Multivariable Calculus Quiz: Plane Equations And Normals
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Plane Equations And NormalsQuestion 1 of 12

The line LL is parameterized by r⃗(t)=⟨t,2−t,1+t⟩\vec{r}(t) = \langle t, 2-t, 1+t \rangle. The plane P\mathcal{P} contains the line LL and is orthogonal to the plane x+y+z=1x+y+z=1. Which of the following is an equation for plane P\mathcal{P}?

x+z=1x + z = 1
x−z=−1x - z = -1
x−2y+z=−3x - 2y + z = -3
y+z=3y + z = 3
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Plane Equations And Normals

Practice Plane Equations And Normals in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Plane Equations And Normals, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

The line LL is parameterized by r⃗(t)=⟨t,2−t,1+t⟩\vec{r}(t) = \langle t, 2-t, 1+t \rangle. The plane P\mathcal{P} contains the line LL and is orthogonal to the plane x+y+z=1x+y+z=1. Which of the following is an equation for plane P\mathcal{P}?

  1. x+z=1x + z = 1
  2. x−z=−1x - z = -1 (correct answer)
  3. x−2y+z=−3x - 2y + z = -3
  4. y+z=3y + z = 3
Explanation: Let the desired plane be P\mathcal{P} and the given plane be P1\mathcal{P}_1. Let n⃗\vec{n} be the normal vector to P\mathcal{P} and n⃗1\vec{n}_1 be the normal vector to P1\mathcal{P}_1. From the equation of P1\mathcal{P}_1, we have n⃗1=⟨1,1,1⟩\vec{n}_1 = \langle 1, 1, 1 \rangle. The direction vector of the line LL is d⃗=⟨1,−1,1⟩\vec{d} = \langle 1, -1, 1 \rangle. Since the plane P\mathcal{P} contains the line LL, its normal vector n⃗\vec{n} must be orthogonal to the direction vector d⃗\vec{d}. Since plane P\mathcal{P} is orthogonal to plane P1\mathcal{P}_1, their normal vectors are orthogonal. Therefore, n⃗\vec{n} must be orthogonal to both d⃗\vec{d} and n⃗1\vec{n}_1. We can find n⃗\vec{n} by computing the cross product: n⃗=d⃗×n⃗1=∣ijk1−11111∣=i(−1−1)−j(1−1)+k(1−(−1))=⟨−2,0,2⟩.\vec{n} = \vec{d} \times \vec{n}_1 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & 1 & 1 \end{vmatrix} = \mathbf{i}(-1-1) - \mathbf{j}(1-1) + \mathbf{k}(1-(-1)) = \langle -2, 0, 2 \rangle. We can use a simpler normal vector parallel to this, such as n′⃗=⟨−1,0,1⟩\vec{n'} = \langle -1, 0, 1 \rangle. To find the equation of the plane, we need a point on it. Since it contains the line LL, we can use any point from the line. For t=0t=0, we get the point P0(0,2,1)P_0(0, 2, 1). Using the point-normal form: −1(x−0)+0(y−2)+1(z−1)=0-1(x-0) + 0(y-2) + 1(z-1) = 0, which simplifies to −x+z−1=0-x + z - 1 = 0, or x−z=−1x - z = -1.

Question 2

A plane P\mathcal{P} passes through the point (1,2,−3)(1, 2, -3) and is perpendicular to the line of intersection of the planes x−y+2z=4x - y + 2z = 4 and 2x+y−z=12x + y - z = 1. Which of the following is an equation for the plane P\mathcal{P}?

