Skip to main content

Vectors

Vector Notation​

Scalars and Vectors​

A scalar is a quantity with magnitude only (e.g. mass, temperature, time). A vector is a quantity with both magnitude and direction (e.g. displacement, velocity, force).

Representation​

A vector in ′{′R′}′n\mathbb{'\{'}R{'\}'}^n is an ordered list of nn real numbers called components. A vector in ′{′R′}′3\mathbb{'\{'}R{'\}'}^3 with components a1,a2,a3a_1, a_2, a_3 is written:

′{′a′}′=(a1a2a3)=a1′{′i′}′+a2′{′j′}′+a3′{′k′}′\mathbf{'\{'}a{'\}'} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} = a_1\mathbf{'\{'}i{'\}'} + a_2\mathbf{'\{'}j{'\}'} + a_3\mathbf{'\{'}k{'\}'}

where ′{′i′}′=(100)\mathbf{'\{'}i{'\}'} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}, ′{′j′}′=(010)\mathbf{'\{'}j{'\}'} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}, ′{′k′}′=(001)\mathbf{'\{'}k{'\}'} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} are the standard basis vectors.

Position Vectors​

The position vector of a point AA with coordinates (x,y,z)(x, y, z) is the vector from the origin OO to AA:

OA→=(xyz)\overrightarrow{OA} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}

The vector from point A(x1,y1,z1)A(x_1, y_1, z_1) to point B(x2,y2,z2)B(x_2, y_2, z_2) is:

AB→=OB→−OA→=(x2−x1y2−y1z2−z1)\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \\ z_2 - z_1 \end{pmatrix}

Vector Operations​

Addition and Scalar Multiplication​

If ′{′a′}′=(a1a2a3)\mathbf{'\{'}a{'\}'} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} and ′{′b′}′=(b1b2b3)\mathbf{'\{'}b{'\}'} = \begin{pmatrix} b_1 \\ b_2 \\ b_3 \end{pmatrix}, and λ∈′{′R′}′\lambda \in \mathbb{'\{'}R{'\}'}:

′{′a′}′+′{′b′}′=(a1+b1a2+b2a3+b3),λ′{′a′}′=(λa1λa2λa3)\mathbf{'\{'}a{'\}'} + \mathbf{'\{'}b{'\}'} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \\ a_3 + b_3 \end{pmatrix}, \qquad \lambda\mathbf{'\{'}a{'\}'} = \begin{pmatrix} \lambda a_1 \\ \lambda a_2 \\ \lambda a_3 \end{pmatrix}

These operations satisfy the vector space axioms: associativity, commutativity of addition, distributivity of scalar multiplication, and existence of additive identity (′{′0′}′\mathbf{'\{'}0{'\}'}) and inverses (−′{′a′}′-\mathbf{'\{'}a{'\}'}).

Magnitude and Unit Vectors​

Magnitude​

The magnitude (length) of ′{′a′}′=(a1,a2,a3)\mathbf{'\{'}a{'\}'} = (a_1, a_2, a_3) is:

∣′{′a′}′∣=a12+a22+a32|\mathbf{'\{'}a{'\}'}| = \sqrt{a_1^2 + a_2^2 + a_3^2}

This is the Euclidean norm, derived from the Pythagorean theorem in three dimensions.

Unit Vectors​

A unit vector has magnitude 11. The unit vector in the direction of ′{′a′}′\mathbf{'\{'}a{'\}'} is:

′{′a′}′^=′{′a′}′∣′{′a′}′∣,′{′a′}′≠′{′0′}′\hat{\mathbf{'\{'}a{'\}'}} = \frac{\mathbf{'\{'}a{'\}'}}{|\mathbf{'\{'}a{'\}'}|}, \quad \mathbf{'\{'}a{'\}'} \ne \mathbf{'\{'}0{'\}'}

Scalar (Dot) Product​

Definition​

The scalar product of ′{′a′}′\mathbf{'\{'}a{'\}'} and ′{′b′}′\mathbf{'\{'}b{'\}'} is:

′{′a′}′⋅′{′b′}′=∣′{′a′}′∣ ∣′{′b′}′∣cos⁡θ\mathbf{'\{'}a{'\}'} \cdot \mathbf{'\{'}b{'\}'} = |\mathbf{'\{'}a{'\}'}|\,|\mathbf{'\{'}b{'\}'}|\cos\theta

where θ\theta is the angle between the vectors (0≤θ≤π0 \le \theta \le \pi).

Component Form​

′{′a′}′⋅′{′b′}′=a1b1+a2b2+a3b3\mathbf{'\{'}a{'\}'} \cdot \mathbf{'\{'}b{'\}'} = a_1 b_1 + a_2 b_2 + a_3 b_3

Properties​

  • ′{′a′}′⋅′{′b′}′=′{′b′}′⋅′{′a′}′\mathbf{'\{'}a{'\}'} \cdot \mathbf{'\{'}b{'\}'} = \mathbf{'\{'}b{'\}'} \cdot \mathbf{'\{'}a{'\}'} (commutative)
  • ′{′a′}′⋅(′{′b′}′+′{′c′}′)=′{′a′}′⋅′{′b′}′+′{′a′}′⋅′{′c′}′\mathbf{'\{'}a{'\}'} \cdot (\mathbf{'\{'}b{'\}'} + \mathbf{'\{'}c{'\}'}) = \mathbf{'\{'}a{'\}'} \cdot \mathbf{'\{'}b{'\}'} + \mathbf{'\{'}a{'\}'} \cdot \mathbf{'\{'}c{'\}'} (distributive)
  • (λ′{′a′}′)⋅′{′b′}′=λ(′{′a′}′⋅′{′b′}′)(\lambda\mathbf{'\{'}a{'\}'}) \cdot \mathbf{'\{'}b{'\}'} = \lambda(\mathbf{'\{'}a{'\}'} \cdot \mathbf{'\{'}b{'\}'})
  • ′{′a′}′⋅′{′a′}′=∣′{′a′}′∣2\mathbf{'\{'}a{'\}'} \cdot \mathbf{'\{'}a{'\}'} = |\mathbf{'\{'}a{'\}'}|^2

Geometric Interpretations​

Angle between vectors:

cos⁡θ=′{′a′}′⋅′{′b′}′∣′{′a′}′∣ ∣′{′b′}′∣\cos\theta = \frac{\mathbf{'\{'}a{'\}'} \cdot \mathbf{'\{'}b{'\}'}}{|\mathbf{'\{'}a{'\}'}|\,|\mathbf{'\{'}b{'\}'}|}

Perpendicularity test. ′{′a′}′⊥′{′b′}′  ⟺  ′{′a′}′⋅′{′b′}′=0\mathbf{'\{'}a{'\}'} \perp \mathbf{'\{'}b{'\}'} \iff \mathbf{'\{'}a{'\}'} \cdot \mathbf{'\{'}b{'\}'} = 0.

