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📝 Vector equation of a line in 3D (27 MCQs)

📖 From Calculus • 12. Three Dimensional Space: Vectors • 27 questions available

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What is Vector equation of a line in 3D?

Definition:
The vector equation $\vec{r}(t) = \vec{r}_0 + t\vec{v}$ compactly represents a line, where $\vec{r}_0$ is position vector of a point on line and $\vec{v}$ is direction vector, unifying all coordinates.

Example:
Line with $\vec{r}_0 = \langle 1,2,3 \rangle$ and $\vec{v} = \langle 4,5,6 \rangle$ is $\vec{r}(t) = \langle 1+4t, 2+5t, 3+6t \rangle$ , equivalent to parametric form.

Reason:
Vector notation emphasizes geometric nature over coordinate dependence, simplifying derivations in differential geometry and enabling coordinate-free reasoning in advanced calculus.

Easy

Medium

Hard

📝 All Vector equation of a line in 3D MCQs

Q1. A drone flies along the line $\mathbf{r} = \langle 1,2,3 \rangle + t\langle 4,-1,2 \rangle$ . A sensor at point $P(5,0,7)$ detects the drone when it is closest. Which mathematical condition must be satisfied to find this specific parameter $t$ ?

A.The vector from

P

to the line's position vector must be parallel to the direction vector.

B.The dot product of the direction vector and the vector connecting

P

to a general point on the line must equal zero. ✅

C.The cross product of the position vector of

P

and the direction vector must be minimized.

D.The magnitude of the direction vector must equal the distance from the origin to

P

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Finding the closest point requires minimizing distance, which geometrically corresponds to the perpendicular projection. The vector connecting the external point to the line must be orthogonal to the line's direction vector. Therefore, their dot product must be zero, establishing the necessary scalar equation for $t$ .

Q2. Two lines are given by $L_1: \mathbf{r} = \mathbf{a} + s\mathbf{u}$ and $L_2: \mathbf{r} = \mathbf{b} + t\mathbf{v}$ . If $\mathbf{u} \times \mathbf{v} \neq \mathbf{0}$ but the system of three parametric equations has no solution, what is the precise spatial relationship?

A.The lines are parallel and distinct.

B.The lines intersect at a unique point outside the defined parameter range.

C.The lines are skew. ✅

D.The lines are coincident but parameterized in opposite directions.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Non-parallel direction vectors imply the lines are either intersecting or skew. Since the system lacks a simultaneous solution for all three coordinates, they do not intersect. In three-dimensional space, non-parallel, non-intersecting lines are defined as skew, meaning they lie in different planes.

Q3. A student attempts to find the intersection of $\mathbf{r}_1 = \langle 1,0,1 \rangle + t\langle 2,1,3 \rangle$ and $\mathbf{r}_2 = \langle 0,1,2 \rangle + s\langle 1,-1,0 \rangle$ . They solve the x and y components, finding $t=1, s=-1$ , but fail to check z. What is the fundamental error in this reasoning?

A.Assuming two variables can satisfy three independent constraints without verification. ✅

B.Using the wrong parameter names for each line.

C.Solving for x and y before solving for z violates algebraic order.

D.The direction vectors are orthogonal, making intersection impossible.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: In 3D, two lines generally do not intersect. Solving only two component equations provides a candidate solution that satisfies those specific projections. However, consistency requires the third component to also hold true with the same parameters. Failing to verify leads to assuming an intersection where none exists.

Q4. Consider the vector equation $\mathbf{r}(t) = \langle 3-t, 2+2t, -1+3t \rangle$ . If we reparameterize using $\tau = 2t - 1$ , how does the geometric locus of points change?

A.The line shifts one unit in the negative x-direction.

B.The orientation reverses and the speed of traversal doubles.

C.The geometric set of points remains identical, but the parameterization changes. ✅

D.The line becomes a ray starting at

t=0.5

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Reparameterization via a linear transformation affects only how points are traced relative to the parameter variable. The underlying set of spatial coordinates satisfying the linear relationships remains invariant. While velocity vectors scale, the infinite collection of points defining the line in space is unchanged.

