Multivariable Calculus Quiz: Orientation And Normal Vectors
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Orientation And Normal VectorsQuestion 1 of 7

Let SS be the part of the plane 2x+3y+z=62x + 3y + z = 6 that lies in the first octant. If SS is oriented so that the normal vector has a positive z-component, what is the unit normal vector to SS?

(2,3,1)14\frac{(2, 3, 1)}{\sqrt{14}}
(−2,−3,1)14\frac{(-2, -3, 1)}{\sqrt{14}}
(2,3,−1)14\frac{(2, 3, -1)}{\sqrt{14}}
(−2,−3,−1)14\frac{(-2, -3, -1)}{\sqrt{14}}
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Orientation And Normal Vectors

Practice Orientation And Normal Vectors 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 Orientation And Normal Vectors, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

How to use this quiz

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.

All questions

Question 1

Let SS be the part of the plane 2x+3y+z=62x + 3y + z = 6 that lies in the first octant. If SS is oriented so that the normal vector has a positive z-component, what is the unit normal vector to SS?

  1. (2,3,1)14\frac{(2, 3, 1)}{\sqrt{14}} (correct answer)
  2. (−2,−3,1)14\frac{(-2, -3, 1)}{\sqrt{14}}
  3. (2,3,−1)14\frac{(2, 3, -1)}{\sqrt{14}}
  4. (−2,−3,−1)14\frac{(-2, -3, -1)}{\sqrt{14}}
Explanation: For the plane 2x+3y+z=62x + 3y + z = 6, the normal vector is the gradient of 2x+3y+z−62x + 3y + z - 6, which is (2,3,1)(2, 3, 1). The magnitude is 22+32+12=14\sqrt{2^2 + 3^2 + 1^2} = \sqrt{14}. Since we want the normal with positive z-component, we use (2,3,1)(2, 3, 1) rather than (−2,−3,−1)(-2, -3, -1). Therefore, the unit normal vector is (2,3,1)14\frac{(2, 3, 1)}{\sqrt{14}}.

Question 2

Consider the hemisphere x2+y2+z2=9x^2 + y^2 + z^2 = 9 with z≥0z \geq 0. If this surface is oriented with the outward-pointing normal vectors, what is the z-component of the unit normal vector at the point (0,0,3)(0, 0, 3)?

  1. 11 (correct answer)
  2. −1-1
  3. 13\frac{1}{3}
  4. 00
Explanation: For the sphere x2+y2+z2=9x^2 + y^2 + z^2 = 9, the gradient ∇(x2+y2+z2)=(2x,2y,2z)\nabla(x^2 + y^2 + z^2) = (2x, 2y, 2z) gives the direction of the normal vector. At (0,0,3)(0, 0, 3), this is (0,0,6)(0, 0, 6). The unit normal is (0,0,1)(0, 0, 1). Since we want the outward-pointing normal and the center of the sphere is at the origin, the vector from center to point (0,0,3)(0, 0, 3) is (0,0,3)(0, 0, 3), which points in the positive z-direction. Therefore, the outward normal at (0,0,3)(0, 0, 3) has z-component 11.

Question 3

A surface SS is parameterized by r(u,v)=(u2,v2,u+v)\mathbf{r}(u,v) = (u^2, v^2, u + v) where u,v∈[0,1]u, v \in [0, 1]. At the point (1,1,2)(1, 1, 2) corresponding to u=1,v=1u = 1, v = 1, which statement about the orientation is correct?