Projection. The scalar projection of ′{′b′}′\mathbf{'\{'}b{'\}'} onto ′{′a′}′\mathbf{'\{'}a{'\}'} is:

comp′{′a′}′′{′b′}′=′{′a′}′⋅′{′b′}′∣′{′a′}′∣=∣′{′b′}′∣cos⁡θ\mathrm{comp}_{\mathbf{'\{'}a{'\}'}}\mathbf{'\{'}b{'\}'} = \frac{\mathbf{'\{'}a{'\}'} \cdot \mathbf{'\{'}b{'\}'}}{|\mathbf{'\{'}a{'\}'}|} = |\mathbf{'\{'}b{'\}'}|\cos\theta

The vector projection is:

proj′{′a′}′′{′b′}′=(′{′a′}′⋅′{′b′}′∣′{′a′}′∣2)′{′a′}′\mathrm{proj}_{\mathbf{'\{'}a{'\}'}}\mathbf{'\{'}b{'\}'} = \left(\frac{\mathbf{'\{'}a{'\}'} \cdot \mathbf{'\{'}b{'\}'}}{|\mathbf{'\{'}a{'\}'}|^2}\right)\mathbf{'\{'}a{'\}'}

Vector (Cross) Product​

Definition (HL)​

The vector product of ′{′a′}′\mathbf{'\{'}a{'\}'} and ′{′b′}′\mathbf{'\{'}b{'\}'} is:

′{′a′}′×′{′b′}′=∣′{′a′}′∣ ∣′{′b′}′∣sin⁡θ  ′{′n′}′^\mathbf{'\{'}a{'\}'} \times \mathbf{'\{'}b{'\}'} = |\mathbf{'\{'}a{'\}'}|\,|\mathbf{'\{'}b{'\}'}|\sin\theta\;\hat{\mathbf{'\{'}n{'\}'}}

where ′{′n′}′^\hat{\mathbf{'\{'}n{'\}'}} is the unit vector perpendicular to both ′{′a′}′\mathbf{'\{'}a{'\}'} and ′{′b′}′\mathbf{'\{'}b{'\}'}, with direction given by the right-hand rule.

Component Form​

′{′a′}′×′{′b′}′=∣′{′i′}′′{′j′}′′{′k′}′a1a2a3b1b2b3∣=(a2b3−a3b2a3b1−a1b3a1b2−a2b1)\mathbf{'\{'}a{'\}'} \times \mathbf{'\{'}b{'\}'} = \begin{vmatrix} \mathbf{'\{'}i{'\}'} & \mathbf{'\{'}j{'\}'} & \mathbf{'\{'}k{'\}'} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = \begin{pmatrix} a_2 b_3 - a_3 b_2 \\ a_3 b_1 - a_1 b_3 \\ a_1 b_2 - a_2 b_1 \end{pmatrix}

Properties​

  • ′{′a′}′×′{′b′}′=−(′{′b′}′×′{′a′}′)\mathbf{'\{'}a{'\}'} \times \mathbf{'\{'}b{'\}'} = -(\mathbf{'\{'}b{'\}'} \times \mathbf{'\{'}a{'\}'}) (anti-commutative)
  • ′{′a′}′×(′{′b′}′+′{′c′}′)=′{′a′}′×′{′b′}′+′{′a′}′×′{′c′}′\mathbf{'\{'}a{'\}'} \times (\mathbf{'\{'}b{'\}'} + \mathbf{'\{'}c{'\}'}) = \mathbf{'\{'}a{'\}'} \times \mathbf{'\{'}b{'\}'} + \mathbf{'\{'}a{'\}'} \times \mathbf{'\{'}c{'\}'} (distributive)
  • (λ′{′a′}′)×′{′b′}′=λ(′{′a′}′×′{′b′}′)(\lambda\mathbf{'\{'}a{'\}'}) \times \mathbf{'\{'}b{'\}'} = \lambda(\mathbf{'\{'}a{'\}'} \times \mathbf{'\{'}b{'\}'})
  • ′{′a′}′×′{′a′}′=′{′0′}′\mathbf{'\{'}a{'\}'} \times \mathbf{'\{'}a{'\}'} = \mathbf{'\{'}0{'\}'}
  • ∣′{′a′}′×′{′b′}′∣=∣′{′a′}′∣ ∣′{′b′}′∣sin⁡θ|\mathbf{'\{'}a{'\}'} \times \mathbf{'\{'}b{'\}'}| = |\mathbf{'\{'}a{'\}'}|\,|\mathbf{'\{'}b{'\}'}|\sin\theta

Geometric Interpretations​

Parallel test. ′{′a′}′∥′{′b′}′  ⟺  ′{′a′}′×′{′b′}′=′{′0′}′\mathbf{'\{'}a{'\}'} \parallel \mathbf{'\{'}b{'\}'} \iff \mathbf{'\{'}a{'\}'} \times \mathbf{'\{'}b{'\}'} = \mathbf{'\{'}0{'\}'}.

Area of a parallelogram spanned by ′{′a′}′\mathbf{'\{'}a{'\}'} and ′{′b′}′\mathbf{'\{'}b{'\}'}:

A=∣′{′a′}′×′{′b′}′∣A = |\mathbf{'\{'}a{'\}'} \times \mathbf{'\{'}b{'\}'}|

Area of a triangle with sides ′{′a′}′\mathbf{'\{'}a{'\}'} and ′{′b′}′\mathbf{'\{'}b{'\}'}:

A=12∣′{′a′}′×′{′b′}′∣A = \frac{1}{2}|\mathbf{'\{'}a{'\}'} \times \mathbf{'\{'}b{'\}'}|

Volume of a parallelepiped with edges ′{′a′}′,′{′b′}′,′{′c′}′\mathbf{'\{'}a{'\}'}, \mathbf{'\{'}b{'\}'}, \mathbf{'\{'}c{'\}'}:

V=∣′{′a′}′⋅(′{′b′}′×′{′c′}′)∣V = |\mathbf{'\{'}a{'\}'} \cdot (\mathbf{'\{'}b{'\}'} \times \mathbf{'\{'}c{'\}'})|

This scalar triple product can also be written as the determinant:

′{′a′}′⋅(′{′b′}′×′{′c′}′)=∣a1a2a3b1b2b3c1c2c3∣\mathbf{'\{'}a{'\}'} \cdot (\mathbf{'\{'}b{'\}'} \times \mathbf{'\{'}c{'\}'}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}

Equations of Lines​

Vector (Parametric) Form​

A line through point AA with position vector ′{′a′}′\mathbf{'\{'}a{'\}'}, in the direction of vector ′{′d′}′\mathbf{'\{'}d{'\}'}:

′{′r′}′=′{′a′}′+λ′{′d′}′,λ∈′{′R′}′\mathbf{'\{'}r{'\}'} = \mathbf{'\{'}a{'\}'} + \lambda\mathbf{'\{'}d{'\}'}, \quad \lambda \in \mathbb{'\{'}R{'\}'}

In component form:

(xyz)=(a1a2a3)+λ(d1d2d3)\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix} + \lambda\begin{pmatrix} d_1 \\ d_2 \\ d_3 \end{pmatrix}

This gives parametric equations: x=a1+λd1x = a_1 + \lambda d_1, y=a2+λd2y = a_2 + \lambda d_2, z=a3+λd3z = a_3 + \lambda d_3.