Q5. Given a graph showing two lines in 3D projected onto the xy-plane appearing to cross, and their z-coordinates plotted separately showing different values at that crossing x,y, what conclusion is valid?

A.The lines intersect at the projected point.

B.The lines are parallel.

C.The lines are skew. ✅

D.The graph is misleading; lines in 3D always intersect if projections do.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Visual interpretation of 3D objects on 2D media requires analyzing multiple views. An apparent intersection in one projection combined with differing coordinate values in another dimension confirms non-intersection. Since the projections suggest non-parallelism, the only remaining classification for non-intersecting lines in space is skew.

Q6. Which vector equation correctly models a line passing through $A(1,2,3)$ and perpendicular to both $\mathbf{u}=\langle 1,0,1 \rangle$ and $\mathbf{v}=\langle 0,1,-1 \rangle$ ?

\mathbf{r} = \langle 1,2,3 \rangle + t\langle 1,1,1 \rangle

\mathbf{r} = \langle 1,2,3 \rangle + t\langle -1,1,1 \rangle

✅

\mathbf{r} = \langle 1,2,3 \rangle + t\langle 1,-1,1 \rangle

\mathbf{r} = \langle 1,2,3 \rangle + t\langle 0,0,0 \rangle

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The direction vector must be orthogonal to both given vectors, found via the cross product $\mathbf{u} \times \mathbf{v}$ . Calculating determinant yields $\langle -1, 1, 1 \rangle$ . Option B uses this exact direction vector anchored at point A. Other options use incorrect cross product results or invalid zero vectors.

Q7. If line $L$ is defined by $\mathbf{r} = \mathbf{p} + t\mathbf{d}$ , and a plane contains $L$ , which statement about the plane's normal vector $\mathbf{n}$ is necessarily true?

\mathbf{n}

is parallel to

\mathbf{d}

\mathbf{n} \cdot \mathbf{d} = 0

and

\mathbf{n} \cdot (\mathbf{p} - \mathbf{q}) = 0

for any point

\mathbf{q}

on the plane. ✅

\mathbf{n} = \mathbf{d} \times \mathbf{p}

\mathbf{n}

can be any vector perpendicular to

\mathbf{p}

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For a line to lie entirely within a plane, its direction vector must be orthogonal to the plane's normal. Additionally, any point on the line must satisfy the plane equation. This dual constraint ensures containment rather than mere parallelism or single-point intersection.

Q8. An engineer models a support beam as $\mathbf{r} = \langle 0,0,5 \rangle + t\langle 3,4,0 \rangle$ . A cable runs from origin to the beam perpendicularly. Why can't we simply set $t=0$ to find the connection point?

A.Because

t=0

represents the endpoint, not necessarily the perpendicular foot.

B.Because the beam is horizontal and cables are vertical.

C.Because the origin lies on the line.

D.Because perpendicularity requires minimizing the distance function derivative or dot product. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: The parameter $t=0$ defines a specific fixed point. Perpendicularity is a geometric condition dependent on the relative positions of the external point and the entire line. Unless the fixed point happens to be the projection, arbitrary parameter values won't satisfy orthogonality. One must solve $\mathbf{r}(t) \cdot \mathbf{d} = \mathbf{0}_{relative}$ .

Q9. Compare symmetric equations $\frac{x-1}{2} = \frac{y+3}{0} = \frac{z-2}{-1}$ with vector form. What critical information does the zero denominator convey that might be lost in careless conversion?

A.The line is undefined.

B.The y-coordinate is constant at -3, indicating a direction component of zero. ✅

C.The line is parallel to the y-axis.

D.The equation represents a plane, not a line.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A zero in the denominator of symmetric equations indicates the corresponding direction component is zero. This means the coordinate remains constant along the line. Converting back to vector form requires explicitly setting that component to zero in the direction vector and fixing the position coordinate, avoiding division by zero errors.

