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Geometry and Trigonometry - Intersections among lines and planes (HL)

Grade 11IB_AA

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

🔑Concepts

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The intersection of a line and a plane occurs at a unique point unless the line is parallel to the plane. To find this point, substitute the parametric components of the line into the Cartesian equation of the plane ax+by+cz=dax + by + cz = d and solve for the scalar parameter λ\lambda.

A line piercing a plane at point P.
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Two non-parallel planes intersect in a unique line. This line is perpendicular to both normal vectors n1\mathbf{n_1} and n2\mathbf{n_2}, meaning its direction vector v\mathbf{v} can be found using the cross product v=n1×n2\mathbf{v} = \mathbf{n_1} \times \mathbf{n_2}.

Two planes intersecting along a single straight line.
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The relationship between a line and a plane is determined by the dot product of the line's direction vector v\mathbf{v} and the plane's normal vector n\mathbf{n}. If v⋅n=0\mathbf{v} \cdot \mathbf{n} = 0, the line is either parallel to the plane or lies entirely within it.

Normal vector n perpendicular to a plane and a parallel line v.
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To find the angle ϕ\phi between a line and a plane, calculate the angle θ\theta between the line's direction v\mathbf{v} and the plane's normal n\mathbf{n} using cos⁡θ\cos \theta. Then, ϕ=90∘−θ\phi = 90^\circ - \theta, leading to the formula sin⁡ϕ=∣v⋅n∣∣v∣∣n∣\sin \phi = \frac{|\mathbf{v} \cdot \mathbf{n}|}{|\mathbf{v}||\mathbf{n}|}.

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Systems of linear equations in three variables can represent the intersection of three planes. Possible outcomes include a single point of intersection (unique solution), a line of intersection (infinitely many solutions), or no common intersection (inconsistent system).

📐Formulae

r=a+λv\mathbf{r} = \mathbf{a} + \lambda \mathbf{v}

ax+by+cz=dax + by + cz = d

r⋅n=a⋅n\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}

cos⁡θ=∣v⋅n∣∣v∣∣n∣ (where θ is the angle between the line and the normal vector)\cos \theta = \frac{|\mathbf{v} \cdot \mathbf{n}|}{|\mathbf{v}||\mathbf{n}|} \text{ (where } \theta \text{ is the angle between the line and the normal vector)}

sin⁡ϕ=∣v⋅n∣∣v∣∣n∣ (where ϕ is the angle between the line and the plane)\sin \phi = \frac{|\mathbf{v} \cdot \mathbf{n}|}{|\mathbf{v}||\mathbf{n}|} \text{ (where } \phi \text{ is the angle between the line and the plane)}

💡Examples

Problem 1:

Find the coordinates of the point of intersection between the line r=\mathbf{r} = (101)\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} +λ(21−1) + \lambda \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} and the plane x+2y−z=8x + 2y - z = 8.

Solution:

  1. Write the parametric equations of the line: x=1+2λx = 1 + 2\lambda y=λy = \lambda z=1−λz = 1 - \lambda
  2. Substitute these into the plane equation: (1+2λ)+2(λ)−(1−λ)=8(1 + 2\lambda) + 2(\lambda) - (1 - \lambda) = 8 1+2λ+2λ−1+λ=81 + 2\lambda + 2\lambda - 1 + \lambda = 8 5λ=8  ⟹  λ=1.65\lambda = 8 \implies \lambda = 1.6
  3. Substitute λ=1.6\lambda = 1.6 back into the line equation: x=1+2(1.6)=4.2x = 1 + 2(1.6) = 4.2 y=1.6y = 1.6 z=1−1.6=−0.6z = 1 - 1.6 = -0.6 The intersection point is (4.2,1.6,−0.6)(4.2, 1.6, -0.6).

Explanation:

We use the parametric form of the line to represent any point on the line in terms of λ\lambda. By substituting these into the plane's equation, we find the specific value of λ\lambda where the point also lies on the plane.

Problem 2:

Determine if the lines L1:r=L_1: \mathbf{r} = (123)\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} +λ+ \lambda (10−1)\begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix} and L2:r=L_2: \mathbf{r} = (222)\begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} + μ\mu (011)\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} intersect.

Solution:

Equate the components: x:1+λ=2  ⟹  λ=1x: 1 + \lambda = 2 \implies \lambda = 1 y:2=2+μ  ⟹  μ=0y: 2 = 2 + \mu \implies \mu = 0 z:3−λ=2+μz: 3 - \lambda = 2 + \mu Check the zz component with λ=1\lambda = 1 and μ=0\mu = 0: 3−(1)=23 - (1) = 2 and 2+(0)=22 + (0) = 2. Since 2=22 = 2, the system is consistent. The lines intersect at the point where λ=1\lambda = 1: x=1+1=2,y=2,z=3−1=2x = 1 + 1 = 2, y = 2, z = 3 - 1 = 2. The intersection point is (2,2,2)(2, 2, 2).