  1. x−5y−3z=0x - 5y - 3z = 0 (correct answer)
  2. x+5y−3z=20x + 5y - 3z = 20
  3. 3x+z=03x + z = 0
  4. x−y+2z=−7x - y + 2z = -7
Explanation: Let the two given planes be P1\mathcal{P}_1 and P2\mathcal{P}_2. Their normal vectors are n⃗1=⟨1,−1,2⟩\vec{n}_1 = \langle 1, -1, 2 \rangle and n⃗2=⟨2,1,−1⟩\vec{n}_2 = \langle 2, 1, -1 \rangle. The direction vector, d⃗\vec{d}, of the line of intersection is perpendicular to both normal vectors. We can find d⃗\vec{d} by computing the cross product: d⃗=n⃗1×n⃗2=⟨(−1)(−1)−(2)(1),(2)(2)−(1)(−1),(1)(1)−(−1)(2)⟩=⟨−1,5,3⟩\vec{d} = \vec{n}_1 \times \vec{n}_2 = \langle (-1)(-1) - (2)(1), (2)(2) - (1)(-1), (1)(1) - (-1)(2) \rangle = \langle -1, 5, 3 \rangle. The desired plane P\mathcal{P} is perpendicular to this line, so its normal vector n⃗\vec{n} is parallel to d⃗\vec{d}. We can choose n⃗=⟨−1,5,3⟩\vec{n} = \langle -1, 5, 3 \rangle. The plane passes through the point (1,2,−3)(1, 2, -3), so its equation is given by −1(x−1)+5(y−2)+3(z−(−3))=0-1(x - 1) + 5(y - 2) + 3(z - (-3)) = 0. This simplifies to −x+1+5y−10+3z+9=0-x + 1 + 5y - 10 + 3z + 9 = 0, which is −x+5y+3z=0-x + 5y + 3z = 0. Multiplying by -1 gives the equivalent equation x−5y−3z=0x - 5y - 3z = 0.

Question 3

Three planes have equations x+2y−z=4x + 2y - z = 4, 2x−y+3z=12x - y + 3z = 1, and 3x+y+2z=k3x + y + 2z = k. For what value of kk do the three planes intersect in exactly one point, and what happens when k=7k = 7?

  1. Any k≠7k \neq 7; when k=7k = 7, the planes intersect in a line
  2. Any k≠5k \neq 5; when k=7k = 7, the planes have no common intersection (correct answer)
  3. k=5k = 5 only; when k=7k = 7, the planes intersect in a line
  4. Any k≠7k \neq 7; when k=7k = 7, the planes have no common intersection
Explanation: Set up the augmented matrix and row reduce. The coefficient matrix has determinant ∣12−12−13312∣=1(−2−3)−2(4−9)−1(2+3)=−5+10−5=0 \begin{vmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \\ 3 & 1 & 2 \end{vmatrix} = 1(-2-3) - 2(4-9) - 1(2+3) = -5 + 10 - 5 = 0. Since the coefficient matrix is singular, the system either has no solution or infinitely many solutions. Row reducing the augmented matrix shows that the system is consistent (has solutions) when k=5k = 5, giving infinitely many solutions (intersection in a line). For k≠5k \neq 5, the system is inconsistent (no solutions). When k=7k = 7, the system is inconsistent, so the planes have no common intersection point.

Question 4

A tetrahedron has vertices at A(1,0,0)A(1, 0, 0), B(0,2,0)B(0, 2, 0), C(0,0,3)C(0, 0, 3), and D(1,1,1)D(1, 1, 1). What is the equation of the plane containing face ABCABC, and what is the distance from vertex DD to this plane?

  1. 6x+3y+2z=66x + 3y + 2z = 6; distance = 57\frac{5}{7} (correct answer)
  2. 6x+3y+2z=66x + 3y + 2z = 6; distance = 37\frac{3}{7}
  3. 3x+6y+2z=63x + 6y + 2z = 6; distance = 57\frac{5}{7}
  4. 6x+3y+2z=66x + 3y + 2z = 6; distance = 17\frac{1}{7}
Explanation: For plane through A(1,0,0)A(1,0,0), B(0,2,0)B(0,2,0), C(0,0,3)C(0,0,3), use vectors AB⃗=(−1,2,0)\vec{AB} = (-1, 2, 0) and AC⃗=(−1,0,3)\vec{AC} = (-1, 0, 3). The normal vector is AB⃗×AC⃗=(−1,2,0)×(−1,0,3)=(6,3,2)\vec{AB} \times \vec{AC} = (-1, 2, 0) \times (-1, 0, 3) = (6, 3, 2). Using point A(1,0,0)A(1, 0, 0): 6(x−1)+3(y−0)+2(z−0)=06(x-1) + 3(y-0) + 2(z-0) = 0 gives 6x+3y+2z=66x + 3y + 2z = 6. The distance from D(1,1,1)D(1, 1, 1) to this plane is ∣6(1)+3(1)+2(1)−6∣62+32+22=∣6+3+2−6∣36+9+4=549=57\frac{|6(1) + 3(1) + 2(1) - 6|}{\sqrt{6^2 + 3^2 + 2^2}} = \frac{|6 + 3 + 2 - 6|}{\sqrt{36 + 9 + 4}} = \frac{5}{\sqrt{49}} = \frac{5}{7}.