Q10. Three points $A, B, C$ are collinear. If $\vec{AB} = k \vec{AC}$ , and $k < 0$ , what does this imply about the arrangement of points?

A.B lies between A and C.

B.C lies between A and B.

C.A lies between B and C. ✅

D.The points form a triangle.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Scalar multiples relate displacement vectors. A positive $k$ implies same direction from A. A negative $k$ implies $\vec{AB}$ and $\vec{AC}$ point in opposite directions relative to A. Therefore, A must be situated between B and C on the infinite line containing them.

Q11. A line passes through $(2,1,0)$ with direction $\langle 1,2,3 \rangle$ . Another line passes through $(0,0,0)$ with direction $\langle 2,4,6 \rangle$ . Without calculation, classify their relationship.

A.Intersecting at origin.

B.Skew.

C.Parallel and distinct. ✅

D.Coincident.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Direction vectors are scalar multiples ( $\langle 2,4,6 \rangle = 2\langle 1,2,3 \rangle$ ), confirming parallelism. Checking containment: substituting origin into first line's equation yields no valid $t$ . Thus, they share direction but not points, making them parallel and distinct rather than coincident.

Q12. In modeling a particle trajectory $\mathbf{r}(t)$ , why is the vector form superior to describing the path as intersection of two planes for kinematic analysis?

A.Vector form directly provides position as a function of time, enabling velocity/acceleration derivation. ✅

B.Plane intersections are computationally heavier.

C.Vector forms cannot represent curves.

D.Plane equations lose directional information entirely.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Kinematics requires temporal evolution. Vector equations explicitly map scalar time to spatial position, allowing differentiation for dynamics. Implicit plane intersections define geometry statically without inherent parameterization. Recovering motion from implicit surfaces requires arbitrary parameter selection, whereas vector form encodes physical motion naturally.

Q13. Given lines $L_1: \mathbf{r} = \mathbf{a} + t\mathbf{u}$ and $L_2: \mathbf{r} = \mathbf{a} + s\mathbf{v}$ where $\mathbf{u}$ and $\mathbf{v}$ are non-parallel, what geometric object is formed by all points $\mathbf{a} + t\mathbf{u} + s\mathbf{v}$ ?

A.A line bisecting the angle between

L_1

and

L_2

B.A plane containing both lines. ✅

C.A hyperboloid of one sheet.

D.Two intersecting planes.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Linear combinations of two non-parallel vectors anchored at a common point span a two-dimensional subspace. Adding all possible scalar multiples of $\mathbf{u}$ and $\mathbf{v}$ to base point $\mathbf{a}$ generates every point in the unique plane defined by the intersecting lines, not just the lines themselves.

Q14. A student writes the line through $(1,2,3)$ parallel to $\langle 2,4,6 \rangle$ as $\mathbf{r} = \langle 1,2,3 \rangle + t\langle 1,2,3 \rangle$ . Is this correct?

A.No, the direction vector must be exactly

\langle 2,4,6 \rangle

B.Yes, because direction vectors can be scaled by any non-zero constant. ✅

C.No, the point should also be scaled.

D.Yes, but only if

t

is restricted to even integers.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Direction vectors define orientation, not magnitude. Any non-zero scalar multiple represents the same line direction. Simplifying $\langle 2,4,6 \rangle$ to $\langle 1,2,3 \rangle$ is mathematically valid and often preferred for simplicity. The resulting parametric equations trace the identical geometric locus.

Q15. If the shortest distance between two skew lines is zero, what must be true?

A.They are parallel.

B.They intersect.

C.They are coincident.

D.Distance cannot be zero for skew lines by definition. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Skew lines are defined as non-parallel, non-intersecting lines in 3D. Non-intersection implies strictly positive separation. If calculated distance is zero, the lines actually intersect, contradicting the skew classification. Thus, zero distance proves they are not skew, highlighting importance of verifying assumptions before applying formulas.