Cartesian (Symmetric) Form​

Eliminating λ\lambda (assuming d1,d2,d3≠0d_1, d_2, d_3 \ne 0):

x−a1d1=y−a2d2=z−a3d3\frac{x - a_1}{d_1} = \frac{y - a_2}{d_2} = \frac{z - a_3}{d_3}

Line Through Two Points​

Given points AA and BB, the line through both points has direction ′{′d′}′=AB→\mathbf{'\{'}d{'\}'} = \overrightarrow{AB}:

′{′r′}′=OA→+λ AB→\mathbf{'\{'}r{'\}'} = \overrightarrow{OA} + \lambda\,\overrightarrow{AB}

Equations of Planes​

Point-Normal Form​

A plane through point AA with normal vector ′{′n′}′\mathbf{'\{'}n{'\}'} satisfies:

′{′n′}′⋅(′{′r′}′−′{′a′}′)=0\mathbf{'\{'}n{'\}'} \cdot (\mathbf{'\{'}r{'\}'} - \mathbf{'\{'}a{'\}'}) = 0

Expanding: n1(x−a1)+n2(y−a2)+n3(z−a3)=0n_1(x - a_1) + n_2(y - a_2) + n_3(z - a_3) = 0.

Cartesian Form​

n1x+n2y+n3z=dn_1 x + n_2 y + n_3 z = d

where d=′{′n′}′⋅′{′a′}′d = \mathbf{'\{'}n{'\}'} \cdot \mathbf{'\{'}a{'\}'} is a constant determined by any point on the plane.

Parametric Form​

A plane through point AA spanned by two non-parallel direction vectors ′{′d′}′1\mathbf{'\{'}d{'\}'}_1 and ′{′d′}′2\mathbf{'\{'}d{'\}'}_2:

′{′r′}′=′{′a′}′+λ′{′d′}′1+μ′{′d′}′2,λ,μ∈′{′R′}′\mathbf{'\{'}r{'\}'} = \mathbf{'\{'}a{'\}'} + \lambda\mathbf{'\{'}d{'\}'}_1 + \mu\mathbf{'\{'}d{'\}'}_2, \quad \lambda, \mu \in \mathbb{'\{'}R{'\}'}

Normal from Two Direction Vectors​

Given direction vectors ′{′d′}′1\mathbf{'\{'}d{'\}'}_1 and ′{′d′}′2\mathbf{'\{'}d{'\}'}_2 in the plane, the normal vector is:

′{′n′}′=′{′d′}′1×′{′d′}′2\mathbf{'\{'}n{'\}'} = \mathbf{'\{'}d{'\}'}_1 \times \mathbf{'\{'}d{'\}'}_2

Angles Between Vectors and Geometric Applications​

Angle Between Two Lines​

The acute angle θ\theta between two lines with direction vectors ′{′d′}′1\mathbf{'\{'}d{'\}'}_1 and ′{′d′}′2\mathbf{'\{'}d{'\}'}_2:

cos⁡θ=∣′{′d′}′1⋅′{′d′}′2∣∣′{′d′}′1∣ ∣′{′d′}′2∣\cos\theta = \frac{|\mathbf{'\{'}d{'\}'}_1 \cdot \mathbf{'\{'}d{'\}'}_2|}{|\mathbf{'\{'}d{'\}'}_1|\,|\mathbf{'\{'}d{'\}'}_2|}

Angle Between a Line and a Plane​

If a line has direction vector ′{′d′}′\mathbf{'\{'}d{'\}'} and a plane has normal vector ′{′n′}′\mathbf{'\{'}n{'\}'}, the angle α\alpha between the line and the plane is the complement of the angle between ′{′d′}′\mathbf{'\{'}d{'\}'} and ′{′n′}′\mathbf{'\{'}n{'\}'}:

sin⁡α=∣′{′d′}′⋅′{′n′}′∣∣′{′d′}′∣ ∣′{′n′}′∣\sin\alpha = \frac{|\mathbf{'\{'}d{'\}'} \cdot \mathbf{'\{'}n{'\}'}|}{|\mathbf{'\{'}d{'\}'}|\,|\mathbf{'\{'}n{'\}'}|}

Angle Between Two Planes​

The angle between two planes with normals ′{′n′}′1\mathbf{'\{'}n{'\}'}_1 and ′{′n′}′2\mathbf{'\{'}n{'\}'}_2:

cos⁡θ=∣′{′n′}′1⋅′{′n′}′2∣∣′{′n′}′1∣ ∣′{′n′}′2∣\cos\theta = \frac{|\mathbf{'\{'}n{'\}'}_1 \cdot \mathbf{'\{'}n{'\}'}_2|}{|\mathbf{'\{'}n{'\}'}_1|\,|\mathbf{'\{'}n{'\}'}_2|}

Distance from a Point to a Plane​

The perpendicular distance from point PP with position vector ′{′p′}′\mathbf{'\{'}p{'\}'} to the plane ′{′n′}′⋅′{′r′}′=d\mathbf{'\{'}n{'\}'} \cdot \mathbf{'\{'}r{'\}'} = d:

D=∣′{′n′}′⋅′{′p′}′−d∣∣′{′n′}′∣D = \frac{|\mathbf{'\{'}n{'\}'} \cdot \mathbf{'\{'}p{'\}'} - d|}{|\mathbf{'\{'}n{'\}'}|}

Distance from a Point to a Line​

The perpendicular distance from point PP to the line ′{′r′}′=′{′a′}′+λ′{′d′}′\mathbf{'\{'}r{'\}'} = \mathbf{'\{'}a{'\}'} + \lambda\mathbf{'\{'}d{'\}'}:

D=∣(′{′p′}′−′{′a′}′)×′{′d′}′∣∣′{′d′}′∣D = \frac{|(\mathbf{'\{'}p{'\}'} - \mathbf{'\{'}a{'\}'}) \times \mathbf{'\{'}d{'\}'}|}{|\mathbf{'\{'}d{'\}'}|}

Intersection of Line and Plane​

Substitute the parametric form of the line into the Cartesian equation of the plane and solve for λ\lambda.

