This TSA checks parametric, scalar product and Cartesian equations of planes, special coordinate planes, point-plane checks, line-plane intersections, foot of perpendicular, perpendicular distance, projection onto a plane, line-plane relationships, angles between lines and planes, relationships between planes, line of intersection of planes, reflections, and three-plane intersections.
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Equations of Planes
1. Which is a valid parametric equation of the plane passing through \[ A=\begin{pmatrix}1\\2\\-1\end{pmatrix} \] with two non-parallel direction vectors \[ \begin{pmatrix}1\\0\\2\end{pmatrix} \quad\text{and}\quad \begin{pmatrix}0\\1\\-1\end{pmatrix}? \]
Tested fundamental: Recognising the parametric vector equation of a plane.
Explanation: A plane needs one fixed point and two non-parallel direction vectors lying in the plane. Hence the fixed point is \(\begin{pmatrix}1\\2\\-1\end{pmatrix}\), with the two given direction vectors.
Common wrong choice: C gives only one direction vector, so it describes a line, not a plane.
2. Find the Cartesian equation of the plane with normal vector \[ \begin{pmatrix}2\\-1\\3\end{pmatrix} \] passing through \[ A=\begin{pmatrix}1\\2\\-1\end{pmatrix}. \]
Answer format: Write your answer in the form ax+by+cz=d.
Answer: \(2x-y+3z=-3\)
Tested fundamental: Forming the Cartesian equation of a plane from a normal vector and a point.
Explanation: Since the normal vector is \(\begin{pmatrix}2\\-1\\3\end{pmatrix}\), the plane is \(2x-y+3z=d\). Substituting \(A(1,2,-1)\) gives \(d=2(1)-2+3(-1)=-3\).
Common error: Using the point as the normal vector.
3. A plane has normal vector \[ \begin{pmatrix}2\\-1\\3\end{pmatrix} \] and contains the point \[ A=\begin{pmatrix}1\\0\\-2\end{pmatrix}. \] Which is the correct scalar product form of the plane?
Tested fundamental: Writing the scalar product form of a plane.
Explanation: The plane has form \(\mathbf r\cdot\mathbf n=D\). Here \(D=\begin{pmatrix}1\\0\\-2\end{pmatrix}\cdot\begin{pmatrix}2\\-1\\3\end{pmatrix}=2-6=-4\).
Common wrong choice: A uses the point as the normal vector.
4. A plane contains the points \[ A=\begin{pmatrix}1\\0\\2\end{pmatrix}, \qquad B=\begin{pmatrix}3\\1\\0\end{pmatrix}, \] and is parallel to the vector \[ \begin{pmatrix}0\\1\\1\end{pmatrix}. \] Which is a valid scalar product equation of the plane?
Tested fundamental: Forming a plane equation from two points and a vector parallel to the plane.
Explanation: \(\overrightarrow{AB}=\begin{pmatrix}2\\1\\-2\end{pmatrix}\). A normal vector is \(\overrightarrow{AB}\times\begin{pmatrix}0\\1\\1\end{pmatrix}=\begin{pmatrix}3\\-2\\2\end{pmatrix}\). Using \(A\), \(D=3(1)-2(0)+2(2)=7\).
Common wrong choice: B uses a vector parallel to the plane as though it were the normal vector.
5. The plane \[ 2x-y+3z=6 \] has normal vector \[ \begin{pmatrix}a\\b\\c\end{pmatrix}. \] Find \((a,b,c)\).
Answer format: Write your answer in the form (a,b,c).
Answer: \(\begin{pmatrix}2\\-1\\3\end{pmatrix}\)
Tested fundamental: Reading the normal vector from a Cartesian plane equation.
Explanation: For \(ax+by+cz=d\), a normal vector is \(\begin{pmatrix}a\\b\\c\end{pmatrix}\). Hence the normal vector is \(\begin{pmatrix}2\\-1\\3\end{pmatrix}\).
Common error: Ignoring the negative sign in the \(y\)-coefficient.
6. Which equation represents the \(xz\)-plane?
Tested fundamental: Recognising special coordinate planes.
Explanation: The \(xz\)-plane contains all points with \(y=0\).
Common wrong choice: B represents the \(yz\)-plane, not the \(xz\)-plane.
7. The point \(P(k,1,2)\) lies on the plane \[ 2x-y+3z=5. \] Find \(k\).
Answer format: Give your answer as an integer.
Answer: \(0\)
Tested fundamental: Checking whether a point lies on a plane.