Q16. Consider $\mathbf{r} = \langle t, t^2, t^3 \rangle$ . Why is this NOT a valid vector equation of a line despite being in vector form?

A.It uses too many parameters.

B.Components are not linear functions of the parameter. ✅

C.It doesn't pass through origin.

D.Vector equations must use unit direction vectors.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Lines are characterized by constant direction, requiring all components to be linear (degree 1) in the parameter. Quadratic and cubic terms indicate curvature, representing a space curve rather than a straight line. Form alone doesn't guarantee linearity; functional dependence determines geometric classification.

Q17. Two lines have direction vectors $\mathbf{d}_1 = \langle 1,2,3 \rangle$ and $\mathbf{d}_2 = \langle -2,-4,-6 \rangle$ . A student claims they are skew because parameters differ. Analyze this error.

A.Correct; different parameters prevent intersection.

B.Incorrect; antiparallel vectors still indicate parallel lines, precluding skewness. ✅

C.Incorrect; they must be perpendicular to be skew.

D.Correct; negative scaling changes the line's identity.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Skewness requires non-parallel directions. Here, $\mathbf{d}_2 = -2\mathbf{d}_1$ , confirming parallelism regardless of sign. Parameter naming is irrelevant to geometric classification. The student confused parametric representation with intrinsic geometry. Parallel lines cannot be skew; they either coincide or remain equidistant.

Q18. A navigation system gives aircraft path as intersection of planes $x+y+z=6$ and $2x-y=3$ . Convert to vector form efficiently.

A.Find two points satisfying both equations, then compute direction via subtraction.

B.Cross the normal vectors for direction, then find one particular solution point. ✅

C.Use elimination to express two variables in terms of the third as parameter.

D.All methods work equally well; choice is purely aesthetic.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Cross product of normals $\langle 1,1,1 \rangle \times \langle 2,-1,0 \rangle$ instantly yields direction vector parallel to intersection. Finding one point via substitution completes the equation. This avoids solving full systems twice for two points or managing messy fractions from elimination, demonstrating optimal method selection.

Q19. If line $L$ has equation $\mathbf{r} = \mathbf{a} + t\mathbf{d}$ , and we replace $\mathbf{d}$ with $\mathbf{d}' = \mathbf{d} + \mathbf{a}$ , what happens?

A.Line translates by vector

\mathbf{a}

B.Line rotates around origin.

C.Line changes completely unless

\mathbf{a}

is parallel to

\mathbf{d}

. ✅

D.Nothing changes; addition doesn't affect direction.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Direction determines line orientation. Adding position vector $\mathbf{a}$ to direction $\mathbf{d}$ alters direction unless $\mathbf{a}$ is already parallel to $\mathbf{d}$ . New direction generally differs, creating an entirely different line through same base point. Only special alignment preserves original geometry.

Q20. Graph shows line segment from $A$ to $B$ . Vector equation $\mathbf{r} = A + t(B-A)$ models infinite line. How to restrict to segment?

A.Set

t \in [0,1]

. ✅

B.Set

t > 0

C.Normalize

B-A

first.

D.Impossible; vector equations only model infinite lines.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Parameter $t=0$ yields $A$ ; $t=1$ yields $B$ . Linear interpolation between endpoints corresponds exactly to $t$ in unit interval. Values outside extend beyond endpoints. Domain restriction converts infinite line representation to bounded segment without altering underlying vector structure.

Q21. Why can't we write symmetric equations for a line perpendicular to the z-axis?

A.Such lines don't exist in 3D.

B.We can; the z-term simply has zero numerator.

C.Division by zero occurs in the z-denominator. ✅

D.Perpendicular lines have undefined slope.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Lines perpendicular to z-axis have direction vector with z-component zero. Symmetric form divides by direction components. Zero denominator is undefined, requiring special notation like $z=c$ alongside other ratios. This limitation highlights symmetric form's inadequacy for certain orientations compared to robust vector form.

