Question

The plane containing the line
x-1/1=y-2/2=z-3/3 and parallel to the line x/1=y/1=z/4 passes through the point
A.(1,0,5)
B.(0,3,-5)
C.(-1,-3, 0)
(1,-2, 5)

Answer 1

here is my reasoning:
the general equation of a plane is:
\[ax + by + cz = d,\quad a,b,c,d \in \mathbb{R}\]

Answer 2

so, we have to compute the parameters \(a,\;b,\;c,\;d\)

Answer 3

from general theory, we know that the vector \((a,b,c)\), is orthogonal with respect to the plane:

Answer 4

Yep....

Answer 5

next:
the parametric equations of parallel line, are:
\[\left\{ {\begin{array}{*{20}{c}}
{x = t} \\
{y = t} \\
{z = 4t}
\end{array}} \right.\]
so, its direction vector is:
\((1,1,4)\)
and we can write:
\[\left( {a,b,c} \right) \cdot \left( {1,1,4} \right) = 0 \Rightarrow a + b + 4c = 0\;\left( * \right)\]

Answer 6

Yep right...

Answer 7

next:
the contained line has the subsequent parametric equations:
\[\left\{ {\begin{array}{*{20}{c}}
{x = 1 + t} \\
{y = 2t + 2} \\
{z = 3t + 3}
\end{array}} \right.\]

Answer 8

the contained line does belong to the plane, if and only if, the subsequent conditions hold:
\[a\left( {1 + t} \right) + b\left( {2t + 2} \right) + c\left( {3t + 3} \right) = d\]

Answer 9

from which I get:
\[\left( {a + 2b + 3c} \right)t + a + 2b + 3c = d\]

Answer 10

therefore, I can write:
\[\left\{ \begin{gathered}
a + 2b + 3c = 0 \hfill \\
\hfill \\
a + 2b + 3c = d \hfill \\
\end{gathered} \right.\]
which tell us that \(d=0\), and:
\[a + 2b + 3c = 0\;\left( {**} \right)\]

Answer 11

now, I have to solve the system of equations \((*)\) and \((**)\):
\[\left\{ \begin{gathered}
a + b + 4c = 0 \hfill \\
a + 2b + 3c = 0\; \hfill \\
\end{gathered} \right.\]

Answer 12

one of the corresponding solution (please note that we have \(\infty\) solutions), is:
\[a = - 5,\;b = 1,\;c = 1\]

Answer 13

and the requested plane is:
\[ - 5x + y + z = 0\]

Answer 14

so, what can you conclude?

Answer 15

First option....:)

Answer 16

that's right!! :)

Answer 17

Thanks sir!!!

Answer 18

:)