Question

Consider the line which passes through the point P(5, -5, 4), and which is parallel to the line

Answer 1

Answer:

intersection points are:

the intersection point at xy=plane : $\left(-1,-8,0\right)$

the intersection point at xz=plane : $\left(15,0,\dfrac{32}{3}\right)$

the intersection point at yz=plane : $\left(0,-\dfrac{15}{2},\dfrac{2}{3}\right)$

Step-by-step explanation:

a general equation of a line can be written as:

$\vec{r}=\vec{r_0}+t\vec{d}$

Here,  

$\vec{r_0}$: is the position vector

$\vec{d}$: is the direction vector

to find the equation of the line we need a point, and a direction vector.

since the line is parallel to the the line with components:

$x = 1 + 6t\\y=2+3t\\z=3+4t$

all we need to find is the direction vector of this line!  

we can see that it resembles our general line equation if we represent the same equations in vector form.

$\vec{r}=\vec{r_0}+t\vec{d}$

$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}+t\begin{bmatrix}6\\3\\4\end{bmatrix}$

by comparing the two equations above it is clear that the direction vector of the second line is:

$\vec{d}=\begin{bmatrix}6\\3\\4\end{bmatrix}$

and since both lines are parallel, this is the direction vector for our line as well.

the position vector of the line is also provided as the point P.

hence our equation is:

$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}5\\-5\\4\end{bmatrix}+t\begin{bmatrix}6\\3\\4\end{bmatrix}$

break them down into components

$x = 5 + 6t\\y=-5+3t\\z=4+4t$

Intersection points:

1) at the xy-plane, the coordinates are (x,y,0). so we can find the value of 't' when z = 0,

$z=4+4t$

$0=4+4t$

$t=-1$

we can use this value of t to find both the x and y coordinates:

$x =5+6t = 5+6(-1) = -1$

$y =-5+3t = -5+3(-1) = -8$

the intersection point at xy=plane is: $\left(-1,-8,0\right)$

2) at the xz-plane, the coordinates are (x,0,z). so we can find the value of 't' when y = 0,

$y=-5+3t$

$0=-5+3t$

$t=\dfrac{5}{3}$

we can use this value of t to find both the x and y coordinates:

$x =5+6t = 5+6\left(\dfrac{5}{3}\right) = 15$

$z =4+4t = 4+4\left(\dfrac{5}{3}\right) = \dfrac{32}{3}$

the intersection point at xz=plane is: $\left(15,0,\dfrac{32}{3}\right)$

3) at the yz-plane, the coordinates are (0,y,z). so we can find the value of 't' when x = 0,

$x=5+6t$

$0=5+6t$

$t=\dfrac{-5}{6}$

we can use this value of t to find both the x and y coordinates:

$y =-5+3t = -5+3\left(-\dfrac{5}{6}\right) = -\dfrac{15}{2}$

$z =4+4t = 4+4\left(-\dfrac{5}{6}\right) = \dfrac{2}{3}$

the intersection point at yz=plane is: $\left(0,-\dfrac{15}{2},\dfrac{2}{3}\right)$