  1. The normal vector ru×rv=(2,2,−4)\mathbf{r}_u \times \mathbf{r}_v = (2, 2, -4) points in the direction of decreasing z
  2. The normal vector ru×rv=(−2,−2,4)\mathbf{r}_u \times \mathbf{r}_v = (-2, -2, 4) points in the direction of increasing z (correct answer)
  3. The normal vector ru×rv=(−2,−2,4)\mathbf{r}_u \times \mathbf{r}_v = (-2, -2, 4) points in the direction of decreasing z
  4. The normal vector ru×rv=(2,2,4)\mathbf{r}_u \times \mathbf{r}_v = (2, 2, 4) points in the direction of increasing z
Explanation: When you encounter a surface parameterization problem asking about orientation, you need to find the normal vector using the cross product of the partial derivatives, then determine which direction it points. First, find the partial derivatives of r(u,v)=(u2,v2,u+v)\mathbf{r}(u,v) = (u^2, v^2, u + v):
  • ru=(2u,0,1)\mathbf{r}_u = (2u, 0, 1)
  • rv=(0,2v,1)\mathbf{r}_v = (0, 2v, 1)
At the point where u=1,v=1u = 1, v = 1:
  • ru=(2,0,1)\mathbf{r}_u = (2, 0, 1)
  • rv=(0,2,1)\mathbf{r}_v = (0, 2, 1)
Calculate the cross product ru×rv\mathbf{r}_u \times \mathbf{r}_v: ru×rv=∣ijk201021∣=(0−2)i−(2−0)j+(4−0)k=(−2,−2,4)\mathbf{r}_u \times \mathbf{r}_v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = (0-2)\mathbf{i} - (2-0)\mathbf{j} + (4-0)\mathbf{k} = (-2, -2, 4) Since the z-component is positive (4), this normal vector points in the direction of increasing z. Therefore, answer B is correct. Answer A uses the wrong cross product rv×ru\mathbf{r}_v \times \mathbf{r}_u (which reverses the orientation) and incorrectly states the direction. Answer C has the correct normal vector but wrongly claims it points toward decreasing z when the positive z-component clearly indicates increasing z. Answer D contains computational errors in both the y and z components of the normal vector. Remember: the cross product ru×rv\mathbf{r}_u \times \mathbf{r}_v follows the right-hand rule, and the sign of the z-component tells you the vertical orientation—positive means increasing z, negative means decreasing z.

Question 4

Consider the surface SS defined by x2+y2=z2x^2 + y^2 = z^2 for 1≤z≤31 \leq z \leq 3. At the point (2,2,22)(2, 2, 2\sqrt{2}), if the surface is oriented with normal vectors pointing away from the z-axis, what is the direction of the unit normal vector?

  1. (2,2,−22)4\frac{(2, 2, -2\sqrt{2})}{4}
  2. (2,2,22)4\frac{(2, 2, 2\sqrt{2})}{4}
  3. (−2,−2,22)4\frac{(-2, -2, 2\sqrt{2})}{4}
  4. (1,1,−2)2\frac{(1, 1, -\sqrt{2})}{2} (correct answer)
Explanation: For the surface x2+y2−z2=0x^2 + y^2 - z^2 = 0, the normal vector is ∇(x2+y2−z2)=(2x,2y,−2z)\nabla(x^2 + y^2 - z^2) = (2x, 2y, -2z). At (2,2,22)(2, 2, 2\sqrt{2}), this gives (4,4,−42)(4, 4, -4\sqrt{2}). The magnitude is 16+16+32=64=8\sqrt{16 + 16 + 32} = \sqrt{64} = 8. To check orientation: the vector from the z-axis to point (2,2,22)(2, 2, 2\sqrt{2}) is (2,2,0)(2, 2, 0). The dot product (4,4,−42)⋅(2,2,0)=16>0(4, 4, -4\sqrt{2}) \cdot (2, 2, 0) = 16 > 0, so this normal points away from the z-axis as desired. The unit normal is (4,4,−42)8=(1,1,−2)2\frac{(4, 4, -4\sqrt{2})}{8} = \frac{(1, 1, -\sqrt{2})}{2}.

Question 5

Consider the surface SS parameterized by r(s,t)=(scos⁡t,ssin⁡t,s2)\mathbf{r}(s,t) = (s\cos t, s\sin t, s^2) for s>0s > 0. At which of the following points does the normal vector point in the direction that makes the smallest angle with the positive z-axis?

  1. (1,0,1)(1, 0, 1) corresponding to s=1,t=0s = 1, t = 0
  2. (2,0,4)(2, 0, 4) corresponding to s=2,t=0s = 2, t = 0
  3. (12,0,14)(\frac{1}{2}, 0, \frac{1}{4}) corresponding to s=12,t=0s = \frac{1}{2}, t = 0 (correct answer)
  4. (22,22,12)\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}, \frac{1}{2}\right) corresponding to s=1,t=π4s = 1, t = \frac{\pi}{4}
Explanation: The normal vector is rs×rt\mathbf{r}_s \times \mathbf{r}_t where rs=(cos⁡t,sin⁡t,2s)\mathbf{r}_s = (\cos t, \sin t, 2s) and rt=(−ssin⁡t,scos⁡t,0)\mathbf{r}_t = (-s\sin t, s\cos t, 0). Computing the cross product: rs×rt=(−2s2cos⁡t,−2s2sin⁡t,s)\mathbf{r}_s \times \mathbf{r}_t = (-2s^2\cos t, -2s^2\sin t, s). The angle with the positive z-axis depends on the z-component relative to the magnitude. The unit normal has z-component s4s4+s2=14s2+1\frac{s}{\sqrt{4s^4 + s^2}} = \frac{1}{\sqrt{4s^2 + 1}}. This is maximized when ss is smallest, which occurs at s=12s = \frac{1}{2}, giving the smallest angle with the z-axis.

