Find The Point At Which The Given Lines Intersect.

We thoroughly check each answer to a question to provide you with the most correct answers. Found a mistake? Let us know about it through the REPORT button at the bottom of the page.

Find the point at which the given lines intersect: r=<1, 1, 0>+t<1, -1, 2>, r=<2, 0, 2>+s<-1, 1, 0>

Answer: (2, 0, 2)

Explanation

Principles:

The vector equation of the two lines represents the set of all the points, through which the line passes through, where different points would have different values for the parametric constant.

Since the two lines intersect then there exists a point such that it satisfies both vector equations, which means that there exists some parametric constant $t$ and $s$ such that it would give the same vector, as we shall see in the calculation section.

Thus, in order to find the point of intersection, we solve both lines equations together in order to find the point of intersection.

But since the vector equation of the line is given, it would be easier to represent it in parametric equation form, where if we have our vector equation in terms of the parametric constant t, as follows:

< a + bt, c + dt, e + ft >

Then the parametric form of the equation for this line would be x = a + bt, y = c + dt, e + ft

 

Givens:

Representing the two given equations, in the parametric form we would have the following equation, for the first line we have

x = 1 + t, y = 1 – t, z = 2t

For the second line, we have

x = 2 – s, y = s, z = 2

In order to find the values of the parametric constant t and s, which would give the same point when substituted in their equations, we will simply equate each component and try to solve for t and s.

 

Calculations:

It’s pretty obvious, that we can solve both equations together by examining the z-component, we find that by equating both equations we get 2t = 2

Thus, the value of the parametric constant t is 1, and from the y-component substituting the value of the parametric constant t to be 1 and equating both equations we get 1 – 1 = s

Thus, the parametric constant s is equal to zero.

Substituting in the vector equation of each line, we get

< 1 + t, 1 – t, 2t >=<2, 0, 2>

And,

<2 – s, 0 + s, 2 >=<2, 0, 2>

Thus, this is the point of intersection between the two lines (2, 0, 2) where the vector between the origin and this point is the vector < 2, 0, 2 >.

Was this helpful?