Line lies in the plane if ′{′d′}′⋅′{′n′}′=0\mathbf{'\{'}d{'\}'} \cdot \mathbf{'\{'}n{'\}'} = 0 and ′{′a′}′\mathbf{'\{'}a{'\}'} satisfies the plane equation.

Line is parallel to the plane (no intersection) if ′{′d′}′⋅′{′n′}′=0\mathbf{'\{'}d{'\}'} \cdot \mathbf{'\{'}n{'\}'} = 0 but ′{′a′}′\mathbf{'\{'}a{'\}'} does not satisfy the plane equation.

Intersection of Two Planes​

Two planes ′{′n′}′1⋅′{′r′}′=d1\mathbf{'\{'}n{'\}'}_1 \cdot \mathbf{'\{'}r{'\}'} = d_1 and ′{′n′}′2⋅′{′r′}′=d2\mathbf{'\{'}n{'\}'}_2 \cdot \mathbf{'\{'}r{'\}'} = d_2 intersect in a line (provided ′{′n′}′1≠k′{′n′}′2\mathbf{'\{'}n{'\}'}_1 \ne k\mathbf{'\{'}n{'\}'}_2). The direction of the line of intersection is:

′{′d′}′=′{′n′}′1×′{′n′}′2\mathbf{'\{'}d{'\}'} = \mathbf{'\{'}n{'\}'}_1 \times \mathbf{'\{'}n{'\}'}_2

Find a point on the line by setting one variable to zero (if possible) and solving the resulting system.

Common Pitfalls​

  1. Confusing position vectors and direction vectors. A position vector specifies a point relative to the origin; a direction vector specifies a direction. They play fundamentally different roles in the equation of a line.

  2. Normal vector direction. The normal to a plane is perpendicular to every direction vector in the plane, not to the plane itself (a plane does not have a single direction).

  3. Cross product order. ′{′a′}′×′{′b′}′≠′{′b′}′×′{′a′}′\mathbf{'\{'}a{'\}'} \times \mathbf{'\{'}b{'\}'} \ne \mathbf{'\{'}b{'\}'} \times \mathbf{'\{'}a{'\}'}. Swapping the order negates the result.

  4. Zero direction components in symmetric form. When a direction component is zero, the symmetric form is undefined for that coordinate. Write that coordinate as a constant instead.

  5. Absolute values in angle formulas. For the acute angle between lines or planes, use the absolute value of the dot product. Without it, you obtain the obtuse supplement.

  6. Forgetting that ′{′a′}′×′{′b′}′\mathbf{'\{'}a{'\}'} \times \mathbf{'\{'}b{'\}'} is a vector. The dot product produces a scalar; the cross product produces a vector. The magnitude of the cross product gives area, not the cross product itself.

Practice Problems​

Problem 1

Given ′{′a′}′=2′{′i′}′−′{′j′}′+3′{′k′}′\mathbf{'\{'}a{'\}'} = 2\mathbf{'\{'}i{'\}'} - \mathbf{'\{'}j{'\}'} + 3\mathbf{'\{'}k{'\}'} and ′{′b′}′=′{′i′}′+4′{′j′}′−2′{′k′}′\mathbf{'\{'}b{'\}'} = \mathbf{'\{'}i{'\}'} + 4\mathbf{'\{'}j{'\}'} - 2\mathbf{'\{'}k{'\}'}, find: (a) ′{′a′}′⋅′{′b′}′\mathbf{'\{'}a{'\}'} \cdot \mathbf{'\{'}b{'\}'} (b) ∣′{′a′}′∣|\mathbf{'\{'}a{'\}'}| (c) the angle between ′{′a′}′\mathbf{'\{'}a{'\}'} and ′{′b′}′\mathbf{'\{'}b{'\}'}

Problem 2

Find the vector equation of the line through A(1,−2,3)A(1, -2, 3) and B(4,1,−1)B(4, 1, -1).

Problem 3

Find the Cartesian equation of the plane through (1,0,2)(1, 0, 2), (3,−1,1)(3, -1, 1), and (2,1,−1)(2, 1, -1).

Problem 4

Find the distance from the point P(3,1,−2)P(3, 1, -2) to the plane 2x−y+3z=42x - y + 3z = 4.

Problem 5

Find the area of the triangle with vertices A(1,2,3)A(1, 2, 3), B(4,1,0)B(4, 1, 0), C(2,−1,5)C(2, -1, 5).

Problem 6

Find the shortest distance between the parallel lines ′{′r′}′=(102)+λ(21−1)\mathbf{'\{'}r{'\}'} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} and ′{′r′}′=(320)+μ(21−1)\mathbf{'\{'}r{'\}'} = \begin{pmatrix} 3 \\ 2 \\ 0 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}.

Problem 7

Determine whether the line ′{′r′}′=(1,2,−1)+t(3,−1,2)\mathbf{'\{'}r{'\}'} = (1, 2, -1) + t(3, -1, 2) intersects the plane x+2y−z=5x + 2y - z = 5. If so, find the point of intersection.

Problem 8

Find the acute angle between the planes 2x+y−2z=12x + y - 2z = 1 and x+3y+z=4x + 3y + z = 4.

Answers to Selected Problems

Problem 1: (a) ′{′a′}′⋅′{′b′}′=2(1)+(−1)(4)+3(−2)=2−4−6=−8\mathbf{'\{'}a{'\}'} \cdot \mathbf{'\{'}b{'\}'} = 2(1) + (-1)(4) + 3(-2) = 2 - 4 - 6 = -8 (b) ∣′{′a′}′∣=4+1+9=14|\mathbf{'\{'}a{'\}'}| = \sqrt{4 + 1 + 9} = \sqrt{14} (c) cos⁡θ=−814⋅21=−876\cos\theta = \dfrac{-8}{\sqrt{14}\cdot\sqrt{21}} = \dfrac{-8}{7\sqrt{6}}, so θ≈122∘\theta \approx 122^\circ

Problem 3: Direction vectors: AB→=(2,−1,−1)\overrightarrow{AB} = (2, -1, -1) and AC→=(1,1,−3)\overrightarrow{AC} = (1, 1, -3). Normal: ′{′n′}′=AB→×AC→=(3−(−1),  (−1)(−3)−(−1)(1),  2−(−1))=(4,4,3)\mathbf{'\{'}n{'\}'} = \overrightarrow{AB} \times \overrightarrow{AC} = (3 - (-1),\; (-1)(-3) - (-1)(1),\; 2 - (-1)) = (4, 4, 3). Plane: 4(x−1)+4(y−0)+3(z−2)=04(x - 1) + 4(y - 0) + 3(z - 2) = 0, i.e. 4x+4y+3z=104x + 4y + 3z = 10.

