Answer:
The point of intersection is [tex]\left[\begin{array}{c}4&-17&3\end{array}\right][/tex]
Step-by-step explanation:
In order to find the intersection point we will need to find the vector equation of the line.
The plane equation is [tex]y=-17[/tex]
Now to find the line that is perpendicular to this plane, we can do the following reasoning :
The plane and the line will be perpendicular between each other if and only if the direction vector of the line is a scalar multiple from the normal vector of the plane.
The plane equation is [tex]y=-17[/tex]. Then any vector with the following form :
[tex]\left[\begin{array}{c}0&a&0\end{array}\right][/tex]
With [tex]a[/tex] ∈ IR , [tex]a\neq 0[/tex] will be perpendicular to the plane.
Let's choose [tex]a=1[/tex] to simplify the calculations.
The normal vector of the plane is [tex]\left[\begin{array}{c}0&1&0\end{array}\right][/tex]
Now the direction vector of the line must be a scalar multiple of this vector ⇒ [tex]V=b\left[\begin{array}{c}0&1&0\end{array}\right][/tex] , [tex]b[/tex] ∈ IR
Let's also choose [tex]b=1[/tex] in order to simplify the calculations.
The vectorial equation of the line is :
[tex]\left[\begin{array}{c}x&y&z\end{array}\right]=\alpha V+P[/tex] with [tex]\alpha[/tex] ∈ IR
Where [tex]V[/tex] is the direction vector and [tex]P[/tex] is any point where the line passes through.
Using the data from the question we complete our equation with :
[tex]IL=\left[\begin{array}{c}x&y&z\end{array}\right]=\alpha\left[\begin{array}{c}0&1&0\end{array}\right]+\left[\begin{array}{c}4&8&3\end{array}\right][/tex] with [tex]\alpha[/tex] ∈ IR
Now let's find the intersection point between the plane [tex]y=-17[/tex] and the line [tex]IL[/tex]
This intersection point must belong to the line and also to the plane.
In order to belong to the plane its second component must be [tex]-17[/tex]
Knowing this and using the vector equation of the line we obtain this equation :
[tex]\alpha +8=-17[/tex] ⇒ [tex]\alpha =-25[/tex]
Using [tex]\alpha =-25[/tex] in the vector equation of the line we find that the point is
[tex]\left[\begin{array}{c}4&-17&3\end{array}\right][/tex]
This point belongs to the plane (because its second component is [tex]-17[/tex]) and also to the line (because it can be obtained using the vector line equation with [tex]\alpha =-25[/tex]).
