Chapter 9.6, Problem 36E

a.

To determine

If the line is perpendicular to the plane, find the relationship between $vandn$ .

Answer to Problem 36E

The vector $v$ and $n$ will be parallel to each other.

Explanation of Solution

Given information:

A line is parallel to the vector $v$ , and a plane has normal vector $n$ .

Calculation:

Hence, it is given that vector $v$ is parallel to a line and the vector $n$ is normal to the plane.

If the line is perpendicular to the plane, the vector $v$ and the vector $n$ will be in same direction.

Hence the vector $v$ and the vector $n$ will be parallel to each other.

b.

To determine

If the line is parallel to the plane, find the relationship between $vandn$ .

Answer to Problem 36E

Vector $v$ and $n$ will be perpendicular to each other.

Explanation of Solution

Given information:

A line is parallel to the vector $v$ , and a plane has normal vector $n$ .

Calculation:

We know that the vector $v$ is parallel to the line and the vector $n$ is perpendicular to the plane.

Hence, if the line is parallel to the plane, then the vector $v$ and the vector $n$ will be perpendicular to each other.

c.

To determine

Find the parallel and perpendicular line to the plane.

Answer to Problem 36E

The vector $p$ is parallel to the vector $s$ .

The vectors $r\&s$ are perpendicular to each other.

Explanation of Solution

Given information:

A line is parallel to the vector $v$ , and a plane has normal vector $n$ .

Prove line is parallel to the plane $x-y+4z=6$ as per parametric equations.

  $\begin{array}{l}Line1:x=2t,y=3-2t,z=4+8t\\ Line2:x=-2t,y=5+2t,z=3+t\end{array}$

Calculation:

Consider ,

  $x=2t,y=3-2t,z=4+8t,and,x=-2t,y=5+2t,z=3+t$

We know that a line passing through the point $P(x_0,y_0,z_0)$ and parallel to the vector $<a,b,c>$ is given by the parametric equations

  $\begin{array}{l}x=x_0+at,\\ y=y_0+bt,\&\\ z=z_0+ct.\end{array}$

Compare the line

  $\begin{array}{l}x=2t,\\ y=3-2t,\&\\ z=4+8t,\end{array}$

With standard parametric equations of a line, find the line is passing through $(0,3,4)$ and parallel to the vector $p=<2,-2,8>$

On comparing the line

  $\begin{array}{l}x=-2t\\ y=5+2t,\&\\ z=3+t\end{array}$

With standard parametric equations of a line, find the line is passing through $(0,5,3)$ and parallel to the vector $r=<-2,-2,1>$

Compare the plane $x-y+4z=6$ with standard equation of the plane, passing through the point $\left(x_0,y_0,z_0\right)$ and perpendicular to the vector $<a,b,c>$

  $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ ,we find the plane $x-y+4z=6$ is perpendicular to the vector $s=<1,-1,4>$

Two vectors are said to be parallel, if one vector is the scalar multiply of other vector.

We see that

  $\begin{array}{l}p=<2,-2,8>\\ =2<1,-1,4>\\ =2s\end{array}$

Hence, the vector $p$ is parallel to the vector $s$ .

Since, line $x=2t,y=3-2t\&z=4+8t$ is parallel to the plane $x-y+4z=6$ .

Two vectors are said to be perpendicular, if their dot product is zero.

The dot product of vectors $a=\langle a_1,a_2,a_3\rangle \&b=\langle b_1,b_2,b_3\rangle$ is given by

  $\begin{array}{l}a.b=\langle a_1,a_2,a_3\rangle .\langle b_1,b_2,b_3\rangle \\ =a_1{b}_1+a_2{b}_2+a_3{b}_3.\end{array}$

The dot product of vectors $r=\langle -2,2,1\rangle \&s\langle 1,-1,4\rangle$ is

  $\begin{array}{l}r.s=\langle -2,2,1\rangle .\langle 1,-1,4\rangle \\ =(-2)(1)+2(-1)+1(4)\\ =-2-2+4\\ =0\end{array}$

Hence, the vectors $r\&s$ are perpendicular to each other.

Hence, the line $x=-2t,y=5+2t\&z=3+t$ is perpendicular to the

plane $x-y+4z=6$ .