This TSA checks vector and Cartesian equations of lines, parameters, coordinate axes, points on lines, foot of perpendicular, shortest distance, reflection in a line, relationships between lines, angles between lines, and projections onto or perpendicular to a line.
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Vector and Cartesian Equations of Lines
1. The line \(l\) passes through \[ A=\begin{pmatrix}1\\2\\-1\end{pmatrix} \quad\text{and}\quad B=\begin{pmatrix}4\\-1\\5\end{pmatrix}. \] Which of the following is a valid vector equation of \(l\)?
Tested fundamental: Forming the vector equation of a line from two points.
Explanation: \(\overrightarrow{AB}=\begin{pmatrix}3\\-3\\6\end{pmatrix}=3\begin{pmatrix}1\\-1\\2\end{pmatrix}\). A valid equation may use \(B\) as the fixed point and \(\begin{pmatrix}1\\-1\\2\end{pmatrix}\) as the direction vector.
Common wrong choice: D uses a valid direction vector, but restricts \(\lambda\) to integers, so it does not describe the whole line.
2. A line passes through the fixed point \[ \begin{pmatrix}2\\-1\\3\end{pmatrix} \] and has direction vector \[ \begin{pmatrix}1\\0\\-2\end{pmatrix}. \] Which is the most complete vector equation of the line?
Tested fundamental: Recognising the complete form of a vector equation of a line.
Explanation: The fixed point comes first, followed by a scalar parameter multiplying the direction vector. The parameter should also be stated as \(\lambda\in\mathbb R\).
Common wrong choice: C looks almost identical, but it declares \(\Lambda\in\mathbb R\) while the equation uses \(\lambda\).
3. True or False?
If a line has vector equation
\[
\mathbf r=\mathbf a+\lambda\mathbf d,\quad \lambda\in\mathbb R,
\]
then replacing \(\mathbf d\) with \(3\mathbf d\) gives the same line.
Tested fundamental: Understanding that scalar multiples of a direction vector describe the same direction.
Explanation: Replacing \(\mathbf d\) with \(3\mathbf d\) only changes the parameter scale. The set of points on the line remains the same.
Common wrong choice: False usually comes from thinking the direction vector must be written in only one exact form.
4. The line \[ \mathbf r=\begin{pmatrix}1\\-2\\3\end{pmatrix} +\lambda\begin{pmatrix}2\\0\\-1\end{pmatrix},\quad \lambda\in\mathbb R \] is written in Cartesian form as:
Tested fundamental: Converting a vector equation to Cartesian form when one direction component is zero.
Explanation: Since the \(y\)-component of the direction vector is \(0\), \(y\) is fixed at \(-2\). We cannot divide by \(0\).
Common wrong choice: A incorrectly writes a denominator of \(0\).
5. Which vector equation represents the Cartesian equation \[ \frac{x-2}{3}=y+1=\frac{z-4}{-2}? \]
Tested fundamental: Converting Cartesian form to vector form.
Explanation: The fixed point is \((2,-1,4)\), and the direction vector is \(\begin{pmatrix}3\\1\\-2\end{pmatrix}\).
Common wrong choice: D treats \(y+1\) as though the \(y\)-direction component is \(-1\), but it is \(1\).
6. The point \(P(k,5,0)\) lies on the line \[ \mathbf r=\begin{pmatrix}2\\-1\\3\end{pmatrix} +\lambda\begin{pmatrix}1\\2\\-1\end{pmatrix},\quad \lambda\in\mathbb R. \] Find \(k\).
Answer format: Give your answer as an integer.
Answer: \(5\)
Tested fundamental: Checking whether a point lies on a line by using one common parameter value.
Explanation: From the \(y\)-coordinate, \(-1+2\lambda=5\), so \(\lambda=3\). Then \(z=3-3=0\), which is consistent, and \(x=2+3=5\). Hence \(k=5\).
Common error: Solving each coordinate with a different value of \(\lambda\).
7. Which vector equation represents the \(y\)-axis?
Tested fundamental: Recognising vector equations of coordinate axes.