Explanation: Substituting \(P(k,1,2)\) gives \(2k-1+6=5\), so \(2k=0\), hence \(k=0\).
Common error: Substituting into the wrong coordinate position.
Line and Plane
8. A line has equation \[ \mathbf r=\mathbf a+\lambda\mathbf d,\quad \lambda\in\mathbb R, \] and a plane has equation \[ \mathbf n\cdot\mathbf r=p. \] Which condition means the line lies in the plane?
Tested fundamental: Identifying when a line lies in a plane.
Explanation: The line direction must be parallel to the plane, so \(\mathbf n\cdot\mathbf d=0\). Also, a point on the line must lie in the plane, so \(\mathbf n\cdot\mathbf a=p\).
Common wrong choice: B describes a line parallel to the plane but not lying in it.
9. The line \[ \mathbf r=\begin{pmatrix}1\\0\\2\end{pmatrix} +\lambda\begin{pmatrix}1\\2\\-1\end{pmatrix},\quad \lambda\in\mathbb R \] intersects the plane \[ x+y+z=7. \] Find the point of intersection.
Answer format: Write your answer in the form (a,b,c).
Answer: \((3,4,0)\)
Tested fundamental: Finding the intersection between a line and a plane.
Explanation: A point on the line is \((1+\lambda,2\lambda,2-\lambda)\). Substituting into \(x+y+z=7\) gives \(3+2\lambda=7\), so \(\lambda=2\). Hence the point is \((3,4,0)\).
Common error: Substituting the direction vector into the plane instead of the full line expression.
10. The line \[ \mathbf r=\begin{pmatrix}1\\0\\0\end{pmatrix} +\lambda\begin{pmatrix}1\\2\\-1\end{pmatrix},\quad \lambda\in\mathbb R \] and the plane \[ x+2y+5z=3 \] are related as follows:
Tested fundamental: Classifying the relationship between a line and a plane.
Explanation: The normal vector is \(\begin{pmatrix}1\\2\\5\end{pmatrix}\), and \(\begin{pmatrix}1\\2\\5\end{pmatrix}\cdot\begin{pmatrix}1\\2\\-1\end{pmatrix}=0\). So the line is parallel to the plane. The fixed point \((1,0,0)\) does not satisfy the plane equation, so the line does not lie in the plane.
Common wrong choice: B checks only the parallel condition but forgets to test whether the fixed point lies in the plane.
11. The plane is \[ x+y+z=3. \] Find the foot of the perpendicular from \[ P=\begin{pmatrix}2\\3\\4\end{pmatrix} \] to the plane.
Answer format: Write your answer in the form (a,b,c).
Answer: \((0,1,2)\)
Tested fundamental: Finding the foot of perpendicular from a point to a plane.
Explanation: The normal vector is \(\begin{pmatrix}1\\1\\1\end{pmatrix}\). The perpendicular line through \(P\) is \(\mathbf r=\begin{pmatrix}2\\3\\4\end{pmatrix}+t\begin{pmatrix}1\\1\\1\end{pmatrix}\). Substituting into \(x+y+z=3\) gives \(9+3t=3\), so \(t=-2\). Hence the foot is \((0,1,2)\).
Common error: Using a direction vector lying in the plane instead of the normal vector.
12. Using the same point and plane in Question 11, find the perpendicular distance from \(P\) to the plane.
Answer format: Give your answer exactly using p sqrt(q), for example 5sqrt(2).
Answer: \(2\sqrt3\)
Tested fundamental: Finding the perpendicular distance from a point to a plane.
Explanation: From Question 11, the foot is \(F=(0,1,2)\). Thus \(\overrightarrow{FP}=\begin{pmatrix}2\\2\\2\end{pmatrix}\), so the distance is \(\sqrt{2^2+2^2+2^2}=2\sqrt3\).
Common error: Finding the distance to an arbitrary point on the plane instead of the foot of perpendicular.
13. Find the perpendicular distance from the origin to the plane \[ 2x-y+2z=6. \]
Answer format: Give your answer as an integer.
Answer: \(2\)
Tested fundamental: Finding the distance from the origin to a plane.
Explanation: The distance is \(\frac{|6|}{\sqrt{2^2+(-1)^2+2^2}}=\frac6{3}=2\).
Common error: Forgetting to divide by the magnitude of the normal vector.
14. Find the length of projection of \[ \mathbf v=\begin{pmatrix}3\\4\\12\end{pmatrix} \] onto the plane \[ x=0. \]
Answer format: Give your answer exactly using p sqrt(q), for example 5sqrt(2).