Question 5

Find the equation of the plane that contains the line x=1+2t,y=−1+3t,z=4tx = 1 + 2t, y = -1 + 3t, z = 4t and the point P(2,1,1)P(2, 1, 1).

  1. 5x−2y−z=75x - 2y - z = 7 (correct answer)
  2. x−6y+4z=0x - 6y + 4z = 0
  3. 5x+2y−z=115x + 2y - z = 11
  4. x+2y+z=5x + 2y + z = 5
Explanation: First, identify the direction vector of the line, v⃗=⟨2,3,4⟩\vec{v} = \langle 2, 3, 4 \rangle. Next, find a point on the line by setting t=0t=0, which gives P0(1,−1,0)P_0(1, -1, 0). To find the normal vector of the plane, we need two non-parallel vectors that lie in the plane. One is v⃗\vec{v}, and another can be the vector connecting the point on the line P0P_0 to the given point PP, which is w⃗=P0P⃗=⟨2−1,1−(−1),1−0⟩=⟨1,2,1⟩\vec{w} = \vec{P_0P} = \langle 2-1, 1-(-1), 1-0 \rangle = \langle 1, 2, 1 \rangle. The normal vector n⃗\vec{n} to the plane is perpendicular to both v⃗\vec{v} and w⃗\vec{w}, so we can find it using the cross product: n⃗=v⃗×w⃗=⟨(3)(1)−(4)(2),(4)(1)−(2)(1),(2)(2)−(3)(1)⟩=⟨−5,2,1⟩\vec{n} = \vec{v} \times \vec{w} = \langle (3)(1) - (4)(2), (4)(1) - (2)(1), (2)(2) - (3)(1) \rangle = \langle -5, 2, 1 \rangle. Using the point-normal form with point P(2,1,1)P(2, 1, 1) and normal n⃗\vec{n}, the equation of the plane is −5(x−2)+2(y−1)+1(z−1)=0-5(x-2) + 2(y-1) + 1(z-1) = 0, which simplifies to −5x+10+2y−2+z−1=0-5x + 10 + 2y - 2 + z - 1 = 0, or −5x+2y+z=−7-5x + 2y + z = -7. Multiplying by -1 gives 5x−2y−z=75x - 2y - z = 7.

Question 6

The plane x+4y+8z=16x + 4y + 8z = 16 intersects the positive coordinate axes at points AA, BB, and CC. What is the area of the triangle ABCABC?

  1. 99
  2. 3232
  3. 3636 (correct answer)
  4. 7272
Explanation: First, find the points where the plane intersects the coordinate axes. The x-intercept is found by setting y=0y=0 and z=0z=0, which gives x=16x=16. So, A=(16,0,0)A=(16,0,0). The y-intercept is found by setting x=0x=0 and z=0z=0, giving 4y=164y=16, so y=4y=4. Thus, B=(0,4,0)B=(0,4,0). The z-intercept is found by setting x=0x=0 and y=0y=0, giving 8z=168z=16, so z=2z=2. Thus, C=(0,0,2)C=(0,0,2). Now, we can form two vectors on the sides of the triangle, for example, AB⃗=⟨0−16,4−0,0−0⟩=⟨−16,4,0⟩\vec{AB} = \langle 0-16, 4-0, 0-0 \rangle = \langle -16, 4, 0 \rangle and AC⃗=⟨0−16,0−0,2−0⟩=⟨−16,0,2⟩\vec{AC} = \langle 0-16, 0-0, 2-0 \rangle = \langle -16, 0, 2 \rangle. The area of the triangle ABCABC is half the magnitude of the cross product of these two vectors. AB⃗×AC⃗=∣ijk−1640−1602∣=i(8−0)−j(−32−0)+k(0−(−64))=⟨8,32,64⟩.\vec{AB} \times \vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -16 & 4 & 0 \\ -16 & 0 & 2 \end{vmatrix} = \mathbf{i}(8-0) - \mathbf{j}(-32-0) + \mathbf{k}(0 - (-64)) = \langle 8, 32, 64 \rangle. The magnitude is ∣∣⟨8,32,64⟩∣∣=82+322+642=64+1024+4096=5184=72||\langle 8, 32, 64 \rangle|| = \sqrt{8^2 + 32^2 + 64^2} = \sqrt{64 + 1024 + 4096} = \sqrt{5184} = 72. The area of the triangle is 12∣∣AB⃗×AC⃗∣∣=12(72)=36\frac{1}{2} ||\vec{AB} \times \vec{AC}|| = \frac{1}{2}(72) = 36.

