A line in 3D is determined when we know any point on the line and the direction of the line

What two things are needed in order to describe a line?

If given a point on the line L and a vector parallel to the line then the triangle law shows how the position vector for any other point on the line can be found

How is the triangle law used to determine the equation of a line?

Change the vector equation of the line into Cartesian form and isolate each basis component (x,y,z). These equations give the Cartesian components of a point on the line depending on the value of the parameter t

How are the parametric equations of of a line found and what is their significance?

Identify the given point as r₀ and the parallel vector as v and plug into the vector equation of a line and simplify. To find additional points on the line simply assign random numbers to the parameter t

Find a vector equation and parametric equations for the line that passes through

the point (5,1,3) and is parallel to the vector

i + 4j – 2k

Also, find another point on the line

the point (5,1,3) and is parallel to the vector

i + 4j – 2k

Also, find another point on the line

The symmetric equations of a line are found by isolating t in each of the parametric equations and then eliminated t by setting them each equal to each other. This shows that each component (x,y,z) on the line is proportional to the others just as are the components of any other vector that is parallel to the line

How are the symmetric equations of a line found and what is their significance?

Use the displacement vector between points a and b as the direction vector then choose one of the given points and create the parametric and symmetric equations with those

Find the parametric equations and symmetric equations of the line that passes through the points

A(2,4,⁻3)

B(3,⁻1,1)

A(2,4,⁻3)

B(3,⁻1,1)

Use the given parametric equations to find the symmetric equations then insert 0 for z and solve for x and y

At what point does the following line intersect the

xy plane?

x = 2 + t

y = 4 – 5t

z = ⁻3 + 4t

xy plane?

x = 2 + t

y = 4 – 5t

z = ⁻3 + 4t

The line segment from r₀ to r₁ is given by this vector equation

vector equation of a line segment (equation)

First compare the direction numbers (coefficients of t) of the two lines, if they are scalar multiples of each other then the lines are parallel. Next, set each corresponding component equation equal to each other, solve for 2 of the 3 equations and if the t values found satisfy the third equation then the lines intersect at that point. Since neither of these are true in this case the lines are skew

Show that the lines L₁ and L₂ are skew lines

L₁:

x = 1 + t

y = ⁻2 + 3t

z = 4 – t

L₂:

x = 2s

y = 3 + s

z = ⁻3 + 4s

L₁:

x = 1 + t

y = ⁻2 + 3t

z = 4 – t

L₂:

x = 2s

y = 3 + s

z = ⁻3 + 4s

False, only a vector perpendicular to a plane fully describes its direction

A vector parallel to a plane is enough to describe the direction of the plane

(T/F)

(T/F)

A normal vector is a vector that is orthogonal to a plane and that fully describes the direction of a plane

Normal vector (definition)

Since the definition of the dot product says that if there is 90° between two vectors their dot product will be 0, the equation of a plane can be described as the dot product of a vector on its surface to the normal vector orthogonal to it all being equal to 0

Equation of a plane (vector and scalar equations)

Use the scalar equation of a plane derived from use of dot products, plug in the components of the normal vector and the point on the plane, and simplify

Find an equation of the plane through the point,

(2,4,⁻1)

with normal vector

n = <2,3,4>

(2,4,⁻1)

with normal vector

n = <2,3,4>

To ﬁnd the x-intercept we set y and z equal to 0 in this equation and obtain x = 6. Similarly, the y-intercept is 4 and the z-intercept is 3

Find the intercepts of the plane,

2x + 3y + 4z = 12

and sketch the plane

2x + 3y + 4z = 12

and sketch the plane

The linear equation of a plane is a simplified version of the scalar equation of a plane

Equation of a plane (linear)

Find an equation of the plane that passes through the points

P(1,3,2)

Q(3,⁻1,6)

R(5,2,0)

P(1,3,2)

Q(3,⁻1,6)

R(5,2,0)

Two planes are parallelif their normal vectors are parallel. For instance, the planes,

x + 2y – 3z = 4 and

2x + 4y – 6z = 3

are parallel because their normal vectors

n₁ = <1,2,⁻3> and

n₂ = <2,4,⁻6>

are scalar multiples of eachother,

n₂ = 2n₁

x + 2y – 3z = 4 and

2x + 4y – 6z = 3

are parallel because their normal vectors

n₁ = <1,2,⁻3> and

n₂ = <2,4,⁻6>

are scalar multiples of eachother,

n₂ = 2n₁

How can it be proven that two planes are parallel?

Use the component coefficients of the plane equations to determine their normal vectors, then use the definition of the dot product to find the angle between those vectors which will also be the angle between the planes

Find the angle between the planes

x + y + z = 1 and

x – 2y + 3z = 1

x + y + z = 1 and

x – 2y + 3z = 1

First find a point on L. For instance, to find the point where the line intersects the xy-plane set z = 0 in both plane equations to get the equations,

x + y =1 and

x – 2y = 1

whose solution is

x = 1 and y = 0

So the point (1,0,0) lies on line L.

Since L lies in both planes it is perpendicular to both normal vectors so use the cross product of the normal vectors and create symmetric equations

x + y =1 and

x – 2y = 1

whose solution is

x = 1 and y = 0

So the point (1,0,0) lies on line L.

Since L lies in both planes it is perpendicular to both normal vectors so use the cross product of the normal vectors and create symmetric equations

Find the symmetric equations for the line of intersection L of the planes

x + y + z = 1

x – 2y + 3z = 1

x + y + z = 1

x – 2y + 3z = 1

Create a triangle between the normal vector of the plane and a random vector on the plane. The distance of the point at the end of the normal vector from the plane will be equal to the component of the triangle’s hypotenuse vector onto the normal vector. This is derived from the definition of the dot product

Find a formula for the distance D from a point,

P₁(x₁,y₁,z₁)

to the plane,

ax + by + cz + d = 0

P₁(x₁,y₁,z₁)

to the plane,

ax + by + cz + d = 0

In order for planes to be parallel their normal vectors must be parallel. by setting 2 components of the equation of one plane to 0 and solving for the other component it is possible to find a point on the that plane. Use this point on one plane and the normal vector of the other in the distance from a plane formula

Find the distance between the parallel planes,

10x + 2y – 2z = 5 and

5x + y – z = 1

10x + 2y – 2z = 5 and

5x + y – z = 1