Source: Math 126 Midterm 1 Practice Exam
Difficulty: Intermediate to Advanced | Prerequisites: Dot product, cross product, vector-valued functions, basic differentiation rules (chain rule, product rule).
Lines and planes in 3D are the geometry toolkit of multivariable calculus. You will use them to describe tangent lines to curves, normal lines to surfaces, and the regions you integrate over. This topic pulls together derivatives of vector functions (for tangent lines), the cross product (for finding normal vectors to planes), and algebraic problem-solving (for constructing planes through given lines). If you can differentiate a vector function and compute a cross product, you are ready.
A tangent line to a space curve comes from differentiating the vector function and evaluating at the right parameter value. Equations of planes come from knowing a point on the plane and a normal vector. The cross product is your tool for producing normal vectors when you have two direction vectors. These three ideas, combined, handle most of the lines-and-planes problems on the midterm.
Tangent vector
The derivative r'(t) = ⟨x'(t), y'(t), z'(t)⟩ of a vector function. It points in the direction the curve is heading at parameter value t. In simple terms, it is the velocity of a particle tracing the curve.
Parametric equations of a line
A line through point P₀ = (x₀, y₀, z₀) in the direction of vector d = ⟨a, b, c⟩ is given by x = x₀ + at, y = y₀ + bt, z = z₀ + ct. Think of it as starting at a point and walking in a fixed direction.
Normal vector to a plane
A vector perpendicular to every vector lying in the plane. For the plane ax + by + cz = d, the normal vector is n = ⟨a, b, c⟩. In simple terms, it points straight out of the plane like a flagpole on flat ground.
Cross product
For vectors u = ⟨u₁, u₂, u₃⟩ and v = ⟨v₁, v₂, v₃⟩, the cross product u x v = ⟨u₂v₃ - u₃v₂, u₃v₁ - u₁v₃, u₁v₂ - u₂v₁⟩. It produces a vector perpendicular to both u and v. Think of it as a way to manufacture a normal direction from two directions that lie in a plane.
Skew lines
Two lines in 3D that are neither parallel nor intersecting. They pass each other at different heights, like a road and a bridge. There is no single plane that contains both.
Parallel planes
Planes that share the same normal vector but have different constants. They never intersect. ax + by + cz = d₁ and ax + by + cz = d₂ are parallel whenever d₁ ≠ d₂.
To find the tangent line to r(t) at a given point:
Find the parameter value t₀ by solving r(t₀) = the given point.
Compute r'(t) and evaluate at t₀ to get the direction vector.
Write the parametric equations: x = x₀ + x'(t₀)s, y = y₀ + y'(t₀)s, z = z₀ + z'(t₀)s (using a new parameter s to avoid confusion).
Let r(t) = ⟨√(3t² + 1), ln(2t²), t³⟩. Find parametric equations for the tangent line at (2, ln 2, -1).
Step 1: Find t₀
From the third component: t³ = -1, so t₀ = -1.
Check: √(3(1) + 1) = √4 = 2 ✓ and ln(2(1)) = ln 2 ✓.
Step 2: Compute r'(t)
r'(t) = ⟨3t/√(3t² + 1), 4t/(2t²) , 3t²⟩ = ⟨3t/√(3t² + 1), 2/t, 3t²⟩
Note: for ln(2t²), the derivative is (1/(2t²)) · 4t = 2/t.
Step 3: Evaluate at t₀ = -1
r'(-1) = ⟨-3/√4, 2/(-1), 3(1)⟩ = ⟨-3/2, -2, 3⟩
Step 4: Write the parametric equations
x = 2 - (3/2)s, y = ln 2 - 2s, z = -1 + 3s
When two lines are skew, each lies in its own plane, and those two planes are parallel. To find both planes:
Find the direction vectors of both lines.
Cross the two direction vectors to get the shared normal vector.
Use each line's known point to write each plane's equation.
L₁ is perpendicular to -2x + 3y - 5z = 8 through (1, 4, -2). L₂ is the intersection of x - y = 4 and -x - y - z = 2. Find the two parallel planes containing L₁ and L₂.
Step 1: Direction vector of L₁
L₁ is perpendicular to the plane -2x + 3y - 5z = 8, so its direction is the plane's normal: d₁ = ⟨-2, 3, -5⟩.
Step 2: Direction vector of L₂
L₂ is the intersection of x - y = 4 and -x - y - z = 2. Add the two equations: -2y - z = 6, so z = -2y - 6. From x - y = 4: x = y + 4.
Parametrise with y = t: x = t + 4, y = t, z = -2t - 6.
Direction vector: d₂ = ⟨1, 1, -2⟩.
Step 3: Normal vector (cross product)
n = d₁ x d₂ = ⟨(3)(-2) - (-5)(1), (-5)(1) - (-2)(-2), (-2)(1) - (3)(1)⟩
= ⟨-6 + 5, -5 - 4, -2 - 3⟩ = ⟨-1, -9, -5⟩
Simplify (multiply by -1): n = ⟨1, 9, 5⟩.
