Question

The point of intersection of the line through the point (2, 1, 0) parallel to the vector u= (1, -1, 2) with the plane with equation r + y + 2z = 23 is:

Answer 1

Answer:

The point of intersection is P(7,-4,10).

Step-by-step explanation:

The idea with this exercise is to find the parametric equation of the line and then substitute it into the equation of the plane, which gives us a linear equation on $t$. After we solve it, we substitute the value of $t$ into the equation of the line and get the wanted point. This "path" may seem to be complicated, but it is not difficult at all.

First step: To find the (parametric) equation of the line. Recall that the parametric equation of a line has the form

$r(t) = \begin{cases}x&=x_0+v_1t\\y&=y_0+v_2t\\z&=z_0+v_3t\end{cases}$

where $(x_0,y_0,z_0)$ is any point of the given line and $(v_1,v_2,v_3)$ is the direction vector.

In this problem is easy to see that the we can choose $(x_0,y_0,v_0)= (2,1,0)$, because is a point on the line. Notice that the line is parallel to the vector $u=(1,-1,2)$, this, means that the direction vector of the line is parallel to $u$. Thus, we can use $u$ as direction vector. Therefore, the parametric equation of the line is

$r(t) = \begin{cases}x&=2+t\\y&=1-t\\z&=0+2t\end{cases}$.

Second step: Substitute $r(t)$ into the equation of the plane.

Notice that in the parametric equation we have an expression for each coordinate of the line. So, we substitute this expressions into the equation of the plane (here we are assuming that the plane is given by x + y + 2z=23):

$(2+t) + (1-t) + 2(0+2t)=23$.

Then, we get the linear equation:

$3+4t=23$.

From the above equation we obtain the solution $t=5$.

Third step: Find the point.

As we have a value fot $t$ we substitute it into the parametric equation of the line:

$r(5) = \begin{cases}x&=2+5\\y&=1-5\\z&=2\cdot 5\end{cases}$.

This gives us the point P(7,-4,10) that is on the line. Also, notice that the point P satisfies the equation of the plane:

$7-4+2\cdot 10 =23$.