Answer to Question #240783 in Analytic Geometry for Sakshi

Question

Give parametric equation (point-direction form) of the line which lies on both of the planes:

x + y + z = 1 and −x + 2y + 10z = 2. What is the direction d of this line?

b) Let n1 and n2 be the normal vectors to the two given planes. Without actual computation,

describe the relationship between d and n1 × n2.

Answer 1

ANSWER: a)The parametric equation of the line passing through $P_{ 0 }=(0,1,0)$ and parallel to the vector $\overrightarrow { d }$ :$x=\frac { 8 }{ \sqrt { 194 } } t,\ y=1-\frac { 11 }{ \sqrt { 194 } } t,\ z=\frac { 3 }{ \sqrt { 194 } } t$ . $\overrightarrow { d } =\frac { 1 }{ \sqrt { 194 } } \left< 8,\ -11,3 \right> \quad ,\left\| \overrightarrow { d } \right\| =1$

b)the vector $\overrightarrow { d }$ is perpendicular to both normal vector , $\overrightarrow { d } \quad \parallel \overrightarrow { \ n1 } \times \overrightarrow { n2 }$

EXPLANATION:

a)

The point $(a,b,c)$ lies on both of the planes, then $\begin{cases} a+b+c=1 \\ -a+2b+10c=2 \end{cases}$ . Let $c=0$ , we have

$\begin{cases} a+b=1 \\ -a+2b=2 \end{cases}\quad \Leftrightarrow \quad \begin{cases} a+b=1 \\ -a+a+b+2b=3 \end{cases}\Leftrightarrow \begin{cases} a\ =0 \\ \ b\ =1 \end{cases}\quad$ . Let $a=1$ , then $b+c=0$ and $-1+8c=2$ . Thus the points $(0,1,0),$ $(1,-3/8,3/8)$ lie on both planes . Therefore the vector $\overrightarrow { D } =\ \ \left( 1,\ -\frac { 3 }{ 8 } ,\frac { 3 }{ 8 } \right) -\left( 0,1,0 \right) =\left< 1,\ -\frac { 11 }{ 8 } ,\frac { 3 }{ 8 } \right>$ determines the direction

of the line of intersection of these planes .Since $\left\| \overrightarrow { D } \right\| =\sqrt { 1+\frac { 121 }{ 64 } +\frac { 9 }{ 64 } \quad } =\ \sqrt { \frac { 194 }{ 64 } }$ , then $\overrightarrow { d } =\frac { \overrightarrow { D } }{ \left\| \overrightarrow { D } \right\| } =\frac { 8 }{ \sqrt { 194 } } \left< 1,\ -\frac { 11 }{ 8 } ,\frac { 3 }{ 8 } \right> =\frac { 1 }{ \sqrt { 194 } } \left< 8,\ -11,3 \right> \ \ , \ \left\| \overrightarrow { d } \right\| =1.$ So, the parametric equations of the line passing through the point $P_{ 0 }=(0,1,0)$ and parallel to d the vector $\overrightarrow { d } \ :\quad$ $x=\frac { 8 }{ \sqrt { 194 } } t,\ \ y=1-\frac { 11 }{ \sqrt { 194 } } t,\ z=\frac { 3 }{ \sqrt { 194 } } t.$

b)

The plane $x+y+z=1$ have normal vector $\overrightarrow { n1 } =(1,1,1)$ and the plane $-x+2y+10z=2$ have normal vector $\overrightarrow { n2 } =(-1,2,10).$ The line is on both planes and thus the line (and the vector $\overrightarrow { d } )$ is perpendicular to both normal vectors.

$(verification:\ \overrightarrow { n1 } \cdot \overrightarrow { d } =\frac { 1 }{ \sqrt { 194 } } (8-11+3)=0,\ \overrightarrow { n2 } \cdot \overrightarrow { d } =\frac { 1 }{ \sqrt { 194 } } (-8-22+30)=0) \\$

So, $\overrightarrow { d } \ \parallel \overrightarrow { \ n1 } \times \overrightarrow { n2 }$ .