This TSA checks 3D vector notation, position vectors, magnitude, unit vectors, Cartesian component form, ratio theorem, midpoint theorem, collinearity, scalar product, projections, vector product, perpendicular distance from a point to a vector, and area using vector product.
Choose an option for MCQ or True/False questions. Enter a short answer for SAQ questions. Once you submit an answer, it will be locked. The timer starts only when you click Begin. You can still review the questions after time is up.
Three-Dimensional Vectors and Vector Geometry
These questions check 3D vector notation, position vectors, ratio theorem, midpoint theorem and collinearity.
1. Points \(A(1,-2,3)\) and \(B(4,0,-1)\) are given. Find the length of \(AB\).
Answer format: Give your answer exactly using sqrt(), for example sqrt(5).
Answer: \(\sqrt{29}\)
Tested fundamental: Finding the magnitude of a vector between two points.
Explanation: \(\overrightarrow{AB}=\begin{pmatrix}3\\2\\-4\end{pmatrix}\), so \(AB=\sqrt{3^2+2^2+(-4)^2}=\sqrt{29}\).
Common error: Subtracting the coordinates in the wrong order does not affect the length here, but forgetting to square all three components does.
2. Find the unit vector in the direction of \[ \begin{pmatrix}0\\3\\4\end{pmatrix}. \]
Answer format: Write your answer in the form (a,b,c).
Answer: \(\begin{pmatrix}0\\ \frac35\\ \frac45\end{pmatrix}\)
Tested fundamental: Finding a unit vector in a given direction.
Explanation: The magnitude is \(\sqrt{0^2+3^2+4^2}=5\). Hence the unit vector is \(\frac15\begin{pmatrix}0\\3\\4\end{pmatrix}=\begin{pmatrix}0\\ \frac35\\ \frac45\end{pmatrix}\).
Common error: Giving the original vector instead of dividing by its magnitude.
3. Which column vector represents \[ 2\mathbf{i}-\mathbf{j}+3\mathbf{k}? \]
Tested fundamental: Converting Cartesian component form to column vector form.
Explanation: The coefficients of \(\mathbf{i},\mathbf{j},\mathbf{k}\) are \(2,-1,3\), so the column vector is \(\begin{pmatrix}2\\-1\\3\end{pmatrix}\).
Common wrong choice: A ignores the negative sign in \(-\mathbf{j}\).
4. Point \(P\) divides \(AB\) internally in the ratio \(AP:PB=2:3\), where \[ A=\begin{pmatrix}1\\0\\2\end{pmatrix},\qquad B=\begin{pmatrix}6\\10\\-3\end{pmatrix}. \] Find the position vector of \(P\).
Answer format: Write your answer in the form (a,b,c).
Answer: \(\begin{pmatrix}3\\4\\0\end{pmatrix}\)
Tested fundamental: Applying the ratio theorem for internal division.
Explanation: Since \(AP:PB=2:3\), \(P=\frac{3A+2B}{5}\). Thus \(P=\frac{3\begin{pmatrix}1\\0\\2\end{pmatrix}+2\begin{pmatrix}6\\10\\-3\end{pmatrix}}5=\begin{pmatrix}3\\4\\0\end{pmatrix}\).
Common error: Reversing the weights and using \(\frac{2A+3B}{5}\).
5. Point \(P\) divides \(AB\) externally in the ratio \(AP:PB=2:1\), where \[ A=\begin{pmatrix}1\\2\\0\end{pmatrix},\qquad B=\begin{pmatrix}4\\-1\\6\end{pmatrix}. \] Find the position vector of \(P\).
Answer format: Write your answer in the form (a,b,c).
Answer: \(\begin{pmatrix}7\\-4\\12\end{pmatrix}\)
Tested fundamental: Applying the ratio theorem for external division.
Explanation: Since \(AP:PB=2:1\), \(P=\frac{2B-1A}{2-1}=2B-A\). Hence \(P=2\begin{pmatrix}4\\-1\\6\end{pmatrix}-\begin{pmatrix}1\\2\\0\end{pmatrix}=\begin{pmatrix}7\\-4\\12\end{pmatrix}\).
Common error: Treating external division as internal division.