Explanation:

To check for intersection, we solve for the parameters using two coordinates and verify with the third. Consistency across all three coordinates confirms an intersection.

Problem 3:

Find the Cartesian equation of the line of intersection of the two planes Π1:x+y+z=3\Pi_1: x + y + z = 3 and Π2:2x−y+3z=4\Pi_2: 2x - y + 3z = 4.

Solution:

  1. Find the direction vector v\mathbf{v} of the line using the cross product of the normals n1=\mathbf{n}_1 = (111)\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} and n2=(2−13)\mathbf{n}_2 = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix}: v=n1×n2=∣ijk1112−13∣=i(3−(−1))−j(3−2)+k(−1−2)=(4−1−3)\mathbf{v} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ 2 & -1 & 3 \end{vmatrix} = \mathbf{i}(3 - (-1)) - \mathbf{j}(3 - 2) + \mathbf{k}(-1 - 2) = \begin{pmatrix} 4 \\ -1 \\ -3 \end{pmatrix}.
  2. Find a point on the line by setting z=0z = 0: x+y=3x + y = 3 2x−y=42x - y = 4 Adding the equations: 3x=7  ⟹  x=733x = 7 \implies x = \frac{7}{3}. Substituting back: 73+y=3  ⟹  y=23\frac{7}{3} + y = 3 \implies y = \frac{2}{3}. Point: (73,23,0)(\frac{7}{3}, \frac{2}{3}, 0).
  3. Vector equation: r=\mathbf{r} = (7/32/30)\begin{pmatrix} 7/3 \\ 2/3 \\ 0 \end{pmatrix} + λ\lambda (4−1−3)\begin{pmatrix} 4 \\ -1 \\ -3 \end{pmatrix}.

Explanation:

The line of intersection is perpendicular to both normal vectors, hence the use of the cross product. We then find a specific point that satisfies both plane equations to define the position vector.

Problem 4:

Find the point of intersection between the line L:r=(2−13)+λ(12−1)L: \mathbf{r} = \begin{pmatrix} 2 \\ -1 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} and the plane Π:3x−y+2z=11\Pi: 3x - y + 2z = 11.

A line passing through a rectangular plane representation with the intersection point labeled.

Solution:

  1. Express the line in parametric form: x=2+λx = 2 + \lambda, y=−1+2λy = -1 + 2\lambda, z=3−λz = 3 - \lambda.
  2. Substitute these into the plane equation: 3(2+λ)−(−1+2λ)+2(3−λ)=113(2 + \lambda) - (-1 + 2\lambda) + 2(3 - \lambda) = 11.
  3. Expand and simplify: 6+3λ+1−2λ+6−2λ=116 + 3\lambda + 1 - 2\lambda + 6 - 2\lambda = 11.
  4. Combine terms: 13−λ=11  ⟹  λ=213 - \lambda = 11 \implies \lambda = 2.
  5. Find the point: x=2+2=4x = 2 + 2 = 4, y=−1+2(2)=3y = -1 + 2(2) = 3, z=3−2=1z = 3 - 2 = 1.
  6. Point of intersection is (4,3,1)(4, 3, 1).

Explanation:

To find where a line meets a plane, we find the specific value of the parameter λ\lambda that satisfies the plane's equation. This represents the unique point shared by both objects.

Problem 5:

Determine if the line L:r=(111)+t(2−11)L: \mathbf{r} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + t \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} intersects the plane x+2y=5x + 2y = 5.

A line drawn parallel above a plane to illustrate no intersection.

Solution:

  1. Parametric form: x=1+2tx = 1 + 2t, y=1−ty = 1 - t, z=1+tz = 1 + t.
  2. Substitute into plane equation: (1+2t)+2(1−t)=5(1 + 2t) + 2(1 - t) = 5.
  3. Simplify: 1+2t+2−2t=5  ⟹  3=51 + 2t + 2 - 2t = 5 \implies 3 = 5.
  4. Since 3=53 = 5 is a contradiction, there is no value of tt that satisfies the equation.
  5. Conclusion: The line is parallel to the plane and does not intersect it.

Explanation:

If the substitution leads to a contradiction, the line and plane are parallel and distinct. If it led to an identity (e.g., 5=55 = 5), the line would lie entirely within the plane.