Step 4: Write each plane equation
Plane through L₁ (point (1, 4, -2)): 1(x - 1) + 9(y - 4) + 5(z + 2) = 0 → x + 9y + 5z = 27
Plane through L₂ (point (4, 0, -6), from t = 0): 1(x - 4) + 9(y - 0) + 5(z + 6) = 0 → x + 9y + 5z = -26
Result: x + 9y + 5z = 27 and x + 9y + 5z = -26.
Parametric line through P₀ in direction d:
x = x_0 + at,\quad y = y_0 + bt,\quad z = z_0 + ctEquation of a plane with normal n = ⟨a, b, c⟩ through point (x₀, y₀, z₀):
a(x - x_0) + b(y - y_0) + c(z - z_0) = 0Cross product:
\mathbf{u} \times \mathbf{v} = \langle u_2 v_3 - u_3 v_2,\; u_3 v_1 - u_1 v_3,\; u_1 v_2 - u_2 v_1 \rangleDerivative of a vector function:
\mathbf{r}'(t) = \langle x'(t),\; y'(t),\; z'(t) \rangleTangent lines to curves describe instantaneous direction of motion, which is essential in robotics (tool path planning) and physics (velocity vectors). The cross product is used in computer graphics to compute surface normals for lighting. Parallel planes appear in structural engineering when analysing the spacing between parallel load-bearing walls.
Students often confuse the direction vector of a line with the normal vector of a plane. A line perpendicular to a plane has its direction equal to the plane's normal. A line lying in a plane has its direction perpendicular to the plane's normal.
Students sometimes forget to find the parameter value t₀ before differentiating. You need to know which point on the curve you are at before you can evaluate the derivative there.
Students frequently get the cross product wrong because they mix up the signs. Use the determinant formula or the "cover-up" method and double-check by verifying that the result is perpendicular to both input vectors (dot product with each should be zero).
Students sometimes write the tangent line using the same parameter t as the original curve. Use a different letter (s, u, or similar) to avoid confusion.
⚠️ Tangent line problems always require you to find t₀ first. Do not start differentiating until you know which parameter value gives the specified point.
⚠️ For "find two parallel planes containing two skew lines," the method is always the same: cross the direction vectors, then plug in one point from each line.
⚠️ When a line is described as "perpendicular to the plane ax + by + cz = d," its direction vector is ⟨a, b, c⟩. This is the fastest free piece of information in any problem.
⚠️ Always verify your cross product by checking that n · d₁ = 0 and n · d₂ = 0. This takes seconds and catches sign errors.
True or False: The direction vector of a line perpendicular to 3x - y + 2z = 7 is ⟨3, -1, 2⟩. (True.)
Fill in the blank: Two planes are parallel if and only if their normal vectors are ___. (parallel, i.e. scalar multiples of each other.)
True or False: The cross product u x v is perpendicular to both u and v. (True.)
Fill in the blank: To find the tangent line to r(t) at a point, you first find t₀, then compute ___. (r'(t₀), the derivative evaluated at t₀.)
True or False: Two skew lines can be contained in a single plane. (False. Each lies in its own parallel plane.)
Q: Find parametric equations for the tangent line to r(t) = ⟨t², e^t, sin t⟩ at t = 0.
A: r(0) = ⟨0, 1, 0⟩. r'(t) = ⟨2t, e^t, cos t⟩, so r'(0) = ⟨0, 1, 1⟩. Parametric equations: x = 0, y = 1 + s, z = s.
Q: Find the equation of the plane containing the point (2, -1, 3) with normal vector ⟨4, 1, -2⟩.
A: 4(x - 2) + 1(y + 1) - 2(z - 3) = 0 → 4x + y - 2z = 1.
Q: Lines L₁ and L₂ have direction vectors ⟨1, 0, 2⟩ and ⟨0, 3, 1⟩. Find the normal to a plane containing both directions.
A: n = ⟨1, 0, 2⟩ x ⟨0, 3, 1⟩ = ⟨0·1 - 2·3, 2·0 - 1·1, 1·3 - 0·0⟩ = ⟨-6, -1, 3⟩. Check: ⟨-6, -1, 3⟩ · ⟨1, 0, 2⟩ = -6 + 0 + 6 = 0 ✓ and ⟨-6, -1, 3⟩ · ⟨0, 3, 1⟩ = 0 - 3 + 3 = 0 ✓.
Q: The line of intersection of x + y = 1 and y + z = 2 has what direction vector?
A: From x + y = 1: x = 1 - y. From y + z = 2: z = 2 - y. Let y = t: x = 1 - t, y = t, z = 2 - t. Direction: ⟨-1, 1, -1⟩.
Tangent lines are the first step toward arc length and curvature, both of which appear later in Math 126. The cross product used here for normals reappears in torque problems and in computing the area of parallelograms spanned by two vectors. Planes are the building blocks of linear approximation (tangent planes to surfaces), which is one of the central topics of multivariable calculus.
Tags: tangent line, tangent vector, space curve, parametric equations, line in 3D, plane equation, normal vector, cross product, skew lines, parallel planes, intersection of planes, direction vector, derivative of vector function, Math 126, Calculus III, UW, midterm 1, vector-valued functions