Question 7

Two planes are tangent to the sphere x2+(y−1)2+(z+2)2=4x^2 + (y-1)^2 + (z+2)^2 = 4 and are parallel to the plane 2x−2y+z=52x - 2y + z = 5. What is the distance between these two tangent planes?

  1. 22
  2. 33
  3. 44 (correct answer)
  4. 1212
Explanation: The equation of the sphere is x2+(y−1)2+(z+2)2=4x^2 + (y-1)^2 + (z+2)^2 = 4. Its center is C(0,1,−2)C(0, 1, -2) and its radius is r=4=2r = \sqrt{4} = 2. The two tangent planes are parallel to the plane 2x−2y+z=52x - 2y + z = 5, so they share the same normal vector n⃗=⟨2,−2,1⟩\vec{n} = \langle 2, -2, 1 \rangle. The two tangent planes will be located on opposite sides of the sphere's center, each at a distance equal to the radius. The total distance between these two parallel planes is therefore the diameter of the sphere, which is 2r=2(2)=42r = 2(2) = 4. Alternatively, one can find the equations of the planes. They are of the form 2x−2y+z=D2x - 2y + z = D. The distance from the center C(0,1,−2)C(0, 1, -2) to such a plane must be the radius, 2. Using the distance formula: ∣2(0)−2(1)+1(−2)−D∣22+(−2)2+12=2\frac{|2(0) - 2(1) + 1(-2) - D|}{\sqrt{2^2 + (-2)^2 + 1^2}} = 2 ∣−4−D∣9=2  ⟹  ∣D+4∣=6\frac{|-4 - D|}{\sqrt{9}} = 2 \implies |D+4| = 6 This gives two values for DD: D+4=6  ⟹  D=2D+4 = 6 \implies D=2 and D+4=−6  ⟹  D=−10D+4 = -6 \implies D=-10. The two planes are 2x−2y+z=22x - 2y + z = 2 and 2x−2y+z=−102x - 2y + z = -10. The distance between two parallel planes Ax+By+Cz=D1Ax+By+Cz=D_1 and Ax+By+Cz=D2Ax+By+Cz=D_2 is ∣D1−D2∣A2+B2+C2\frac{|D_1 - D_2|}{\sqrt{A^2+B^2+C^2}}. So the distance is ∣2−(−10)∣22+(−2)2+12=123=4\frac{|2 - (-10)|}{\sqrt{2^2+(-2)^2+1^2}} = \frac{12}{3} = 4.

Question 8

The angle between the planes x+y=1x+y=1 and y+z=2y+z=2 is θ\theta. What is the value of cos⁡(θ)\cos(\theta)?

