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Geometry and Trigonometry - Vector Equations of Lines and Planes

Grade 12IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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The Vector Equation of a Line: A line in 3D space is defined by a position vector a\mathbf{a} of a known point on the line and a direction vector b\mathbf{b}. The equation is given by r=a+tb\mathbf{r} = \mathbf{a} + t\mathbf{b}, where tt is a scalar parameter. Visually, this represents a point moving along a straight path; a\mathbf{a} acts as the 'starting point' from the origin, and tbt\mathbf{b} allows you to move infinitely in either direction along the line's orientation.

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The Vector Equation of a Plane: A plane is a flat, infinite 2D surface in 3D space. It can be defined using a point a\mathbf{a} on the plane and two non-parallel direction vectors u\mathbf{u} and v\mathbf{v} that lie within the plane. The equation is r=a+λu+μv\mathbf{r} = \mathbf{a} + \lambda \mathbf{u} + \mu \mathbf{v}. Geometrically, the two parameters λ\lambda and μ\mu allow you to 'sweep out' the entire flat surface from the reference point a\mathbf{a}.

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The Normal Vector (n\mathbf{n}): This is a vector that is perpendicular (orthogonal) to every vector lying in a specific plane. It is crucial for the scalar product form of a plane. Visually, if you imagine a flat table, any vector pointing straight up or down from the surface is a normal vector. The orientation of the plane is uniquely determined by the direction of its normal.

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Scalar Product Form and Cartesian Equation of a Plane: Using the normal vector n=(abc)\mathbf{n} = \begin{pmatrix} a \\ b \\ c \end{pmatrix}, a plane can be described by r⋅n=a⋅n\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}, which simplifies to the Cartesian form ax+by+cz=dax + by + cz = d. In this form, the coefficients of x,y,x, y, and zz are the components of the normal vector. This is often the most useful form for finding intersections and distances.

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Intersection of a Line and a Plane: To find where a line r=a+tb\mathbf{r} = \mathbf{a} + t\mathbf{b} meets a plane ax+by+cz=dax + by + cz = d, substitute the parametric expressions for x,y,x, y, and zz from the line equation into the plane equation and solve for tt. Visually, this is the single point where the 'skewer' (the line) passes through the 'sheet' (the plane). If no solution for tt exists, the line is parallel to the plane.

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Angles Between Geometric Elements: The angle θ\theta between two lines is the angle between their direction vectors. The angle between two planes is the angle between their normal vectors. However, the angle ϕ\phi between a line and a plane is found using sin⁡ϕ=∣b⋅n∣∣b∣∣n∣\sin \phi = \frac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}| |\mathbf{n}|}. Visually, this ϕ\phi is the angle between the line and its orthogonal projection onto the plane's surface.

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Distance from a Point to a Plane: This is the shortest (perpendicular) distance from a point P(x0,y0,z0)P(x_0, y_0, z_0) to the plane ax+by+cz=dax + by + cz = d. Visually, this is the length of the normal segment dropped from the point to the plane surface. It is calculated by finding the magnitude of the projection of the vector from the plane to the point onto the normal vector.

📐Formulae

Vector Line: r=a+tb\mathbf{r} = \mathbf{a} + t\mathbf{b}

Cartesian Line: x−x1b1=y−y1b2=z−z1b3\frac{x - x_1}{b_1} = \frac{y - y_1}{b_2} = \frac{z - z_1}{b_3}

Parametric Plane: r=a+λu+μv\mathbf{r} = \mathbf{a} + \lambda \mathbf{u} + \mu \mathbf{v}

Scalar Product Plane: r⋅n=d\mathbf{r} \cdot \mathbf{n} = d

Cartesian Plane: ax+by+cz=dax + by + cz = d

Angle between lines: cos⁡θ=∣b1⋅b2∣∣b1∣∣b2∣\cos \theta = \frac{|\mathbf{b}_1 \cdot \mathbf{b}_2|}{|\mathbf{b}_1| |\mathbf{b}_2|}

Angle between line and plane: sin⁡ϕ=∣b⋅n∣∣b∣∣n∣\sin \phi = \frac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}| |\mathbf{n}|}

Distance from Point to Plane: D=∣ax0+by0+cz0−d∣a2+b2+c2D = \frac{|ax_0 + by_0 + cz_0 - d|}{\sqrt{a^2 + b^2 + c^2}}

💡Examples

Problem 1:

Find the point of intersection between the line r=(1−12)+λ(21−1)\mathbf{r} = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} and the plane 3x−y+2z=143x - y + 2z = 14.

Solution:

  1. Express the coordinates of a general point on the line in terms of λ\lambda: x=1+2λx = 1 + 2\lambda, y=−1+λy = -1 + \lambda, z=2−λz = 2 - \lambda.
  2. Substitute these into the plane equation: 3(1+2λ)−(−1+λ)+2(2−λ)=143(1 + 2\lambda) - (-1 + \lambda) + 2(2 - \lambda) = 14.
  3. Expand and simplify: 3+6λ+1−λ+4−2λ=143 + 6\lambda + 1 - \lambda + 4 - 2\lambda = 14.
  4. Combine like terms: 3λ+8=14  ⟹  3λ=6  ⟹  λ=23\lambda + 8 = 14 \implies 3\lambda = 6 \implies \lambda = 2.
  5. Substitute λ=2\lambda = 2 back into the line equation: x=1+2(2)=5x = 1 + 2(2) = 5, y=−1+2=1y = -1 + 2 = 1, z=2−2=0z = 2 - 2 = 0.
  6. The point of intersection is (5,1,0)(5, 1, 0).

Explanation:

To find the intersection, we treat the line as a set of parametric coordinates and find the specific parameter value λ\lambda that satisfies the constraints of the plane's Cartesian equation.

Problem 2:

Find the Cartesian equation of the plane containing the points A(1,0,2)A(1, 0, 2), B(2,1,0)B(2, 1, 0), and C(0,2,1)C(0, 2, 1).

Solution:

  1. Find two vectors in the plane: AB⃗=(2−11−00−2)=(11−2)\vec{AB} = \begin{pmatrix} 2-1 \\ 1-0 \\ 0-2 \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ -2 \end{pmatrix} and AC⃗=(0−12−01−2)=(−12−1)\vec{AC} = \begin{pmatrix} 0-1 \\ 2-0 \\ 1-2 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ -1 \end{pmatrix}.
  2. Calculate the normal vector n\mathbf{n} using the cross product AB⃗×AC⃗\vec{AB} \times \vec{AC}: n=∣ijk11−2−12−1∣=i(−1−(−4))−j(−1−2)+k(2−(−1))=3i+3j+3k\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & -2 \\ -1 & 2 & -1 \end{vmatrix} = \mathbf{i}(-1 - (-4)) - \mathbf{j}(-1 - 2) + \mathbf{k}(2 - (-1)) = 3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}.
  3. Simplify the normal vector (optional but recommended): n=(111)\mathbf{n} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}.
  4. Use the Cartesian form 1x+1y+1z=d1x + 1y + 1z = d. Substitute point A(1,0,2)A(1, 0, 2): 1(1)+1(0)+1(2)=31(1) + 1(0) + 1(2) = 3. Thus d=3d = 3.
  5. The equation is x+y+z=3x + y + z = 3.

Explanation:

A plane is defined by its normal vector. By taking the cross product of two vectors lying within the plane (formed by the three points), we find a vector perpendicular to the surface. We then solve for the constant dd using any of the given points.