Explanation: The \(y\)-axis passes through the origin and has direction vector \(\begin{pmatrix}0\\1\\0\end{pmatrix}\).
Common wrong choice: D passes through a point on the \(y\)-axis but moves in the \(x\)-direction.
Foot of Perpendicular, Distance and Reflection
These questions check the perpendicular condition, foot of perpendicular, shortest distance and reflection in a line.
8. A line \(l\) has direction vector \(\mathbf d\). \(F\) is the foot of the perpendicular from point \(P\) to \(l\). Which condition must be satisfied?
Tested fundamental: Setting up the foot of perpendicular from a point to a line.
Explanation: \(F\) must lie on the line, and the vector from \(F\) to \(P\) must be perpendicular to the line direction. Hence \(\overrightarrow{FP}\cdot\mathbf d=0\).
Common wrong choice: A and B use position vectors instead of the vector joining the foot to the point.
9. The line \(l\) is given by \[ \mathbf r=\begin{pmatrix}1\\0\\2\end{pmatrix} +\lambda\begin{pmatrix}1\\2\\2\end{pmatrix},\quad \lambda\in\mathbb R. \] Find the foot of the perpendicular from \[ P=\begin{pmatrix}5\\3\\6\end{pmatrix} \] to \(l\).
Answer format: Write your answer in the form (a,b,c).
Answer: \((3,4,6)\)
Tested fundamental: Finding the foot of perpendicular from a point to a line.
Explanation: Let \(F=\begin{pmatrix}1+\lambda\\2\lambda\\2+2\lambda\end{pmatrix}\). Since \(\overrightarrow{FP}\cdot\begin{pmatrix}1\\2\\2\end{pmatrix}=0\), we get \(18-9\lambda=0\), so \(\lambda=2\). Hence \(F=(3,4,6)\).
Common error: Using \(\overrightarrow{OP}\cdot\mathbf d=0\) instead of \(\overrightarrow{FP}\cdot\mathbf d=0\).
10. Using the same line and point in Question 9, find the shortest distance from \(P\) to \(l\).
Answer format: Give your answer exactly using sqrt().
Answer: \(\sqrt5\)
Tested fundamental: Finding the shortest distance from a point to a line after finding the foot of perpendicular.
Explanation: From Question 9, \(F=(3,4,6)\). Hence \(\overrightarrow{FP}=(2,-1,0)\), so the shortest distance is \(\sqrt{2^2+(-1)^2}=\sqrt5\).
Common error: Finding the distance from \(P\) to the fixed point on the line instead of to the foot of perpendicular.
11. Using the same line and point in Question 9, find the reflection of \(P\) in the line \(l\).
Answer format: Write your answer in the form (a,b,c).
Answer: \((1,5,6)\)
Tested fundamental: Finding the reflection of a point in a line using the foot of perpendicular.
Explanation: Since \(F\) is the midpoint of \(P\) and its reflection \(P’\), \(P’=2F-P=2(3,4,6)-(5,3,6)=(1,5,6)\).
Common error: Using \(P’=P-2F\) instead of \(P’=2F-P\).
Relationships Between Lines
These questions check coincident, intersecting, skew, parallel and perpendicular lines.
12. Consider the two lines \[ l_1:\mathbf r=\begin{pmatrix}1\\0\\2\end{pmatrix} +s\begin{pmatrix}1\\2\\-1\end{pmatrix},\quad s\in\mathbb R, \] \[ l_2:\mathbf r=\begin{pmatrix}3\\4\\0\end{pmatrix} +t\begin{pmatrix}2\\4\\-2\end{pmatrix},\quad t\in\mathbb R. \] Which describes the relationship between \(l_1\) and \(l_2\)?
Tested fundamental: Distinguishing coincident lines from parallel distinct lines.
Explanation: The direction vectors are parallel, and \(\begin{pmatrix}3\\4\\0\end{pmatrix}\) lies on \(l_1\) when \(s=2\). Hence the two lines are coincident.