  1. 00
  2. 32\frac{\sqrt{3}}{2}
  3. 22\frac{\sqrt{2}}{2}
  4. 12\frac{1}{2} (correct answer)
Explanation: When you encounter a question about the angle between two planes, you need to find the angle between their normal vectors. The angle between planes is defined as the acute angle between their normal vectors. To find normal vectors, look at the coefficients of x, y, and z in each plane equation. For the plane x+y=1x + y = 1, we can rewrite this as x+y+0z=1x + y + 0z = 1, giving us normal vector n1⃗=⟨1,1,0⟩\vec{n_1} = \langle 1, 1, 0 \rangle. For the plane y+z=2y + z = 2, we can write 0x+y+z=20x + y + z = 2, giving us normal vector n2⃗=⟨0,1,1⟩\vec{n_2} = \langle 0, 1, 1 \rangle. The cosine of the angle between these vectors uses the dot product formula: cos⁡(θ)=n1⃗⋅n2⃗∣n1⃗∣∣n2⃗∣\cos(\theta) = \frac{\vec{n_1} \cdot \vec{n_2}}{|\vec{n_1}||\vec{n_2}|} Calculate each component:
  • n1⃗⋅n2⃗=(1)(0)+(1)(1)+(0)(1)=1\vec{n_1} \cdot \vec{n_2} = (1)(0) + (1)(1) + (0)(1) = 1
  • ∣n1⃗∣=12+12+02=2|\vec{n_1}| = \sqrt{1^2 + 1^2 + 0^2} = \sqrt{2}
  • ∣n2⃗∣=02+12+12=2|\vec{n_2}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}
Therefore: cos⁡(θ)=12⋅2=12\cos(\theta) = \frac{1}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2} Choice (A) 00 would mean perpendicular planes, but our dot product isn't zero. Choice (B) 32\frac{\sqrt{3}}{2} corresponds to a 30° angle, which doesn't match our calculation. Choice (C) 22\frac{\sqrt{2}}{2} corresponds to a 45° angle and likely comes from incorrectly using just the magnitude 2\sqrt{2} without proper division. Remember: always extract normal vectors from the coefficients, then apply the dot product formula systematically. The angle between planes always uses their normal vectors, not direction vectors.

Question 9

What are the coordinates of the reflection of the point Q(3,1,0)Q(3, 1, 0) across the plane x−y+z=5x - y + z = 5?

  1. (4,0,1)(4, 0, 1)
  2. (5,−1,2)(5, -1, 2) (correct answer)
  3. (1,3,−2)(1, 3, -2)
  4. (9,−5,6)(9, -5, 6)
Explanation: Let Q=(3,1,0)Q=(3,1,0). The normal vector to the plane is n⃗=⟨1,−1,1⟩\vec{n} = \langle 1, -1, 1 \rangle. Let's find a point P0P_0 on the plane, for instance, by setting y=0,z=0y=0, z=0, we get x=5x=5, so P0=(5,0,0)P_0=(5,0,0). Let v⃗=P0Q⃗=⟨3−5,1−0,0−0⟩=⟨−2,1,0⟩\vec{v} = \vec{P_0Q} = \langle 3-5, 1-0, 0-0 \rangle = \langle -2, 1, 0 \rangle. The vector from the reflection point QreflQ_{refl} to QQ is twice the projection of the vector QP0⃗\vec{QP_0} onto n⃗\vec{n}, or −2projn⃗v⃗-2\text{proj}_{\vec{n}}\vec{v}. The projection vector is given by: projn⃗v⃗=v⃗⋅n⃗∣∣n⃗∣∣2n⃗\text{proj}_{\vec{n}}\vec{v} = \frac{\vec{v} \cdot \vec{n}}{||\vec{n}||^2} \vec{n} The dot product is v⃗⋅n⃗=(−2)(1)+(1)(−1)+(0)(1)=−3\vec{v} \cdot \vec{n} = (-2)(1) + (1)(-1) + (0)(1) = -3. The squared magnitude is ∣∣n⃗∣∣2=12+(−1)2+12=3||\vec{n}||^2 = 1^2 + (-1)^2 + 1^2 = 3. So, projn⃗v⃗=−33⟨1,−1,1⟩=⟨−1,1,−1⟩\text{proj}_{\vec{n}}\vec{v} = \frac{-3}{3} \langle 1, -1, 1 \rangle = \langle -1, 1, -1 \rangle. The position vector of the reflection is OQ⃗refl=OQ⃗−2projn⃗v⃗\vec{OQ}_{refl} = \vec{OQ} - 2 \text{proj}_{\vec{n}}\vec{v}. So, OQ⃗refl=⟨3,1,0⟩−2⟨−1,1,−1⟩=⟨3,1,0⟩−⟨−2,2,−2⟩=⟨5,−1,2⟩\vec{OQ}_{refl} = \langle 3, 1, 0 \rangle - 2\langle -1, 1, -1 \rangle = \langle 3, 1, 0 \rangle - \langle -2, 2, -2 \rangle = \langle 5, -1, 2 \rangle. The reflected point is (5,−1,2)(5, -1, 2).