Problem 4: D=∣2(3)−1(1)+3(−2)−4∣4+1+9=∣6−1−6−4∣14=514D = \dfrac{|2(3) - 1(1) + 3(-2) - 4|}{\sqrt{4 + 1 + 9}} = \dfrac{|6 - 1 - 6 - 4|}{\sqrt{14}} = \dfrac{5}{\sqrt{14}}

Problem 5: AB→=(3,−1,−3)\overrightarrow{AB} = (3, -1, -3), AC→=(1,−3,2)\overrightarrow{AC} = (1, -3, 2). Area =12∣AB→×AC→∣=12∣(−11,−9,−8)∣=121+81+642=2662= \dfrac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| = \dfrac{1}{2}|(-11, -9, -8)| = \dfrac{\sqrt{121 + 81 + 64}}{2} = \dfrac{\sqrt{266}}{2}.

Problem 7: Substituting x=1+3tx = 1 + 3t, y=2−ty = 2 - t, z=−1+2tz = -1 + 2t into x+2y−z=5x + 2y - z = 5: (1+3t)+2(2−t)−(−1+2t)=5  ⟹  6−t=5  ⟹  t=1(1 + 3t) + 2(2 - t) - (-1 + 2t) = 5 \implies 6 - t = 5 \implies t = 1. Intersection point: (4,1,1)(4, 1, 1).

Problem 8: cos⁡θ=∣(2)(1)+(1)(3)+(−2)(1)∣9⋅11=3311=111\cos\theta = \dfrac{|(2)(1) + (1)(3) + (-2)(1)|}{\sqrt{9}\cdot\sqrt{11}} = \dfrac{3}{3\sqrt{11}} = \dfrac{1}{\sqrt{11}}, so θ=arccos⁡ ⁣(111)≈72.5∘\theta = \arccos\!\left(\dfrac{1}{\sqrt{11}}\right) \approx 72.5^\circ.

Worked Examples​

Worked Example: Finding the Intersection of Two Lines

Find the point of intersection of the lines ′{′r′}′1=(1,2,−1)+s(3,−2,1)\mathbf{'\{'}r{'\}'}_1 = (1, 2, -1) + s(3, -2, 1) and ′{′r′}′2=(4,0,3)+t(1,1,−2)\mathbf{'\{'}r{'\}'}_2 = (4, 0, 3) + t(1, 1, -2), or show that they are skew.

Solution

Equating components:

1+3s=4+t(1)1 + 3s = 4 + t \quad \mathrm{(1)} 2−2s=0+t(2)2 - 2s = 0 + t \quad \mathrm{(2)} −1+s=3−2t(3)-1 + s = 3 - 2t \quad \mathrm{(3)}

From (2): t=2−2st = 2 - 2s.

Substituting into (1): 1+3s=4+2−2s  ⟹  5s=5  ⟹  s=11 + 3s = 4 + 2 - 2s \implies 5s = 5 \implies s = 1.

Then t=2−2(1)=0t = 2 - 2(1) = 0.

Check (3): LHS =−1+1=0= -1 + 1 = 0, RHS =3−0=3= 3 - 0 = 3. Since 0≠30 \ne 3, the system is inconsistent.

The lines do not intersect. To check if they are parallel: direction vectors (3,−2,1)(3, -2, 1) and (1,1,−2)(1, 1, -2) are not scalar multiples. Therefore the lines are skew.

Worked Example: Distance Between Skew Lines

Find the shortest distance between the skew lines ′{′r′}′1=(0,0,0)+s(1,2,3)\mathbf{'\{'}r{'\}'}_1 = (0, 0, 0) + s(1, 2, 3) and ′{′r′}′2=(1,−1,0)+t(2,3,4)\mathbf{'\{'}r{'\}'}_2 = (1, -1, 0) + t(2, 3, 4).

Solution

The shortest distance between two skew lines is given by:

d=∣(′{′a′}′2−′{′a′}′1)⋅(′{′d′}′1×′{′d′}′2)∣∣′{′d′}′1×′{′d′}′2∣d = \frac{|(\mathbf{'\{'}a{'\}'}_2 - \mathbf{'\{'}a{'\}'}_1) \cdot (\mathbf{'\{'}d{'\}'}_1 \times \mathbf{'\{'}d{'\}'}_2)|}{|\mathbf{'\{'}d{'\}'}_1 \times \mathbf{'\{'}d{'\}'}_2|}

′{′a′}′2−′{′a′}′1=(1,−1,0)\mathbf{'\{'}a{'\}'}_2 - \mathbf{'\{'}a{'\}'}_1 = (1, -1, 0), ′{′d′}′1=(1,2,3)\mathbf{'\{'}d{'\}'}_1 = (1, 2, 3), ′{′d′}′2=(2,3,4)\mathbf{'\{'}d{'\}'}_2 = (2, 3, 4).

′{′d′}′1×′{′d′}′2=∣′{′i′}′′{′j′}′′{′k′}′123234∣=(8−9) ′{′i′}′−(4−6) ′{′j′}′+(3−4) ′{′k′}′=(−1,2,−1)\mathbf{'\{'}d{'\}'}_1 \times \mathbf{'\{'}d{'\}'}_2 = \begin{vmatrix} \mathbf{'\{'}i{'\}'} & \mathbf{'\{'}j{'\}'} & \mathbf{'\{'}k{'\}'} \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{vmatrix} = (8 - 9)\,\mathbf{'\{'}i{'\}'} - (4 - 6)\,\mathbf{'\{'}j{'\}'} + (3 - 4)\,\mathbf{'\{'}k{'\}'} = (-1, 2, -1)

∣′{′d′}′1×′{′d′}′2∣=1+4+1=6|\mathbf{'\{'}d{'\}'}_1 \times \mathbf{'\{'}d{'\}'}_2| = \sqrt{1 + 4 + 1} = \sqrt{6}.

(′{′a′}′2−′{′a′}′1)⋅(′{′d′}′1×′{′d′}′2)=(1)(−1)+(−1)(2)+(0)(−1)=−1−2=−3(\mathbf{'\{'}a{'\}'}_2 - \mathbf{'\{'}a{'\}'}_1) \cdot (\mathbf{'\{'}d{'\}'}_1 \times \mathbf{'\{'}d{'\}'}_2) = (1)(-1) + (-1)(2) + (0)(-1) = -1 - 2 = -3.

d=∣−3∣6=36=62d = \frac{|-3|}{\sqrt{6}} = \frac{3}{\sqrt{6}} = \frac{\sqrt{6}}{2}

Worked Example: Plane Through Three Points with Verification

Find the Cartesian equation of the plane through A(2,1,−1)A(2, 1, -1), B(0,3,2)B(0, 3, 2), C(1,−1,1)C(1, -1, 1), and verify that all three points satisfy the equation.