6. Find the midpoint of \(AB\), where \[ A=\begin{pmatrix}2\\-1\\3\end{pmatrix},\qquad B=\begin{pmatrix}8\\5\\-1\end{pmatrix}. \]
Answer format: Write your answer in the form (a,b,c).
Answer: \(\begin{pmatrix}5\\2\\1\end{pmatrix}\)
Tested fundamental: Using midpoint theorem in 3D.
Explanation: The midpoint is \(\frac12(A+B)=\frac12\begin{pmatrix}10\\4\\2\end{pmatrix}=\begin{pmatrix}5\\2\\1\end{pmatrix}\).
Common error: Averaging only two coordinates and forgetting the third coordinate.
7. Points \(A(1,0,2)\), \(B(3,4,6)\) and \(C(k,10,12)\) are collinear. Find \(k\).
Answer format: Give your answer as an integer.
Answer: \(6\)
Tested fundamental: Using scalar multiples to test collinearity.
Explanation: \(\overrightarrow{AB}=\begin{pmatrix}2\\4\\4\end{pmatrix}\) and \(\overrightarrow{AC}=\begin{pmatrix}k-1\\10\\10\end{pmatrix}\). Since \(10=2.5(4)\), we need \(k-1=2.5(2)=5\), so \(k=6\).
Common error: Comparing only one coordinate and not checking the same multiplier across the vector.
8. Given the position vectors \[ \mathbf{a}=\begin{pmatrix}1\\-2\\3\end{pmatrix},\qquad \mathbf{b}=\begin{pmatrix}4\\0\\-1\end{pmatrix}, \] find \(\overrightarrow{AB}\).
Answer format: Write your answer in the form (a,b,c).
Answer: \(\begin{pmatrix}3\\2\\-4\end{pmatrix}\)
Tested fundamental: Finding a vector between two position vectors.
Explanation: \(\overrightarrow{AB}=\mathbf{b}-\mathbf{a}=\begin{pmatrix}4\\0\\-1\end{pmatrix}-\begin{pmatrix}1\\-2\\3\end{pmatrix}=\begin{pmatrix}3\\2\\-4\end{pmatrix}\).
Common error: Using \(\mathbf{a}-\mathbf{b}\), which gives \(\overrightarrow{BA}\) instead.
Scalar Product
These questions check angle, perpendicularity and sign interpretations using the scalar product.
9. Given \[ \mathbf{a}=\begin{pmatrix}1\\2\\2\end{pmatrix},\qquad \mathbf{b}=\begin{pmatrix}2\\1\\2\end{pmatrix}, \] find \(\cos\theta\), where \(\theta\) is the angle between \(\mathbf{a}\) and \(\mathbf{b}\).
Tested fundamental: Using the scalar product formula for angle between two vectors.
Explanation: \(\mathbf{a}\cdot\mathbf{b}=8\), while \(|\mathbf{a}|=3\) and \(|\mathbf{b}|=3\). Hence \(\cos\theta=\frac{8}{3\cdot3}=\frac89\).
Common wrong choice: B comes from forgetting to divide by both magnitudes.
10. True or False?
For two non-zero vectors \(\mathbf{a}\) and \(\mathbf{b}\), if
\[
\mathbf{a}\cdot\mathbf{b}<0,
\]
then the angle between them is obtuse.
Tested fundamental: Interpreting the sign of the scalar product.
Explanation: Since \(\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta\), a negative scalar product means \(\cos\theta<0\). Hence the angle is obtuse.
Common wrong choice: False usually comes from thinking a negative scalar product only happens when vectors point in exactly opposite directions.
11. Find \(k\) if \[ \begin{pmatrix}k\\4\\-2\end{pmatrix} \] is perpendicular to \[ \begin{pmatrix}3\\-1\\4\end{pmatrix}. \]
Answer format: Give your answer as an integer.
Answer: \(4\)
Tested fundamental: Using scalar product \(0\) for perpendicular vectors.
Explanation: Perpendicular vectors have scalar product \(0\). Hence \(3k+4(-1)+(-2)(4)=0\), so \(3k-12=0\), giving \(k=4\).
Common error: Making the vectors scalar multiples instead of using scalar product \(0\).
