Question

PLEASE help.

https://gyazo.com/46c42540f516eb2ffaa897f7c05d1b3c

I wrote the parametric form of the symmetric form:
x=1+2t
y=-1-t
z=-2+t
Direction vector: (2,-1,1)
So Normal: (1,1,2)
x+y+2z+D=0
Then I'm confused. Any help or hints wil be much appreciated!!!

Answer 1

@Shadow

Answer 2

@Hero

Answer 3

@Angle would you mind checking this out when you get a chance?

Answer 4

Given
\(\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z+2}{1}\)
\(x-6y-2z-47=0\)

Answer 5

Step 1: Change the cartesian eqns into parametric eqns

Answer 6

u will get
\(x=1+2t\)
\(y=1-t\)
\(z=-2+t\)

Answer 7

Step 2 : Sub those parametric eqns into this eqn (\(x-6y-2z-47=0\)) to find the value of t

Answer 8

u will get
\((1+2t)-6(1-t)-2(-2+t)-47=0\)
\(1+2t-6+6t+4-2t-47=0\)

Answer 9

\(6t-48=0\)
\(6t=48\)
\(t=8\)

Answer 10

Step 3: Sub the value of t into the parametric eqns to find the value of \((x,y,z)\)

Answer 11

\(x=1+2(8)=17\)
\(y=1-8=-7\)
\(z=-2+8=6\)

Answer 12

Hope this helps...
u can try to double check my answer with other qc members.