Solution

AB→=(0−2,3−1,2−(−1))=(−2,2,3)\overrightarrow{AB} = (0-2, 3-1, 2-(-1)) = (-2, 2, 3).

AC→=(1−2,−1−1,1−(−1))=(−1,−2,2)\overrightarrow{AC} = (1-2, -1-1, 1-(-1)) = (-1, -2, 2).

Normal vector: ′{′n′}′=AB→×AC→=∣′{′i′}′′{′j′}′′{′k′}′−223−1−22∣=(4−(−6)) ′{′i′}′−(−4−(−3)) ′{′j′}′+(4−(−2)) ′{′k′}′=(10,1,6)\mathbf{'\{'}n{'\}'} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{'\{'}i{'\}'} & \mathbf{'\{'}j{'\}'} & \mathbf{'\{'}k{'\}'} \\ -2 & 2 & 3 \\ -1 & -2 & 2 \end{vmatrix} = (4 - (-6))\,\mathbf{'\{'}i{'\}'} - (-4 - (-3))\,\mathbf{'\{'}j{'\}'} + (4 - (-2))\,\mathbf{'\{'}k{'\}'} = (10, 1, 6).

Using point A(2,1,−1)A(2, 1, -1): 10(x−2)+1(y−1)+6(z−(−1))=010(x - 2) + 1(y - 1) + 6(z - (-1)) = 0.

10x−20+y−1+6z+6=0  ⟹  10x+y+6z=1510x - 20 + y - 1 + 6z + 6 = 0 \implies 10x + y + 6z = 15.

Verification:

A(2,1,−1)A(2, 1, -1): 10(2)+1+6(−1)=20+1−6=1510(2) + 1 + 6(-1) = 20 + 1 - 6 = 15. Confirmed.

B(0,3,2)B(0, 3, 2): 10(0)+3+6(2)=0+3+12=1510(0) + 3 + 6(2) = 0 + 3 + 12 = 15. Confirmed.

C(1,−1,1)C(1, -1, 1): 10(1)+(−1)+6(1)=10−1+6=1510(1) + (-1) + 6(1) = 10 - 1 + 6 = 15. Confirmed.

Worked Example: Line of Intersection of Two Planes

Find the vector equation of the line of intersection of the planes x+y−z=3x + y - z = 3 and 2x−y+3z=12x - y + 3z = 1.

Solution

The direction of the line is ′{′d′}′=′{′n′}′1×′{′n′}′2\mathbf{'\{'}d{'\}'} = \mathbf{'\{'}n{'\}'}_1 \times \mathbf{'\{'}n{'\}'}_2 where ′{′n′}′1=(1,1,−1)\mathbf{'\{'}n{'\}'}_1 = (1, 1, -1) and ′{′n′}′2=(2,−1,3)\mathbf{'\{'}n{'\}'}_2 = (2, -1, 3).

′{′d′}′=∣′{′i′}′′{′j′}′′{′k′}′11−12−13∣=(3−1) ′{′i′}′−(3−(−2)) ′{′j′}′+(−1−2) ′{′k′}′=(2,−5,−3)\mathbf{'\{'}d{'\}'} = \begin{vmatrix} \mathbf{'\{'}i{'\}'} & \mathbf{'\{'}j{'\}'} & \mathbf{'\{'}k{'\}'} \\ 1 & 1 & -1 \\ 2 & -1 & 3 \end{vmatrix} = (3 - 1)\,\mathbf{'\{'}i{'\}'} - (3 - (-2))\,\mathbf{'\{'}j{'\}'} + (-1 - 2)\,\mathbf{'\{'}k{'\}'} = (2, -5, -3)

To find a point on the line, set z=0z = 0:

x+y=3x + y = 3 and 2x−y=12x - y = 1. Adding: 3x=4  ⟹  x=4/33x = 4 \implies x = 4/3, y=3−4/3=5/3y = 3 - 4/3 = 5/3.

A point on the line is (43,53,0)\left(\dfrac{4}{3}, \dfrac{5}{3}, 0\right).

The line of intersection is:

′{′r′}′=(43,53,0)+t(2,−5,−3),t∈′{′R′}′\mathbf{'\{'}r{'\}'} = \left(\frac{4}{3}, \frac{5}{3}, 0\right) + t(2, -5, -3), \quad t \in \mathbb{'\{'}R{'\}'}

Additional Common Pitfalls​

  • Confusing the angle between a line and a plane. The angle α\alpha between a line and a plane satisfies sin⁡α=∣′{′d′}′⋅′{′n′}′∣∣′{′d′}′∣∣′{′n′}′∣\sin\alpha = \dfrac{|\mathbf{'\{'}d{'\}'} \cdot \mathbf{'\{'}n{'\}'}|}{|\mathbf{'\{'}d{'\}'}||\mathbf{'\{'}n{'\}'}|}, not cos⁡α\cos\alpha. The dot product of the direction vector and the normal gives the sine of the angle, not the cosine.

  • Distance formula sign errors. The distance from a point to a plane uses the absolute value in the numerator: D=∣′{′n′}′⋅′{′p′}′−d∣∣′{′n′}′∣D = \dfrac{|\mathbf{'\{'}n{'\}'} \cdot \mathbf{'\{'}p{'\}'} - d|}{|\mathbf{'\{'}n{'\}'}|}. Dropping the absolute value can give a negative distance.

  • Cross product component order. When computing ′{′a′}′×′{′b′}′=(a2b3−a3b2,  a3b1−a1b3,  a1b2−a2b1)\mathbf{'\{'}a{'\}'} \times \mathbf{'\{'}b{'\}'} = (a_2 b_3 - a_3 b_2,\; a_3 b_1 - a_1 b_3,\; a_1 b_2 - a_2 b_1), each component follows a cyclic pattern. Mixing up the indices is the most common arithmetic error in vector problems.

  • Parallel line distance requires cross product. The distance between parallel lines ′{′r′}′1=′{′a′}′1+s′{′d′}′\mathbf{'\{'}r{'\}'}_1 = \mathbf{'\{'}a{'\}'}_1 + s\mathbf{'\{'}d{'\}'} and ′{′r′}′2=′{′a′}′2+t′{′d′}′\mathbf{'\{'}r{'\}'}_2 = \mathbf{'\{'}a{'\}'}_2 + t\mathbf{'\{'}d{'\}'} is ∣(′{′a′}′2−′{′a′}′1)×′{′d′}′∣∣′{′d′}′∣\dfrac{|(\mathbf{'\{'}a{'\}'}_2 - \mathbf{'\{'}a{'\}'}_1) \times \mathbf{'\{'}d{'\}'}|}{|\mathbf{'\{'}d{'\}'}|}, not the magnitude of the difference of the position vectors.

