Respuesta :
Answer:
The vector equation of the line is [tex]\overrightarrow {r}=(11j-8k)+t(4i-4j+6k)[/tex] and parametric equations for the line are [tex]x=4t[/tex], [tex]y=11-4t[/tex], [tex]z=-8+6t[/tex].
Step-by-step explanation:
It is given that the line passes through the point (0,11,-8) and parallel to the line
[tex]x=-1+4t[/tex]
[tex]y=6-4t[/tex]
[tex]z=3+6t[/tex]
The parametric equation are defined as
[tex]x=x_1+at,y=y_1+bt,z=z_1+ct[/tex]
Where, (x₁,y₁,z₁) is point from which line passes through and <a,b,c> is cosine of parallel vector.
From the given parametric equation it is clear that the line passes through the point (-1,6,3) and parallel vector is <4,-4,6>.
The required line is passes through the point (0,11,-8) and parallel vector is <4,-4,6>. So, the parametric equations for the line are
[tex]x=4t[/tex]
[tex]y=11-4t[/tex]
[tex]z=-8+6t[/tex]
Vector equation of a line is
[tex]\overrightarrow {r}=\overrightarrow {r_0}+t\overrightarrow {v}[/tex]
where, [tex]\overrightarrow {r_0}[/tex] is a position vector and [tex]\overrightarrow {v}[/tex] is cosine of parallel vector.
[tex]\overrightarrow {r}=(11j-8k)+t(4i-4j+6k)[/tex]
Therefore the vector equation of the line is [tex]\overrightarrow {r}=(11j-8k)+t(4i-4j+6k)[/tex] and parametric equations for the line are [tex]x=4t[/tex], [tex]y=11-4t[/tex], [tex]z=-8+6t[/tex].