12. For two non-zero parallel vectors \(\mathbf{a}\) and \(\mathbf{b}\) pointing in opposite directions, which statement is true?
Tested fundamental: Scalar product of parallel vectors in opposite directions.
Explanation: Opposite parallel vectors have angle \(180^\circ\), so \(\cos180^\circ=-1\). Hence \(\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|(-1)=-|\mathbf{a}||\mathbf{b}|\).
Common wrong choice: B is for vectors pointing in the same direction.
Projections
These questions check length of projection, vector projection and zero projection.
13. Which expression gives the length of projection of \(\mathbf{a}\) on \(\mathbf{b}\)?
Tested fundamental: Identifying the formula for length of projection.
Explanation: The length of projection of \(\mathbf{a}\) on \(\mathbf{b}\) is \(\frac{|\mathbf{a}\cdot\mathbf{b}|}{|\mathbf{b}|}\). The absolute value is needed because length cannot be negative.
Common wrong choice: C gives the vector projection, not the length of projection.
14. Find the length of projection of \[ \mathbf{a}=\begin{pmatrix}10\\-6\\-8\end{pmatrix} \] on \[ \mathbf{b}=\begin{pmatrix}0\\3\\4\end{pmatrix}. \]
Answer format: Give your answer as an integer.
Answer: \(10\)
Tested fundamental: Computing length of projection.
Explanation: \(\mathbf{a}\cdot\mathbf{b}=10(0)+(-6)(3)+(-8)(4)=-50\), and \(|\mathbf{b}|=5\). The length of projection is \(\frac{|-50|}{5}=10\).
Common error: Giving \(-10\), but length of projection must be non-negative.
15. Find the vector projection of \[ \mathbf{a}=\begin{pmatrix}4\\2\\0\end{pmatrix} \] on \[ \mathbf{b}=\begin{pmatrix}1\\1\\0\end{pmatrix}. \]
Answer format: Write your answer in the form (a,b,c).
Answer: \(\begin{pmatrix}3\\3\\0\end{pmatrix}\)
Tested fundamental: Computing vector projection.
Explanation: \(\operatorname{proj}_{\mathbf{b}}\mathbf{a}=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\mathbf{b}\). Here \(\mathbf{a}\cdot\mathbf{b}=6\) and \(|\mathbf{b}|^2=2\), so the projection is \(3\begin{pmatrix}1\\1\\0\end{pmatrix}=\begin{pmatrix}3\\3\\0\end{pmatrix}\).
Common error: Giving the length of projection instead of a vector.
16. For non-zero vectors \(\mathbf{a}\) and \(\mathbf{b}\), the vector projection of \(\mathbf{a}\) on \(\mathbf{b}\) is the zero vector. What must be true?
Tested fundamental: Connecting zero vector projection to perpendicularity.
Explanation: If the vector projection of \(\mathbf{a}\) on \(\mathbf{b}\) is zero, then \(\mathbf{a}\cdot\mathbf{b}=0\). For non-zero vectors, this means the vectors are perpendicular.
Common wrong choice: A and D describe parallel vectors, not perpendicular vectors.
Vector Product
These questions check cross product calculation, direction, magnitude and interpretation.
17. Find a vector perpendicular to both \[ \begin{pmatrix}2\\-1\\3\end{pmatrix} \quad\text{and}\quad \begin{pmatrix}1\\4\\-2\end{pmatrix} \] using the vector product in the given order.
Answer format: Write your answer in the form (a,b,c).
Answer: \(\begin{pmatrix}-10\\7\\9\end{pmatrix}\)
Tested fundamental: Using the vector product to find a vector perpendicular to two vectors.
Explanation: \(\begin{pmatrix}2\\-1\\3\end{pmatrix}\times\begin{pmatrix}1\\4\\-2\end{pmatrix}=\begin{pmatrix}-10\\7\\9\end{pmatrix}\). This vector is perpendicular to both given vectors.
Common error: Reversing the order gives the opposite vector.
18. True or False?
For any vectors \(\mathbf{a}\) and \(\mathbf{b}\), \(\mathbf{a}\times\mathbf{b}\) and \(\mathbf{b}\times\mathbf{a}\) have the same magnitude but opposite directions.
