Find the cquation of the planc parallcl to line (x)/(1)=(y-7)/(-3)=(z+7)/(2) and containing the lines (x+1)/(-3)=(y-3)/(2)=(z+2)/(1) in vector and Cartesian form also find distance of plane from origin.
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Find the cquation of the planc parallcl to line (x)/(1)=(y-7)/(-3)=(z+7)/(2) and containing the lines (x+1)/(-3)=(y-3)/(2)=(z+2)/(1) in vector and Cartesian form also find distance of plane from origin.
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Step-by-step explanation:
Aloha !
x/1=y-7/-3=z+7/2--->1
x+2/-3=y-3/2=z+2/1--->2
=>> x/1=x+2/-3
=x+2=-3x
=x+2/-3x=1
=x/-3x+2/-3x=1
=-⅓+2/-3x
=-3/3x
=-1=x
:.x= -1
y-7/-3=y-3/2
-3y+9=2y-14
-3y-2y=-14-9
-5y=-23
y=23/5
Thank you
Step-by-step explanation:
[tex] \frac{x}{1} = \frac{y - 7}{ - 3} = \frac{z + 7}{2} ........(1)
\\
\frac{x + 2}{ - 3} = \frac{y - 3}{2} = \frac{z + 2}{1} ......(2) \\ => \frac{x}{1} = \frac{x + 2}{ - 3} \\
= > x+2=-3x \\ = \frac{x}{ - 3x} + \frac{2}{ - 3x} \\
=1
= \frac{1}{3} + \frac{2}{ - 3x}
= \frac{ - 3}{3x} \\
:.x= -1[/tex]
=x+2/-3x=1
=-1=x
y-7/-3=y-3/2
-3y+9=2y-14
-3y-2y=-14-9
-5y=-23
y=23/5