Question 10

A plane P\mathcal{P} contains all points in space that are equidistant from the points A(−2,1,5)A(-2, 1, 5) and B(4,3,−1)B(4, 3, -1). What is the distance from the origin (0,0,0)(0,0,0) to the plane P\mathcal{P}?

  1. 33
  2. 2019\frac{20}{\sqrt{19}}
  3. 1919\frac{\sqrt{19}}{19} (correct answer)
  4. 119\frac{1}{19}
Explanation: The plane P\mathcal{P} is the perpendicular bisector of the segment ABAB. The normal vector to the plane, n⃗\vec{n}, is the vector from AA to BB: n⃗=AB⃗=⟨4−(−2),3−1,−1−5⟩=⟨6,2,−6⟩\vec{n} = \vec{AB} = \langle 4 - (-2), 3 - 1, -1 - 5 \rangle = \langle 6, 2, -6 \rangle. We can use a simpler, parallel vector n′⃗=⟨3,1,−3⟩\vec{n'} = \langle 3, 1, -3 \rangle. The plane passes through the midpoint MM of the segment ABAB: M=(−2+42,1+32,5−12)=(1,2,2)M = \left( \frac{-2+4}{2}, \frac{1+3}{2}, \frac{5-1}{2} \right) = (1, 2, 2). The equation of the plane is 3(x−1)+1(y−2)−3(z−2)=03(x - 1) + 1(y - 2) - 3(z - 2) = 0, which simplifies to 3x−3+y−2−3z+6=03x - 3 + y - 2 - 3z + 6 = 0, or 3x+y−3z+1=03x + y - 3z + 1 = 0. The distance from a point (x0,y0,z0)(x_0, y_0, z_0) to a plane Ax+By+Cz+D=0Ax + By + Cz + D = 0 is given by the formula ∣Ax0+By0+Cz0+D∣A2+B2+C2\frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}. The distance from the origin (0,0,0)(0,0,0) to the plane P\mathcal{P} is d=∣3(0)+1(0)−3(0)+1∣32+12+(−3)2=∣1∣9+1+9=119d = \frac{|3(0) + 1(0) - 3(0) + 1|}{\sqrt{3^2 + 1^2 + (-3)^2}} = \frac{|1|}{\sqrt{9 + 1 + 9}} = \frac{1}{\sqrt{19}}. Rationalizing the denominator gives 1919\frac{\sqrt{19}}{19}.

Question 11

Two points on the line r⃗(t)=⟨1+t,2−t,2t⟩\vec{r}(t) = \langle 1+t, 2-t, 2t \rangle are at a distance of 6\sqrt{6} from the plane x+2y+z=1x+2y+z=1. What is the distance between these two points?

