq}\displaystyle px + qy + rz = s {/eq} and a line given by {eq}\displaystyle \frac{x-x_0}{P}=\frac{y-y_0}{Q}=\frac{z-z_0}{R} {/eq} will be parallel to each other if,

$$\displaystyle \boxed{p \cdot P+q \cdot Q+ r \cdot R=0} $$

Answer and Explanation:


Volume of a parallelepiped having sides {eq}\displaystyle a,b,c {/eq} is given by the formula,

{eq}\displaystyle V=abc\sqrt{1+2\cos(\alpha)\cos(\beta)\cos(\gamma)-\cos^2(\alpha)-\cos^2(\beta)-\cos^2(\gamma)} {/eq}

where, {eq}\displaystyle \alpha, \beta , \gamma {/eq} are the internal angles between the edges {eq}\displaystyle a \text{ and }b {/eq}, {eq}\displaystyle b\text{ and }c {/eq} and {eq}\displaystyle c\text{ and }a {/eq} respectively.


a.

Considering the given plane {eq}\displaystyle 3x - y + 4z = 3 {/eq} and comparing it with the standard form given by, {eq}\displaystyle ax+by+cz=d {/eq} we get {eq}\displaystyle a=3, b=-1, c=4, d=3 {/eq}. Also on comparing the given line {eq}\displaystyle \frac{x - 1}{-2} = \frac{y + 4}{2} =\frac{ z + 3}{2} {/eq} with the standard form {eq}\displaystyle \frac{x-x_0}{A}=\frac{y-y_0}{B}=\frac{z-z_0}{C} {/eq} we get {eq}\displaystyle A=-2, B=2 , C=2 {/eq} and also {eq}\displaystyle x_0=1, y_0=-4, z_0=-3 {/eq}. Now for the line L to be parallel to the plane P the following condition must be satisfied,

$$\displaystyle \boxed{a \cdot A+b \cdot B+ c \cdot C=0} $$

For the given line and plane we have,

{eq}\displaystyle (3)(-2)+(-1)(2)+(4)(2)\\ =-6-2+8\\ =0 {/eq}

Hence the line L is parallel to the plane P.


b.

For the point {eq}\displaystyle A(-3,0,5) {/eq} to lie on the given line {eq}\displaystyle \frac{x - 1}{-2} = \frac{y + 4}{2} =\frac{ z + 3}{2} {/eq} all the ratios of the symmetric equation of line must be satisfied. Let us check this,

{eq}\displaystyle \text{ For x: }\frac{(-3)-1}{-2} =\frac{-4}{-2}=2\\ \displaystyle \text{ For y: }\frac{(0) + 4}{2} =\frac{4}{2}=2\\ \displaystyle \text{ For z: }\frac{ (5) + 3}{2}=\frac{8}{2}=4 {/eq}

As all the ratios are not equal, hence we conclude that the point A does not lie on the line L. Now the vector representing line L will be {eq}\displaystyle \left\langle x_0,y_0,z_0\right\rangle i.e. \left\langle 1,-4,-3\right\rangle {/eq}. Also the vector of point A will be {eq}\displaystyle \left\langle -3,0,5\right\rangle {/eq}. So the distance between the point A and the line L will be given by,

{eq}\displaystyle \begin{align} d&=\frac{\left|(\left\langle -3,0,5 \right\rangle - \left\langle 1,-4,-3\right\rangle) \times \left\langle A,B,C \right\rangle\right| }{\left|\left\langle A,B,C\right\rangle \right|}\\ &=\frac{\left|\left\langle(-3)-(1),(0)-(-4),(5)-(-3)\right\rangle\times \left\langle-2,2,2\right\rangle\right|}{\left\langle -2,2,2\right\rangle } \\ &=\frac{\left|\left\langle -4,4,8\right\rangle\times \left\langle-2,2,2\right\rangle\right|}{\sqrt{(-2)^2+(2)^2+(2)^2}} \\ &=\frac{\left| \left\langle-8,8,0\right\rangle\right|}{2\sqrt{3}}\\ &=\frac{\sqrt{(-8)^2+(8)^2+(0)^2}}{2\sqrt{3}}\\ &=\frac{8\sqrt{2}}{2\sqrt{3}}\\ &=4\sqrt{\frac{2}{3}} \end{align} {/eq}


c.

The normal vector of the given plane will be, {eq}\displaystyle \vec{n}=\left\langle 3,-1,4\right\rangle {/eq}. So the line perpendicular to this plane an passing through the point {eq}\displaystyle A(-3,0,5) {/eq} will be give by,

{eq}\displaystyle \frac{x-(-3)}{3}=\frac{y-0}{-1}=\frac{z-5}{4}\quad \quad \Rightarrow \quad \quad\boxed{\frac{x+3}{3}=\frac{-y}{1}=\frac{z-5}{4}} {/eq}

Now the equation in the parametric form will be given as,

{eq}\displaystyle \frac{x+3}{3}=t\quad \Rightarrow \quad x=3t-3\\ \displaystyle \frac{-y}{1}=t\quad \Rightarrow \quad y=-t\\ \displaystyle \frac{z-5}{4}=t\quad \Rightarrow \quad z=4t+5 {/eq}


d.

To find the point of intersection of the line M and the given plane P, we need to substitute the equation of the line in the plane. So we get,

{eq}\displaystyle \begin{align} &3(3t-3)-(-t)+4(4t+5)=3\\ \therefore \ &9t-9+t+16t+20=3\\ \therefore \ &26t=-13\\ \therefore \ &\boxed{t=\frac{-1}{2}} \end{align} {/eq}

Now the point of intersection will be given by,

{eq}\displaystyle x=3\left(\frac{-1}{2}\right)-3=\frac{-9}{2}\\ \displaystyle y=-\left(\frac{-1}{2} \right)=\frac{1}{2}\\ \displaystyle z=4\left( \frac{-1}{2}\right)+5=3 {/eq}


Learn more about this topic:

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Parametric Equations in Applied Contexts

from Precalculus: High School

Chapter 24 / Lesson 6
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