Common wrong choice: A checks only that the direction vectors are parallel, but ignores the point test.
13. The lines \[ l_1:\mathbf r=\begin{pmatrix}1\\0\\0\end{pmatrix} +s\begin{pmatrix}1\\2\\1\end{pmatrix},\quad s\in\mathbb R \] and \[ l_2:\mathbf r=\begin{pmatrix}0\\1\\1\end{pmatrix} +t\begin{pmatrix}2\\1\\0\end{pmatrix},\quad t\in\mathbb R \] intersect. Find their point of intersection.
Answer format: Write your answer in the form (a,b,c).
Answer: \((2,2,1)\)
Tested fundamental: Finding the intersection point of two lines.
Explanation: Equating coordinates gives \(s=1\) and \(t=1\). Substituting into either line gives the point of intersection \((2,2,1)\).
Common error: Solving with only two coordinates and not checking the third coordinate.
14. Consider the two lines \[ l_1:\mathbf r=s\begin{pmatrix}1\\0\\1\end{pmatrix},\quad s\in\mathbb R, \] \[ l_2:\mathbf r=\begin{pmatrix}0\\1\\0\end{pmatrix} +t\begin{pmatrix}1\\0\\2\end{pmatrix},\quad t\in\mathbb R. \] Which describes the relationship between \(l_1\) and \(l_2\)?
Tested fundamental: Identifying skew lines.
Explanation: The direction vectors are not parallel. Also, \(l_1\) always has \(y=0\), while \(l_2\) always has \(y=1\), so they do not intersect. Hence the lines are skew.
Common wrong choice: D comes from noticing the directions are not parallel but failing to check for intersection.
15. True or False?
If the direction vectors of two lines are perpendicular, then the two lines must intersect.
Tested fundamental: Understanding perpendicular lines in 3D.
Explanation: In 3D, lines with perpendicular direction vectors do not necessarily intersect. They may be skew.
Common wrong choice: True comes from applying a 2D intuition directly to 3D lines.
16. The direction vectors of two lines are \[ \begin{pmatrix}1\\k\\2\end{pmatrix} \quad\text{and}\quad \begin{pmatrix}3\\-1\\1\end{pmatrix}. \] Find \(k\) if the two lines are perpendicular.
Answer format: Give your answer as an integer.
Answer: \(5\)
Tested fundamental: Using scalar product \(0\) for perpendicular direction vectors.
Explanation: \(\begin{pmatrix}1\\k\\2\end{pmatrix}\cdot\begin{pmatrix}3\\-1\\1\end{pmatrix}=0\), so \(3-k+2=0\), giving \(k=5\).
Common error: Making the direction vectors scalar multiples instead of using dot product \(0\).
17. The direction vectors of two lines are \[ \begin{pmatrix}2\\k\\-4\end{pmatrix} \quad\text{and}\quad \begin{pmatrix}1\\3\\-2\end{pmatrix}. \] Find \(k\) if the two lines are parallel.
Answer format: Give your answer as an integer.
Answer: \(6\)
Tested fundamental: Using scalar multiples to identify parallel direction vectors.
Explanation: Since \(\begin{pmatrix}2\\k\\-4\end{pmatrix}=2\begin{pmatrix}1\\3\\-2\end{pmatrix}\), we get \(k=6\).
Common error: Using dot product \(0\), which tests perpendicularity, not parallelism.
18. Two lines have parallel direction vectors. Which extra condition is needed to conclude that the two lines are coincident?
Tested fundamental: Distinguishing parallel distinct lines from coincident lines.
Explanation: Parallel direction vectors alone are not enough. To be coincident, a point on one line must also lie on the other line.
Common wrong choice: B is not sufficient because direction vectors may be different scalar multiples.
Angles Between Lines
These questions check acute angles between two lines and between a line and a coordinate axis.