Q22. Given $\mathbf{r} = \langle 1,2,3 \rangle + t\langle 4,5,6 \rangle$ , a student finds point at $t=2$ as $\langle 9,12,15 \rangle$ . Check validity.

A.Correct arithmetic. ✅

B.Incorrect; should be

\langle 8,10,12 \rangle

C.Incorrect; forgot to add initial point.

D.Incorrect; multiplied t incorrectly.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Substituting $t=2$ : $\langle 1+8, 2+10, 3+12 \rangle = \langle 9,12,15 \rangle$ . Computation follows $\mathbf{p} + t\mathbf{d}$ correctly. Verification reinforces proper order of operations: scale direction first, then translate by base point. Common errors include forgetting base point or mis-scaling.

Q23. Two lines appear parallel in 3D software but numerical check shows direction vectors aren't exact multiples. Best explanation?

A.Software rendering error.

B.Lines are actually skew but nearly parallel.

C.Numerical precision issues; truly parallel lines have proportional components.

D.Both B and C are plausible depending on context. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Visual similarity doesn't guarantee mathematical parallelism. Near-parallel skew lines exist and render similarly at distance. Alternatively, floating-point approximations may obscure exact proportionality. Critical thinking requires distinguishing perceptual artifacts from mathematical reality, checking both numerical ratios and considering computational limitations.

Q24. If $\mathbf{r}(t) = \mathbf{a} + t\mathbf{d}$ describes motion, what does $|\mathbf{d}|$ represent physically?

A.Total distance traveled.

B.Speed of the particle. ✅

C.Acceleration magnitude.

D.Displacement from origin.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Position derivative with respect to time gives velocity vector $\mathbf{d}$ . Its magnitude is instantaneous speed. Constant $\mathbf{d}$ implies uniform motion. Distinguishing vector velocity from scalar speed and understanding parameter's physical meaning connects abstract math to kinematic reality, essential for applied problems.

Q25. Line $L$ intersects plane $\Pi$ at point $Q$ . If $L$ is reflected across $\Pi$ , how is new line's direction related to original?

A.Same direction.

B.Opposite direction.

C.Original direction minus twice its projection onto plane normal. ✅

D.Unpredictable without specific coordinates.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Reflection reverses the normal component while preserving tangential component. Mathematically, $\mathbf{d}_{new} = \mathbf{d} - 2(\mathbf{d} \cdot \mathbf{n})\mathbf{n}/|\mathbf{n}|^2$ . This formula captures geometric reflection precisely. Recognizing this transformation avoids recomputing from scratch, demonstrating deep understanding of vector operations in geometric contexts.

Q26. Student solves for intersection, gets $t=3$ from x-equation and $t=3$ from y-equation, concludes intersection exists. Critique.

A.Valid reasoning; consistent t proves intersection.

B.Incomplete; must verify z-equation yields same t.

C.Invalid; should use different parameters for each line.

D.Both B and C are correct critiques. ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Using same parameter name for different lines is notational error masking potential inconsistency. Even with corrected notation, verifying all three components is mandatory. Consistency in two dimensions doesn't guarantee third. Proper critique addresses both notational rigor and completeness of verification.

Q27. Why is $\mathbf{r} = \mathbf{a} + t\mathbf{d}$ preferred over $\mathbf{r} = \mathbf{b} + s\mathbf{d}$ when $\mathbf{a} \neq \mathbf{b}$ but both represent same line?

A.They aren't equivalent; different points mean different lines.

B.Preference depends on which point simplifies subsequent calculations. ✅

C.First form is always correct; second is wrong.

D.Only unit direction vectors make forms equivalent.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Infinitely many point-direction pairs describe same line. Choice depends on problem context: known point, symmetry, or computational convenience. Recognizing equivalence despite different representations prevents confusion and enables strategic selection. Mathematical objects transcend specific parametrizations.

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