Applications: Lines, Planes and Space Curves | Free Calculus 2 Lesson (4 College Credits) | UPI Study

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Applications: Lines, Planes and Space Curves

Calculus 2 · Chapter 10: Vectors & the Geometry of SpaceSection 10.4 · lesson 54 of 5410 min · free

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In this lesson

Putting both products to work.

  • Write a line from a point and a direction vector
  • Write a plane from a point and a normal vector
  • Decide whether a line meets a plane
  • Read a space curve as three ordinary functions

Transcript

Everything said in the video, section by section (1,492 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 10.4

Lines, Planes and Space Curves

You now have both vector products. So let's put them to work. This section covers four things. Lines in space. Planes in space. Deciding whether a line meets a plane. And space curves. And each one shows a product doing real work — rather than being exercised for its own sake.

Lines

A Point and a Direction

A line in space is determined by two pieces of information. One point on it. And one vector parallel to it. Starting at the position vector r naught, and traveling along the direction v for a scalar amount t, reaches every point of the line. That's the vector form. r of t equals r naught, plus t v. Writing it out component by component gives the parametric form. x equals x naught plus a t. y equals y naught plus b t. And z equals z naught plus c t. Where a, b and c are the components of the direction vector — called the direction numbers. Solving each of those for t, and equating, gives the symmetric form. x minus x naught over a, equals y minus y naught over b, equals z minus z naught over c. But note the restriction. That form requires dividing by each direction number. So it's valid only when none of them is zero. Let's do one. A descent corridor runs from A at one, two, minus three, to B at four, six, nine. The direction vector is B minus A. Three, four, twelve. So x equals one plus three t. y equals two plus four t. And z equals minus three plus twelve t. Check the endpoints. At t equals zero you get one, two, minus three. That's A. At t equals one you get four, six, nine. That's B. So the interval zero to one traces exactly the segment from A to B. And the corridor's total length is the magnitude of the direction vector. Root of nine plus sixteen plus one forty-four. Root one sixty-nine. Thirteen. So the descent covers thirteen hundred meters of straight-line distance.

Planes

A Point and a Perpendicular

A plane is harder to pin down by directions lying within it. Because a plane contains vectors pointing every which way. There's no single direction that characterizes it. So the efficient approach specifies instead the one direction that is perpendicular to all of them. A vector n perpendicular to a plane is called a normal vector. Now let P naught be a point in the plane. A point P lies in the same plane exactly when the vector from P naught to P lies in the plane. Which happens exactly when that vector is perpendicular to n. And perpendicular vectors have zero dot product. So the vector equation is: n dotted with r minus r naught, equals zero. Writing it in components gives the scalar, or point-normal, equation. A times x minus x naught, plus B times y minus y naught, plus C times z minus z naught, equals zero. And expanding gives the general form. A x plus B y plus C z equals D. Here's the fact worth memorizing. The coefficients of x, y and z in the general form are precisely the components of a normal vector. You can read the normal off at a glance. Let's model a sloped ridge. It passes through P naught at one, three, zero. With normal vector one, two, minus one. Substitute into the point-normal form. One times x minus one, plus two times y minus three, plus minus one times z minus zero, equals zero. Distribute. x minus one, plus two y minus six, minus z, equals zero. Collect the constants on the right. Minus one and minus six give minus seven. Add seven to both sides. x plus two y minus z equals seven. Verify. Substituting P naught: one plus six minus zero. Seven. Confirmed. And reading the coefficients back off recovers n equals one, two, minus one. Exactly as expected.

Worked Example 3

Does the Line Meet the Plane?

Now the question that ties them together. Does the descent path clear the ridge? Determine whether the corridor intersects the ridge plane. And locate the crossing point if it exists. The strategy is clean. The line supplies x, y and z as functions of t. And the plane imposes one condition on them. So substitute and solve. Substitute x equals one plus three t, y equals two plus four t, and z equals minus three plus twelve t, into x plus two y minus z equals seven. Expand each grouping. Distribute the two, and the leading minus sign. One, plus three t, plus four, plus eight t, plus three, minus twelve t, equals seven. Collect like terms. The constants give one plus four plus three. Eight. The t-terms give three t plus eight t minus twelve t. Which is minus t. So eight minus t equals seven. Solve. Subtracting eight gives minus t equals minus one. So t equals one. Locate the crossing. Substituting t equals one back into the parametric equations gives x equals four, y equals six, z equals nine. Verify in the plane equation: four plus twelve minus nine. Seven. Confirmed. Now interpret it. The path crosses the ridge surface at four, six, nine. Which is precisely the touchdown reference point B. And since t equals one marks the end of the descent segment, the corridor meets the ridge exactly at its endpoint. And nowhere earlier. So the approach clears the terrain along its entire length. The method generalizes nicely. Had the algebra produced a contradiction — something like eight equals seven — the line would be parallel to the plane, and never meet it. And had it produced an identity — like seven equals seven — the line would lie entirely within the plane.