Answer: \(4\sqrt{10}\)
Tested fundamental: Finding the length of projection of a vector onto a plane.
Explanation: The plane \(x=0\) has normal vector \(\begin{pmatrix}1\\0\\0\end{pmatrix}\). The component of \(\mathbf v\) perpendicular to the plane is \(3\), so the projected length onto the plane is \(\sqrt{|\mathbf v|^2-3^2}=\sqrt{169-9}=\sqrt{160}=4\sqrt{10}\).
Common error: Giving the component perpendicular to the plane instead of the component on the plane.
15. For a plane \[ \mathbf r\cdot\mathbf n=D \] and a point \(P\) with position vector \(\mathbf p\), which expression gives the perpendicular distance from \(P\) to the plane?
Tested fundamental: Selecting the point-to-plane distance formula.
Explanation: The perpendicular distance from \(\mathbf p\) to \(\mathbf r\cdot\mathbf n=D\) is \(\frac{|\mathbf p\cdot\mathbf n-D|}{|\mathbf n|}\).
Common wrong choice: C forgets to divide by the magnitude of the normal vector.
16. A line has direction vector \(\mathbf d\), and a plane has normal vector \(\mathbf n\). If \(\theta\) is the acute angle between the line and the plane, which formula is correct?
Tested fundamental: Choosing the correct angle formula between a line and a plane.
Explanation: The angle between the line and the plane is complementary to the angle between the line direction and the plane normal. Hence \(\sin\theta=\frac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|}\).
Common wrong choice: A gives the cosine of the angle between the line direction and the normal vector, not the line-plane angle.
17. The line has direction vector \[ \begin{pmatrix}1\\2\\2\end{pmatrix} \] and the plane has normal vector \[ \begin{pmatrix}2\\1\\2\end{pmatrix}. \] Find the acute angle between the line and the plane, in degrees.
Answer format: Give your answer in degrees to 1 decimal place.
Answer: \(62.7^\circ\)
Tested fundamental: Finding the acute angle between a line and a plane.
Explanation: \(\sin\theta=\frac{|(1,2,2)\cdot(2,1,2)|}{3\cdot3}=\frac89\). Hence \(\theta=\sin^{-1}\left(\frac89\right)=62.7^\circ\) to 1 decimal place.
Common error: Using \(\cos^{-1}\left(\frac89\right)\), which gives the angle between the line direction and the plane normal.
Two Planes
18. The planes \[ \Pi_1:2x-y+2z=6 \] and \[ \Pi_2:4x-2y+4z=9 \] are:
Tested fundamental: Classifying two planes using their normal vectors and constants.
Explanation: The normal vectors are parallel, since \((4,-2,4)=2(2,-1,2)\). However, multiplying the first equation by \(2\) gives \(4x-2y+4z=12\), not \(9\). Hence the planes are parallel and distinct.
Common wrong choice: A checks only the normal vectors but ignores the constants.
19. Which is a valid vector equation of the line of intersection of the planes \[ x+y+z=3 \] and \[ x-y+z=1? \]
Tested fundamental: Finding the line of intersection of two planes.
Explanation: The point \((2,1,0)\) satisfies both planes. The direction vector \(\begin{pmatrix}1\\0\\-1\end{pmatrix}\) is perpendicular to both normal vectors \(\begin{pmatrix}1\\1\\1\end{pmatrix}\) and \(\begin{pmatrix}1\\-1\\1\end{pmatrix}\), so it is parallel to the line of intersection.
Common wrong choice: B uses a point on the line but a direction vector that does not lie in both planes.
20. Find the acute angle between the planes \[ x+2y+2z=3 \] and \[ 2x+y+2z=5. \]
Answer format: Give your answer in degrees to 1 decimal place.
Answer: \(27.3^\circ\)
Tested fundamental: Finding the acute angle between two planes.
Explanation: The angle between two planes is the angle between their normal vectors. The normal vectors are \(\begin{pmatrix}1\\2\\2\end{pmatrix}\) and \(\begin{pmatrix}2\\1\\2\end{pmatrix}\). Thus \(\cos\theta=\frac8{9}\), so \(\theta=27.3^\circ\) to 1 decimal place.
Common error: Using the line-plane angle formula with sine instead of the plane-plane angle formula with cosine.
21. Find the perpendicular distance between the parallel planes \[ 2x-y+2z=6 \] and \[ 2x-y+2z=12. \]
Answer format: Give your answer as an integer.