  • Scalar triple product for volume. The volume of a parallelepiped is ∣′{′a′}′⋅(′{′b′}′×′{′c′}′)∣|\mathbf{'\{'}a{'\}'} \cdot (\mathbf{'\{'}b{'\}'} \times \mathbf{'\{'}c{'\}'})|. The absolute value is essential since the scalar triple product can be negative. The parentheses matter: ′{′a′}′⋅(′{′b′}′×′{′c′}′)\mathbf{'\{'}a{'\}'} \cdot (\mathbf{'\{'}b{'\}'} \times \mathbf{'\{'}c{'\}'}), not (′{′a′}′⋅′{′b′}′)×′{′c′}′(\mathbf{'\{'}a{'\}'} \cdot \mathbf{'\{'}b{'\}'}) \times \mathbf{'\{'}c{'\}'}.

Exam-Style Problems​

Problem 9

Given ′{′a′}′=3′{′i′}′−′{′j′}′+2′{′k′}′\mathbf{'\{'}a{'\}'} = 3\mathbf{'\{'}i{'\}'} - \mathbf{'\{'}j{'\}'} + 2\mathbf{'\{'}k{'\}'} and ′{′b′}′=2′{′i′}′+′{′j′}′−′{′k′}′\mathbf{'\{'}b{'\}'} = 2\mathbf{'\{'}i{'\}'} + \mathbf{'\{'}j{'\}'} - \mathbf{'\{'}k{'\}'}, find: (a) the vector projection of ′{′b′}′\mathbf{'\{'}b{'\}'} onto ′{′a′}′\mathbf{'\{'}a{'\}'}; (b) the scalar projection of ′{′b′}′\mathbf{'\{'}b{'\}'} onto ′{′a′}′\mathbf{'\{'}a{'\}'}; (c) the component of ′{′b′}′\mathbf{'\{'}b{'\}'} perpendicular to ′{′a′}′\mathbf{'\{'}a{'\}'}.

Problem 10

Find the volume of the parallelepiped with edges OA→=(1,2,3)\overrightarrow{OA} = (1, 2, 3), OB→=(4,−1,0)\overrightarrow{OB} = (4, -1, 0), OC→=(2,1,−2)\overrightarrow{OC} = (2, 1, -2).

Problem 11

Find the acute angle between the line ′{′r′}′=(1,0,−2)+t(1,3,1)\mathbf{'\{'}r{'\}'} = (1, 0, -2) + t(1, 3, 1) and the plane 2x−y+z=52x - y + z = 5.

Problem 12

Points A(1,2,0)A(1, 2, 0), B(3,1,4)B(3, 1, 4), C(0,−1,2)C(0, -1, 2), D(2,0,6)D(2, 0, 6) are given. Show that AB→\overrightarrow{AB} is parallel to CD→\overrightarrow{CD} and find the distance between the parallel lines ABAB and CDCD.

Problem 13

Find the Cartesian equation of the plane that is equidistant from the points P(1,2,3)P(1, 2, 3) and Q(5,−2,7)Q(5, -2, 7) and passes through R(0,1,−1)R(0, 1, -1).

Problem 14

The lines L1:′{′r′}′=(2,−1,3)+λ(1,2,−1)L_1: \mathbf{'\{'}r{'\}'} = (2, -1, 3) + \lambda(1, 2, -1) and L2:′{′r′}′=(4,−5,5)+μ(2,−1,1)L_2: \mathbf{'\{'}r{'\}'} = (4, -5, 5) + \mu(2, -1, 1) intersect. Find the point of intersection and the acute angle between the lines.

Problem 15

A pyramid has a square base ABCDABCD with A=(0,0,0)A = (0, 0, 0), B=(4,0,0)B = (4, 0, 0), C=(4,4,0)C = (4, 4, 0), D=(0,4,0)D = (0, 4, 0), and apex V=(2,2,6)V = (2, 2, 6). Find the angle between the face VABVAB and the base ABCDABCD.

Answers to Additional Problems

Problem 9: (a) ′{′a′}′⋅′{′b′}′=6−1−2=3\mathbf{'\{'}a{'\}'} \cdot \mathbf{'\{'}b{'\}'} = 6 - 1 - 2 = 3. ∣′{′a′}′∣2=9+1+4=14|\mathbf{'\{'}a{'\}'}|^2 = 9 + 1 + 4 = 14. proj′{′a′}′′{′b′}′=314(3,−1,2)=(914,−314,37)\mathrm{proj}_{\mathbf{'\{'}a{'\}'}}\mathbf{'\{'}b{'\}'} = \dfrac{3}{14}(3, -1, 2) = \left(\dfrac{9}{14}, -\dfrac{3}{14}, \dfrac{3}{7}\right). (b) comp′{′a′}′′{′b′}′=314\mathrm{comp}_{\mathbf{'\{'}a{'\}'}}\mathbf{'\{'}b{'\}'} = \dfrac{3}{\sqrt{14}}. (c) Perpendicular component: ′{′b′}′−proj′{′a′}′′{′b′}′=(2,1,−1)−(914,−314,37)=(1914,1714,−107)\mathbf{'\{'}b{'\}'} - \mathrm{proj}_{\mathbf{'\{'}a{'\}'}}\mathbf{'\{'}b{'\}'} = (2, 1, -1) - \left(\dfrac{9}{14}, -\dfrac{3}{14}, \dfrac{3}{7}\right) = \left(\dfrac{19}{14}, \dfrac{17}{14}, -\dfrac{10}{7}\right).

Problem 10: V=∣′{′a′}′⋅(′{′b′}′×′{′c′}′)∣V = |\mathbf{'\{'}a{'\}'} \cdot (\mathbf{'\{'}b{'\}'} \times \mathbf{'\{'}c{'\}'})|. ′{′b′}′×′{′c′}′=∣′{′i′}′′{′j′}′′{′k′}′4−1021−2∣=(2,8,6)\mathbf{'\{'}b{'\}'} \times \mathbf{'\{'}c{'\}'} = \begin{vmatrix} \mathbf{'\{'}i{'\}'} & \mathbf{'\{'}j{'\}'} & \mathbf{'\{'}k{'\}'} \\ 4 & -1 & 0 \\ 2 & 1 & -2 \end{vmatrix} = (2, 8, 6). ′{′a′}′⋅(′{′b′}′×′{′c′}′)=1(2)+2(8)+3(6)=2+16+18=36\mathbf{'\{'}a{'\}'} \cdot (\mathbf{'\{'}b{'\}'} \times \mathbf{'\{'}c{'\}'}) = 1(2) + 2(8) + 3(6) = 2 + 16 + 18 = 36. V=∣36∣=36V = |36| = 36.

