Question

HELp plz

Answer 1

@mathslover

Answer 2

Sorry, I have extra class after 8 minutes. I will try to give you a hint though .

Answer 3

hmmm k

Answer 4

but that's too typical for a 10th grader :) ...

Answer 5

how to solve ..

Answer 6

equations of line are
x-2z+3=0, or x+3=2z or
\[\frac{ x+3 }{ 2 }=z ...\left( 1 \right)\]
and y-z-4=0
or y-4=z ...(2)
from(1) and (2)
\[\frac{ x+3 }{ 2}=\frac{ y-4 }{ 1 }=\frac{ z }{1 } ...(3)\]
\[d.r's of this line are 2,1,1 \]
Any line through (2,1,-1) is
\[\frac{ x-2 }{ a }=\frac{ y-1 }{ b }=\frac{ z+1 }{c }=t \left( say \right) ...\left( 4 \right)\]
because (4) is perpendicular to (3)
2a+b+c=0 ...(5)
any point on (4) is x=at+2,y=bt+1,z=ct-1
If it lies on the plane x+2y+3z-1=0
at+2+2bt+4+3ct-3-1=0
orat+2bt+3ct=0
or a+2b+3c=0 (6)
solving (5) and (6)

from equation of line is
\[\frac{ x-2 }{ 1w }=\frac{ y-1 }{ -5w }=\frac{ z+1 }{ 3w }\]
\[\frac{ x-2 }{ 1 }=\frac{ y-1 }{ -5 }=\frac{ z+1 }{3 }\]

Answer 7

correction 5th line after no. (4)
at+2+2bt+2+3ct-3-1=0