Answer: \(2\)
Tested fundamental: Finding the perpendicular distance between two parallel planes.
Explanation: The distance is \(\frac{|12-6|}{\sqrt{2^2+(-1)^2+2^2}}=\frac6{3}=2\).
Common error: Taking the difference \(12-6=6\) without dividing by the magnitude of the normal vector.
22. Two planes \[ \mathbf r\cdot\mathbf n_1=D_1 \quad\text{and}\quad \mathbf r\cdot\mathbf n_2=D_2 \] are coincident if:
Tested fundamental: Identifying coincident planes.
Explanation: Coincident planes have parallel normal vectors, and the whole plane equation must be in the same ratio, including the constant.
Common wrong choice: D is not sufficient because equal constants alone do not guarantee the same plane.
Parallel Line and Plane
23. The line \[ \mathbf r=\begin{pmatrix}1\\0\\0\end{pmatrix} +\lambda\begin{pmatrix}1\\1\\0\end{pmatrix},\quad \lambda\in\mathbb R \] is parallel to the plane \[ x-y+z=4. \] Find the perpendicular distance between the line and the plane.
Answer format: Give your answer exactly using sqrt().
Answer: \(\sqrt3\)
Tested fundamental: Finding the distance between a parallel line and a plane.
Explanation: Since the line is parallel to the plane, take any point on the line, such as \((1,0,0)\). Its distance to \(x-y+z=4\) is \(\frac{|1-4|}{\sqrt{1^2+(-1)^2+1^2}}=\sqrt3\).
Common error: Using the direction vector of the line as though it were a point.
Reflection
24. Using the same point and plane in Question 11, find the reflection of \(P\) in the plane.
Answer format: Write your answer in the form (a,b,c).
Answer: \((-2,-1,0)\)
Tested fundamental: Finding the reflection of a point in a plane.
Explanation: From Question 11, the foot of perpendicular is \(F=(0,1,2)\). Since \(F\) is the midpoint of \(P\) and its reflection \(P’\), \(P’=2F-P=2(0,1,2)-(2,3,4)=(-2,-1,0)\).
Common error: Using \(P’=P-2F\) instead of \(P’=2F-P\).
25. The line \[ l:\mathbf r=\begin{pmatrix}1\\0\\2\end{pmatrix} +t\begin{pmatrix}1\\1\\-1\end{pmatrix},\quad t\in\mathbb R \] is reflected in the plane \[ z=0. \] Which is a valid vector equation of the reflected line?
Tested fundamental: Reflecting a line in a plane.
Explanation: Reflection in \(z=0\) changes the sign of the \(z\)-coordinate. The point \((1,0,2)\) reflects to \((1,0,-2)\), and the direction vector \((1,1,-1)\) reflects to \((1,1,1)\).
Common wrong choice: B reflects the fixed point but not the direction vector.
Three Planes
26. The three planes \[ \Pi_1:x+y+z=3, \] \[ \Pi_2:x-y+z=1, \] \[ \Pi_3:y=1 \] intersect:
Tested fundamental: Classifying the relationship between three planes.
Explanation: From \(\Pi_3\), \(y=1\). Substituting into \(\Pi_1\) gives \(x+z=2\), and substituting into \(\Pi_2\) also gives \(x+z=2\). Hence there are infinitely many common points along a line.
Common wrong choice: A comes from expecting three planes to usually meet at one point without solving the system.
27. The three planes \[ x+y+z=6, \] \[ x-y=0, \] \[ z=2 \] intersect at a single point. Find the point.
Answer format: Write your answer in the form (a,b,c).
Answer: \((2,2,2)\)
Tested fundamental: Finding the single point of intersection of three planes.
Explanation: From \(z=2\) and \(x-y=0\), we have \(x=y\). Substituting into \(x+y+z=6\) gives \(2x+2=6\), so \(x=2\) and \(y=2\). Hence the point is \((2,2,2)\).
Common error: Stopping after finding \(x=y\) without using the third plane.
28. The three planes \[ x=0,\qquad y=0,\qquad x+y=1 \] have:
Tested fundamental: Identifying when three planes have no common intersection.
Explanation: The first two planes require \(x=0\) and \(y=0\). But then \(x+y=0\), which contradicts the third plane \(x+y=1\). Hence there is no common intersection.
Common wrong choice: A comes from solving only the first two equations and ignoring the contradiction.
Your result is ready.