Problem 11: ′{′d′}′=(1,3,1)\mathbf{'\{'}d{'\}'} = (1, 3, 1), ′{′n′}′=(2,−1,1)\mathbf{'\{'}n{'\}'} = (2, -1, 1). sin⁡α=∣(1)(2)+(3)(−1)+(1)(1)∣∣(1,3,1)∣∣(2,−1,1)∣=∣2−3+1∣11⋅6=066=0\sin\alpha = \dfrac{|(1)(2) + (3)(-1) + (1)(1)|}{|(1,3,1)||(2,-1,1)|} = \dfrac{|2 - 3 + 1|}{\sqrt{11}\cdot\sqrt{6}} = \dfrac{0}{\sqrt{66}} = 0. The angle is 0∘0^\circ, meaning the line is parallel to (lies in) the plane. Verification: the point (1,0,−2)(1, 0, -2) lies on the line and 2(1)−0+(−2)=0≠52(1) - 0 + (-2) = 0 \ne 5, so the line is parallel to but not in the plane.

Problem 12: AB→=(2,−1,4)\overrightarrow{AB} = (2, -1, 4), CD→=(2,1,4)\overrightarrow{CD} = (2, 1, 4). These are not parallel (not scalar multiples). Rechecking: D=(2,0,6)D = (2, 0, 6), C=(0,−1,2)C = (0, -1, 2), so CD→=(2−0,0−(−1),6−2)=(2,1,4)\overrightarrow{CD} = (2 - 0, 0 - (-1), 6 - 2) = (2, 1, 4). Since (2,−1,4)(2, -1, 4) and (2,1,4)(2, 1, 4) are not multiples, the lines are not parallel. The problem statement cannot be verified as stated.

Problem 13: The midpoint of PQPQ is M=(3,0,5)M = (3, 0, 5). The plane is perpendicular to PQ→=(4,−4,4)\overrightarrow{PQ} = (4, -4, 4), so the normal is (4,−4,4)(4, -4, 4), simplified to (1,−1,1)(1, -1, 1). Plane: 1(x−0)−1(y−1)+1(z−(−1))=0  ⟹  x−y+z=−21(x - 0) - 1(y - 1) + 1(z - (-1)) = 0 \implies x - y + z = -2. Check MM: 3−0+5=8≠−23 - 0 + 5 = 8 \ne -2. Wait -- the plane passes through RR, not MM. Plane through R(0,1,−1)R(0, 1, -1) with normal (1,−1,1)(1, -1, 1): (x−0)−(y−1)+(z+1)=0  ⟹  x−y+z+2=0(x - 0) - (y - 1) + (z + 1) = 0 \implies x - y + z + 2 = 0. Distance from PP: ∣1−2+3+2∣/3=4/3|1 - 2 + 3 + 2|/\sqrt{3} = 4/\sqrt{3}. Distance from QQ: ∣5−(−2)+7+2∣/3=16/3≠4/3|5 - (-2) + 7 + 2|/\sqrt{3} = 16/\sqrt{3} \ne 4/\sqrt{3}. This approach is wrong. The correct method: the equidistant plane has normal PQ→\overrightarrow{PQ} and passes through the midpoint M(3,0,5)M(3, 0, 5). So: (x−3)−(y−0)+(z−5)=0  ⟹  x−y+z=8(x-3) - (y-0) + (z-5) = 0 \implies x - y + z = 8. This must also pass through RR: 0−1+(−1)=−2≠80 - 1 + (-1) = -2 \ne 8. No single plane is equidistant from PP, QQ and passes through RR unless RR lies on the perpendicular bisector. Since RR does not satisfy x−y+z=8x - y + z = 8, the problem has no solution.

Problem 14: Equating: 2+λ=4+2μ2 + \lambda = 4 + 2\mu, −1+2λ=−5−μ-1 + 2\lambda = -5 - \mu, 3−λ=5+μ3 - \lambda = 5 + \mu. From equation 1: λ=2+2μ\lambda = 2 + 2\mu. From equation 3: 3−(2+2μ)=5+μ  ⟹  1−2μ=5+μ  ⟹  −3μ=4  ⟹  μ=−4/33 - (2 + 2\mu) = 5 + \mu \implies 1 - 2\mu = 5 + \mu \implies -3\mu = 4 \implies \mu = -4/3. λ=2+2(−4/3)=−2/3\lambda = 2 + 2(-4/3) = -2/3. Check equation 2: −1+2(−2/3)=−1−4/3=−7/3-1 + 2(-2/3) = -1 - 4/3 = -7/3. RHS: −5−(−4/3)=−5+4/3=−11/3-5 - (-4/3) = -5 + 4/3 = -11/3. Since −7/3≠−11/3-7/3 \ne -11/3, the lines do not actually intersect. The problem as stated is incorrect.

Problem 15: Face VABVAB has normal VA→×VB→\overrightarrow{VA} \times \overrightarrow{VB}. VA→=(−2,−2,−6)\overrightarrow{VA} = (-2, -2, -6), VB→=(2,−2,−6)\overrightarrow{VB} = (2, -2, -6). VA→×VB→=(12−12,−(12−12),4−(−4))=(0,0,8)\overrightarrow{VA} \times \overrightarrow{VB} = (12 - 12, -(12 - 12), 4 - (-4)) = (0, 0, 8). Normal to VABVAB: (0,0,1)(0, 0, 1) (simplified). Normal to base: (0,0,1)(0, 0, 1). The angle between the face and the base: cos⁡θ=∣(0)(0)+(0)(0)+(1)(1)∣1⋅1=1\cos\theta = \dfrac{|(0)(0) + (0)(0) + (1)(1)|}{1 \cdot 1} = 1, so θ=0∘\theta = 0^\circ. This indicates face VABVAB is parallel to the base, which is correct since VV is directly above the centre of the square base.

If You Get These Wrong, Revise:​

  • Three-dimensional coordinate geometry → Review ./vectors (sections on Lines and Planes)
  • Trigonometric ratios and inverse trig functions → Review geometry and trigonometry topics
  • Algebraic solving of simultaneous equations → Review ./matrices (section on Solving Systems)
  • Scalar and vector product properties → Review ./vectors (sections on Dot Product and Cross Product)
  • Magnitude and unit vectors → Review ./vectors (section on Magnitude and Unit Vectors)

For the A-Level treatment of this topic, see Vectors.