Question 6

Consider the surface SS given by z=4−x2−y2z = \sqrt{4 - x^2 - y^2} for x2+y2≤4x^2 + y^2 \leq 4. This surface can be oriented in two ways. If we choose the orientation such that the normal vectors point away from the origin, what is the correct expression for the unit normal vector?

  1. (−x,−y,z)2\frac{(-x, -y, z)}{2}
  2. (x,y,z)2\frac{(x, y, z)}{2} (correct answer)
  3. (x,y,−z)2\frac{(x, y, -z)}{2}
  4. (−x,−y,−z)2\frac{(-x, -y, -z)}{2}
Explanation: When finding unit normal vectors to surfaces, you need to determine both the direction and magnitude of the normal vector, then consider which orientation matches the given conditions. The surface z=4−x2−y2z = \sqrt{4 - x^2 - y^2} represents the upper half of a sphere with radius 2 centered at the origin. To find the normal vector, rewrite this as F(x,y,z)=x2+y2+z2−4=0F(x,y,z) = x^2 + y^2 + z^2 - 4 = 0. The gradient ∇F=(2x,2y,2z)\nabla F = (2x, 2y, 2z) gives a normal vector, which simplifies to (x,y,z)(x, y, z) after factoring out the constant. Since we want the normal vectors to point away from the origin, we need the outward normal. For any point (x,y,z)(x, y, z) on this sphere, the vector from the origin to that point is exactly (x,y,z)(x, y, z), which points outward. The magnitude of this vector is x2+y2+z2=4=2\sqrt{x^2 + y^2 + z^2} = \sqrt{4} = 2 (since all points satisfy x2+y2+z2=4x^2 + y^2 + z^2 = 4). Therefore, the unit normal vector is (x,y,z)2\frac{(x, y, z)}{2}, making B correct. Answer A, (−x,−y,z)2\frac{(-x, -y, z)}{2}, points inward in the xx and yy directions, which would be incorrect for the outward orientation. Answer C, (x,y,−z)2\frac{(x, y, -z)}{2}, has the wrong zz-component and would point toward the lower hemisphere. Answer D, (−x,−y,−z)2\frac{(-x, -y, -z)}{2}, points completely inward toward the origin. Remember: for spheres, the outward normal at any point is simply the position vector from the center to that point, normalized by the radius.

Question 7

Let SS be the surface parameterized by r(θ,ϕ)=(sin⁡ϕcos⁡θ,sin⁡ϕsin⁡θ,cos⁡ϕ)\mathbf{r}(\theta, \phi) = (\sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi) for 0≤θ≤2π0 \leq \theta \leq 2\pi and 0≤ϕ≤π20 \leq \phi \leq \frac{\pi}{2}. This parameterization naturally induces an orientation on SS. At the point corresponding to θ=π4,ϕ=π4\theta = \frac{\pi}{4}, \phi = \frac{\pi}{4}, does the normal vector point toward or away from the origin?

  1. Toward the origin, because the z-component of the normal is negative
  2. Away from the origin, because this is the standard orientation for a unit sphere
  3. Toward the origin, because rθ×rϕ\mathbf{r}_{\theta} \times \mathbf{r}_{\phi} points inward for this parameterization (correct answer)
  4. Away from the origin, because the normal vector equals the position vector at each point
Explanation: For the spherical parameterization, rθ=(−sin⁡ϕsin⁡θ,sin⁡ϕcos⁡θ,0)\mathbf{r}_{\theta} = (-\sin\phi\sin\theta, \sin\phi\cos\theta, 0) and rϕ=(cos⁡ϕcos⁡θ,cos⁡ϕsin⁡θ,−sin⁡ϕ)\mathbf{r}_{\phi} = (\cos\phi\cos\theta, \cos\phi\sin\theta, -\sin\phi). The normal vector rθ×rϕ\mathbf{r}_{\theta} \times \mathbf{r}_{\phi} can be computed, but more directly: this parameterization gives rθ×rϕ=−sin⁡ϕ⋅r\mathbf{r}_{\theta} \times \mathbf{r}_{\phi} = -\sin\phi \cdot \mathbf{r}, which points toward the origin (opposite to the position vector). This is the inward normal orientation for this particular parameterization order.