Tested fundamental: Understanding the non-commutative property of vector product.
Explanation: \(\mathbf{b}\times\mathbf{a}=-(\mathbf{a}\times\mathbf{b})\). Hence the two vectors have the same magnitude but opposite directions.
Common wrong choice: False usually comes from thinking vector product behaves like ordinary multiplication.
19. For non-zero vectors \(\mathbf{a}\) and \(\mathbf{b}\), if \[ \mathbf{a}\times\mathbf{b}=\mathbf{0}, \] what must be true?
Tested fundamental: Interpreting zero vector product.
Explanation: For non-zero vectors, \(\mathbf{a}\times\mathbf{b}=\mathbf{0}\) means the vectors are parallel.
Common wrong choice: A or C confuses the cross product condition with the scalar product condition for perpendicular vectors.
20. Given \[ |\mathbf{a}|=6,\qquad |\mathbf{b}|=5, \] and the angle between them is \(30^\circ\), find \[ |\mathbf{a}\times\mathbf{b}|. \]
Answer format: Give your answer as an integer.
Answer: \(15\)
Tested fundamental: Using \(|\mathbf{a}\times\mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin\theta\).
Explanation: \(|\mathbf{a}\times\mathbf{b}|=6(5)\sin30^\circ=30\left(\frac12\right)=15\).
Common error: Using \(\cos30^\circ\), which belongs to scalar product.
21. True or False?
For two non-zero vectors \(\mathbf{a}\) and \(\mathbf{b}\), if
\[
|\mathbf{a}\times\mathbf{b}|=|\mathbf{a}||\mathbf{b}|,
\]
then \(\mathbf{a}\) and \(\mathbf{b}\) are perpendicular.
Tested fundamental: Interpreting the magnitude of a vector product.
Explanation: Since \(|\mathbf{a}\times\mathbf{b}|=|\mathbf{a}||\mathbf{b}|\sin\theta\), equality with \(|\mathbf{a}||\mathbf{b}|\) means \(\sin\theta=1\), so \(\theta=90^\circ\).
Common wrong choice: False usually comes from confusing this with the parallel condition, where \(|\mathbf{a}\times\mathbf{b}|=0\).
Distance and Area Using Vector Product
These questions check perpendicular distance from a point to a vector, and area using vector product.
22. Point \(P\) has position vector \[ \mathbf{p}=\begin{pmatrix}1\\2\\2\end{pmatrix}. \] Find the perpendicular distance from \(P\) to the vector \[ \mathbf{a}=\begin{pmatrix}1\\0\\0\end{pmatrix}. \]
Answer format: Give your answer exactly using sqrt(), for example 3sqrt(2).
Answer: \(2\sqrt2\)
Tested fundamental: Finding perpendicular distance from a point to a vector using vector product.
Explanation: The perpendicular distance is \(\frac{|\mathbf{p}\times\mathbf{a}|}{|\mathbf{a}|}\). Now \(\begin{pmatrix}1\\2\\2\end{pmatrix}\times\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}0\\2\\-2\end{pmatrix}\), so the distance is \(\frac{2\sqrt2}{1}=2\sqrt2\).
Common error: Using \(|\mathbf{p}|\) directly instead of the perpendicular component.
23. Which expression gives the area of the parallelogram formed by two non-parallel vectors \(\mathbf{a}\) and \(\mathbf{b}\)?
Tested fundamental: Finding the area of a parallelogram using vector product.
Explanation: \(|\mathbf{a}\times\mathbf{b}|\) gives the area of the parallelogram formed by the two vectors.
Common wrong choice: B gives the area of the triangle, not the parallelogram.
24. Given that \[ \overrightarrow{AB}\times\overrightarrow{AC} = \begin{pmatrix}2\\-3\\6\end{pmatrix}, \] find the area of triangle \(ABC\).
Answer format: Give your answer as a fraction in the form p/q.
Answer: \(\frac72\)
Tested fundamental: Finding triangle area from a vector product.
Explanation: \(\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|=\sqrt{2^2+(-3)^2+6^2}=7\). Hence the area of triangle \(ABC\) is \(\frac12(7)=\frac72\).
Common error: Giving \(7\), which is the parallelogram area.
Your result is ready.