Reference

Lines Versus Planes

Let's compare lines and planes directly, because the contrast is instructive. A line is determined by a point and a parallel vector. A plane, by a point and a perpendicular vector. The line's vector form is r naught plus t v. The plane's is n dotted with r minus r naught equals zero. The line's scalar form is three parametric equations. The plane's is a single point-normal equation. A line needs one parameter. A plane needs none, in scalar form. And to read the defining vector off the equation: for a line, take the coefficients of t. For a plane, take the coefficients of x, y and z. That contrast is worth holding onto. A line is pinned down by the direction it follows. A plane is pinned down by the direction it avoids.

Space Curves

Three Ordinary Functions of One Variable

Last topic. Space curves. A vector-valued function assigns a vector to each value of a real parameter. Written r of t equals f of t, g of t, h of t. And as t varies, the terminal point traces a path in three dimensions. A space curve. It's the three-dimensional generalization of the parametric curves from Chapter Nine. Let's trace one. The helix r of t equals three cosine t, three sine t, t over two. At t equals zero: three, zero, zero. At t equals pi, using cosine pi equals minus one and sine pi equals zero: minus three, zero, pi over two. And at t equals two pi, the first two components return to their starting values. Giving three, zero, pi. So the curve has completed one full revolution — and risen pi units. That rise is called the pitch of the helix. Look at the two pictures. On the left, viewed from directly above, the x and y components trace a circle of radius three. On the right, the three component functions plotted separately against t. The oscillating cosine and sine generate the circular motion. While the linear component produces the steady climb. And reading a space curve as three ordinary functions of one variable is the central idea carried forward into Calculus Three. Where these functions get differentiated and integrated, one component at a time.

Key Takeaways

What to Carry Forward

Five takeaways. A line needs a point and a parallel vector. Its direction numbers are the coefficients of t. And the symmetric form is valid only when none of them is zero. A plane needs a point and a normal vector. And in the general form, the coefficients of x, y and z are the normal's components. To intersect a line with a plane, substitute the parametric equations in and solve for t. A contradiction means the line is parallel to the plane. An identity means it lies within it. A vector-valued function traces a space curve, read as three ordinary functions of one parameter. And both products earned their place here. The dot product built the plane equation. And the cross product supplies normals, areas and volumes. That completes MAT two sixty-five. From the substitution rule in section one point one, all the way to space curves here. Integration techniques. Applications of integration. Differential equations. Sequences and series. And the geometry of space. Next comes Calculus Three — where these vector-valued functions get differentiated and integrated. Thanks for studying with us.

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FAQ

Questions students ask us

About Applications: Lines, Planes and Space Curves

How do you write the equation of a line in 3D from a point and a direction vector?
Start at a point r_0 and move along a direction vector v, which gives r(t) = r_0 + tv. In components, x = x_0 + at, y = y_0 + bt, z = z_0 + ct, where a, b, c are the direction numbers. For the line through (1,2,-3) and (4,6,9), use v = \langle 3,4,12 \rangle, so x = 1+3t, y = 2+4t, z = -3+12t.
How do you find the equation of a plane given a point and a normal vector?
Use the point-normal equation A(x-x_0) + B(y-y_0) + C(z-z_0) = 0, where \langle A,B,C \rangle is a normal vector and (x_0,y_0,z_0) is a point on the plane. For the point (1,3,0) and normal \langle 1,2,-1 \rangle, this gives (x-1) + 2(y-3) - z = 0, which simplifies to x + 2y - z = 7. In the general form Ax+By+Cz=D, the coefficients are the normal's components.
How do you tell if a line intersects a plane?
Substitute the line's parametric equations into the plane's equation and solve for t. A single solution gives the intersection point once you plug that t back into the line. For x=1+3t, y=2+4t, z=-3+12t and x+2y-z=7, you get 8-t=7, so t=1 and the point is (4,6,9). A contradiction such as 8=7 means the line is parallel to the plane. An identity means the line lies in it.
What is a space curve in calculus?
A space curve is the path traced by a vector-valued function r(t) = \langle f(t), g(t), h(t) \rangle, read as three ordinary functions of one parameter. For example, the helix r(t) = \langle 3\cos t, 3\sin t, t/2 \rangle traces a circle of radius 3 when viewed from above, while its z-component climbs steadily. From t=0 to t=2\pi it moves from (3,0,0) to (3,0,\pi), rising \pi units per revolution.

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