  1. 66
  2. 1212
  3. 666\sqrt{6}
  4. 12612\sqrt{6} (correct answer)
Explanation: A general point on the line is P(t)=(1+t,2−t,2t)P(t) = (1+t, 2-t, 2t). The distance from this point to the plane x+2y+z−1=0x+2y+z-1=0 is given by the formula D=∣Ax0+By0+Cz0+D0∣A2+B2+C2D = \frac{|Ax_0+By_0+Cz_0+D_0|}{\sqrt{A^2+B^2+C^2}}. D=∣(1+t)+2(2−t)+(2t)−1∣12+22+12=∣1+t+4−2t+2t−1∣6=∣t+4∣6D = \frac{|(1+t) + 2(2-t) + (2t) - 1|}{\sqrt{1^2+2^2+1^2}} = \frac{|1+t+4-2t+2t-1|}{\sqrt{6}} = \frac{|t+4|}{\sqrt{6}} We are given that this distance is 6\sqrt{6}. So, ∣t+4∣6=6\frac{|t+4|}{\sqrt{6}} = \sqrt{6}, which implies ∣t+4∣=6|t+4|=6. This gives two values for tt: t+4=6  ⟹  t1=2t+4 = 6 \implies t_1 = 2 and t+4=−6  ⟹  t2=−10t+4 = -6 \implies t_2 = -10. The two points on the line are P1=r⃗(2)=⟨3,0,4⟩P_1 = \vec{r}(2) = \langle 3, 0, 4 \rangle and P2=r⃗(−10)=⟨−9,12,−20⟩P_2 = \vec{r}(-10) = \langle -9, 12, -20 \rangle. The distance between these two points is the magnitude of the vector connecting them. This vector is P2P1⃗=⟨3−(−9),0−12,4−(−20)⟩=⟨12,−12,24⟩\vec{P_2P_1} = \langle 3 - (-9), 0 - 12, 4 - (-20) \rangle = \langle 12, -12, 24 \rangle. The distance is ∣∣⟨12,−12,24⟩∣∣=122+(−12)2+242=144+144+576=864||\langle 12, -12, 24 \rangle|| = \sqrt{12^2 + (-12)^2 + 24^2} = \sqrt{144+144+576} = \sqrt{864}. Simplifying the radical, 864=144×6864 = 144 \times 6, so the distance is 144×6=126\sqrt{144 \times 6} = 12\sqrt{6}. Alternatively, once t1t_1 and t2t_2 are found, the distance is ∣t1−t2∣∣∣d⃗∣∣|t_1 - t_2| ||\vec{d}||, where d⃗=⟨1,−1,2⟩\vec{d}=\langle 1, -1, 2 \rangle is the direction vector of the line. ∣∣d⃗∣∣=12+(−1)2+22=6||\vec{d}||=\sqrt{1^2+(-1)^2+2^2}=\sqrt{6}. The distance is ∣2−(−10)∣6=126|2 - (-10)|\sqrt{6} = 12\sqrt{6}.

Question 12

Which statement best describes the intersection of the three planes defined by the equations x+2y−z=3x + 2y - z = 3, 2x−y+3z=12x - y + 3z = 1, and 8x+y+7z=98x + y + 7z = 9?

  1. The planes intersect at a single point.
  2. The planes intersect in a single line. (correct answer)
  3. The planes form a triangular prism with no common intersection.
  4. Two of the planes are parallel and are intersected by the third plane.
Explanation: First, we examine the normal vectors of the planes: n⃗1=⟨1,2,−1⟩\vec{n}_1 = \langle 1, 2, -1 \rangle, n⃗2=⟨2,−1,3⟩\vec{n}_2 = \langle 2, -1, 3 \rangle, and n⃗3=⟨8,1,7⟩\vec{n}_3 = \langle 8, 1, 7 \rangle. No two vectors are scalar multiples, so no two planes are parallel. To determine if the planes intersect at a point, a line, or not at all (forming a prism), we check if the normal vectors are coplanar by calculating their scalar triple product: det⁡(n⃗1,n⃗2,n⃗3)=∣12−12−13817∣=1(−7−3)−2(14−24)−1(2−(−8))=−10+20−10=0.\det(\vec{n}_1, \vec{n}_2, \vec{n}_3) = \begin{vmatrix} 1 & 2 & -1 \\ 2 & -1 & 3 \\ 8 & 1 & 7 \end{vmatrix} = 1(-7-3) - 2(14-24) - 1(2-(-8)) = -10 + 20 - 10 = 0. Since the scalar triple product is zero, the normal vectors are coplanar. This means the planes either intersect in a single line or form a prism. To distinguish these cases, we check for consistency in the system of linear equations. Notice that n⃗3=2n⃗1+3n⃗2\vec{n}_3 = 2\vec{n}_1 + 3\vec{n}_2. We check if the same linear combination applies to the constant terms of the plane equations written as Ax+By+Cz−D=0Ax+By+Cz-D=0. Let the equations be E1E_1, E2E_2, E3E_3. We check if 2(x+2y−z−3)+3(2x−y+3z−1)=02(x+2y-z-3) + 3(2x-y+3z-1) = 0 corresponds to the third plane. This simplifies to 8x+y+7z−9=08x+y+7z-9=0, which is exactly the third plane's equation. Since the third equation is a linear combination of the first two, the system is dependent and consistent. Therefore, the three planes intersect in a single line.