Answer to Question #240067 in Analytic Geometry for sai

Question #240067

a) 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.


1
Expert's answer
2021-09-29T00:08:35-0400

ANSWER: a)The parametric equation of the line which lies on both of the planes is 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 }" .


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