Find the Intersection Point of Lines $(3,1,-2)$ and $x=-1+t, y=-2+t, z=-1+t$

[mathmari]

Hello!

I have to find the line that passes through $(3, 1, -2)$ and intersects under right angle the line $x=-1+t, y=-2+t, z=-1+t$.

(HINT: If $(x_0, y_0, z_0)$ the intersection point, find the coordinates.)

I have done the following:

The line that passes through $(3, 1, -2)$ is of the form $$\overrightarrow{l}(t)=(3, 1, -2)+t\overrightarrow{u}$$ where $\overrightarrow{u}$ is a vector parallel to the line.

Let $(x_0, y_0, z_0)$ be the intersection point of the line $\overrightarrow{l}$ and the line $x=-1+t, y=-2+t, z=-1+t$, then we have the folowing:

$$x_0=3+tu_1=-1+t, \\ y_0=1+tu_2=-2+t, \\ z_0=-2+tu_3=-1+t$$

Is it correct so far?? (Wondering)

How could I continue?? (Wondering)

Do we have to use the fact that the two lines intersect under right angle?? (Wondering)

 

[Prove It]

Hello!

I have to find the line that passes through $(3, 1, -2)$ and intersects under right angle the line $x=-1+t, y=-2+t, z=-1+t$.

(HINT: If $(x_0, y_0, z_0)$ the intersection point, find the coordinates.)

I have done the following:

The line that passes through $(3, 1, -2)$ is of the form $$\overrightarrow{l}(t)=(3, 1, -2)+t\overrightarrow{u}$$ where $\overrightarrow{u}$ is a vector parallel to the line.

Let $(x_0, y_0, z_0)$ be the intersection point of the line $\overrightarrow{l}$ and the line $x=-1+t, y=-2+t, z=-1+t$, then we have the folowing:

$$x_0=3+tu_1=-1+t, \\ y_0=1+tu_2=-2+t, \\ z_0=-2+tu_3=-1+t$$

Is it correct so far?? (Wondering)

How could I continue?? (Wondering)

Do we have to use the fact that the two lines intersect under right angle?? (Wondering)

If the two lines are at right angles, then so are their direction vectors. So the dot product of those direction vectors must be 0.

 

[mathmari]

If the two lines are at right angles, then so are their direction vectors. So the dot product of those direction vectors must be 0.

Do you mean the following??

$$(u_1, u_2, u_3) \cdot (1, 1, 1)=0$$ (Wondering)