19. The lines \(l_1\) and \(l_2\) are given by \[ l_1:\mathbf r=\begin{pmatrix}1\\0\\0\end{pmatrix} +s\begin{pmatrix}1\\2\\2\end{pmatrix},\quad s\in\mathbb R, \] \[ l_2:\mathbf r=\begin{pmatrix}0\\1\\0\end{pmatrix} +t\begin{pmatrix}2\\1\\2\end{pmatrix},\quad t\in\mathbb R. \] Find \(\cos\theta\), where \(\theta\) is the acute angle between \(l_1\) and \(l_2\).
Answer format: Give your answer as a fraction.
Answer: \(\frac89\)
Tested fundamental: Finding the acute angle between two lines using their direction vectors.
Explanation: The direction vectors are \(\begin{pmatrix}1\\2\\2\end{pmatrix}\) and \(\begin{pmatrix}2\\1\\2\end{pmatrix}\). Their dot product is \(8\), and each has magnitude \(3\). Hence \(\cos\theta=\frac{8}{9}\).
Common error: Using the fixed points instead of the direction vectors.
20. A line has direction vector \[ \begin{pmatrix}2\\1\\2\end{pmatrix}. \] Find \(\cos\theta\), where \(\theta\) is the acute angle between the line and the \(z\)-axis.
Answer format: Give your answer as a fraction.
Answer: \(\frac23\)
Tested fundamental: Finding the angle between a line and a coordinate axis.
Explanation: The \(z\)-axis has direction vector \(\begin{pmatrix}0\\0\\1\end{pmatrix}\). Thus \(\cos\theta=\frac{|2|}{\sqrt{2^2+1^2+2^2}}=\frac23\).
Common error: Using the \(x\)-component or \(y\)-component instead of the \(z\)-component.
Projections Onto and Perpendicular to a Line
These questions check projection of a vector onto a line and onto the perpendicular to a line.
21. A line passes through \(A\) and has direction vector \(\mathbf d\). Which expression gives the length of projection of \(\overrightarrow{AP}\) onto the line?
Tested fundamental: Selecting the formula for length of projection onto a line.
Explanation: Since the line passes through \(A\), the vector being projected is \(\overrightarrow{AP}\). The length of projection onto the line is \(\frac{|\overrightarrow{AP}\cdot\mathbf d|}{|\mathbf d|}\).
Common wrong choice: D uses the correct structure but projects the wrong vector, \(\overrightarrow{OP}\), instead of \(\overrightarrow{AP}\).
22. A line passes through \[ A=\begin{pmatrix}1\\1\\0\end{pmatrix} \] and has direction vector \[ \mathbf d=\begin{pmatrix}3\\4\\0\end{pmatrix}. \] Find the length of projection of \(\overrightarrow{AP}\) onto the line, where \[ P=\begin{pmatrix}5\\4\\0\end{pmatrix}. \]
Answer format: Give your answer as a fraction.
Answer: \(\frac{24}{5}\)
Tested fundamental: Computing the length of projection onto a line.
Explanation: \(\overrightarrow{AP}=\begin{pmatrix}4\\3\\0\end{pmatrix}\). Then \(\overrightarrow{AP}\cdot\mathbf d=4(3)+3(4)=24\), and \(|\mathbf d|=5\). Hence the length of projection is \(\frac{24}{5}\).
Common error: Dividing by \(|\overrightarrow{AP}|\) instead of \(|\mathbf d|\).
23. Using the same line and point in Question 22, find the length of projection of \(\overrightarrow{AP}\) onto the perpendicular to the line.
Answer format: Give your answer as a fraction.
Answer: \(\frac75\)
Tested fundamental: Computing the length of projection onto the perpendicular to a line.
Explanation: \(\overrightarrow{AP}=\begin{pmatrix}4\\3\\0\end{pmatrix}\) and \(\mathbf d=\begin{pmatrix}3\\4\\0\end{pmatrix}\). Since \(|\overrightarrow{AP}\times\mathbf d|=7\) and \(|\mathbf d|=5\), the required length is \(\frac75\).
Common error: Reusing the answer from projection onto the line instead of taking the perpendicular component.
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