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Normal lines and normal vectors to surfaces practice problems
The normal vector and normal line to a surface F(x, y, z) = c at a point. 57 problems with worked solutions; in 57 of them every
equation is proved by a computer algebra system.
20 worked problems
Try each one before opening the solution. Every line is labeled with what checked it. How we verify
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle 6, 0, -1 \right\rangle,\quad \mathbf r(t) = \left\langle 6 t + 1, 0, 1 - t \right\rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle 0, 6, -1 \right\rangle,\quad \mathbf r(t) = \left\langle 0, 6 t + 1, 2 - t \right\rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
∇F at the point is normal to the level surface.✓ Proved
\[ \left[\begin{matrix}\left. 2 t + 1 \right|_{\substack{ t=0 }}\\\left. 2 t - 1 \right|_{\substack{ t=0 }}\\\left. 4 t + 2 \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}1\\-1\\2\end{matrix}\right] \]
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle 2, 2, 4 \right\rangle,\quad \mathbf r(t) = \left\langle 2 t + 1, 2 t - 1, 4 t + 2 \right\rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle 0, 2, 4 \right\rangle,\quad \mathbf r(t) = \left\langle 0, 2 t + 1, 4 t + 2 \right\rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find a normal vector and the normal line to the surface \( \displaystyle x y + x z + y z = 5 \) at \( \displaystyle (2, 1, 1) \).
\[ \left. x y + x z + y z \right|_{\substack{ x=2\\ y=1\\ z=1 }} = 5 \]
The point is on the surface.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial x} \left(x y + x z + y z\right) \right|_{\substack{ x=2\\ y=1\\ z=1 }}\\\left. \frac{\partial}{\partial y} \left(x y + x z + y z\right) \right|_{\substack{ x=2\\ y=1\\ z=1 }}\\\left. \frac{\partial}{\partial z} \left(x y + x z + y z\right) \right|_{\substack{ x=2\\ y=1\\ z=1 }}\end{matrix}\right] = \left[\begin{matrix}2\\3\\3\end{matrix}\right] \]
∇F at the point is normal to the level surface.✓ Proved
\[ \left[\begin{matrix}\left. 2 t + 2 \right|_{\substack{ t=0 }}\\\left. 3 t + 1 \right|_{\substack{ t=0 }}\\\left. 3 t + 1 \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}2\\1\\1\end{matrix}\right] \]
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle 2, 3, 3 \right\rangle,\quad \mathbf r(t) = \left\langle 2 t + 2, 3 t + 1, 3 t + 1 \right\rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find a normal vector and the normal line to the surface \( \displaystyle x y z = -2 \) at \( \displaystyle (1, -1, 2) \).
\[ \left. x y z \right|_{\substack{ x=1\\ y=-1\\ z=2 }} = -2 \]
The point is on the surface.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial x} x y z \right|_{\substack{ x=1\\ y=-1\\ z=2 }}\\\left. \frac{\partial}{\partial y} x y z \right|_{\substack{ x=1\\ y=-1\\ z=2 }}\\\left. \frac{\partial}{\partial z} x y z \right|_{\substack{ x=1\\ y=-1\\ z=2 }}\end{matrix}\right] = \left[\begin{matrix}-2\\2\\-1\end{matrix}\right] \]
∇F at the point is normal to the level surface.✓ Proved
\[ \left[\begin{matrix}\left. 1 - 2 t \right|_{\substack{ t=0 }}\\\left. 2 t - 1 \right|_{\substack{ t=0 }}\\\left. 2 - t \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}1\\-1\\2\end{matrix}\right] \]
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle -2, 2, -1 \right\rangle,\quad \mathbf r(t) = \left\langle 1 - 2 t, 2 t - 1, 2 - t \right\rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
∇F at the point is normal to the level surface.✓ Proved
\[ \left[\begin{matrix}\left. 2 t + 1 \right|_{\substack{ t=0 }}\\\left. - 2 t - 1 \right|_{\substack{ t=0 }}\\\left. 4 t + 2 \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}1\\-1\\2\end{matrix}\right] \]
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle 2, -2, 4 \right\rangle,\quad \mathbf r(t) = \left\langle 2 t + 1, - 2 t - 1, 4 t + 2 \right\rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
∇F at the point is normal to the level surface.✓ Proved
\[ \left[\begin{matrix}\left. 1 - 3 t \right|_{\substack{ t=0 }}\\\left. - t - 1 \right|_{\substack{ t=0 }}\\\left. t + 2 \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}1\\-1\\2\end{matrix}\right] \]
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle -3, -1, 1 \right\rangle,\quad \mathbf r(t) = \left\langle 1 - 3 t, - t - 1, t + 2 \right\rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find a normal vector and the normal line to the surface \( \displaystyle x y z = 1 \) at \( \displaystyle (1, 1, 1) \).
\[ \left. x y z \right|_{\substack{ x=1\\ y=1\\ z=1 }} = 1 \]
The point is on the surface.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial x} x y z \right|_{\substack{ x=1\\ y=1\\ z=1 }}\\\left. \frac{\partial}{\partial y} x y z \right|_{\substack{ x=1\\ y=1\\ z=1 }}\\\left. \frac{\partial}{\partial z} x y z \right|_{\substack{ x=1\\ y=1\\ z=1 }}\end{matrix}\right] = \left[\begin{matrix}1\\1\\1\end{matrix}\right] \]
∇F at the point is normal to the level surface.✓ Proved
\[ \left[\begin{matrix}\left. t + 1 \right|_{\substack{ t=0 }}\\\left. t + 1 \right|_{\substack{ t=0 }}\\\left. t + 1 \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}1\\1\\1\end{matrix}\right] \]
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle 1, 1, 1 \right\rangle,\quad \mathbf r(t) = \left\langle t + 1, t + 1, t + 1 \right\rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle 2, 0, 2 \right\rangle,\quad \mathbf r(t) = \left\langle 2 t + 1, 0, 2 t + 1 \right\rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle 0, 2, -1 \right\rangle,\quad \mathbf r(t) = \left\langle 0, 2 t + 1, 2 - t \right\rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
∇F at the point is normal to the level surface.✓ Proved
\[ \left[\begin{matrix}\left. 6 t + 1 \right|_{\substack{ t=0 }}\\\left. 4 t + 2 \right|_{\substack{ t=0 }}\\\left. 1 - t \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}1\\2\\1\end{matrix}\right] \]
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle 6, 4, -1 \right\rangle,\quad \mathbf r(t) = \left\langle 6 t + 1, 4 t + 2, 1 - t \right\rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
∇F at the point is normal to the level surface.✓ Proved
\[ \left[\begin{matrix}\left. 2 t + 1 \right|_{\substack{ t=0 }}\\\left. - 6 t - 1 \right|_{\substack{ t=0 }}\\\left. 2 - t \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}1\\-1\\2\end{matrix}\right] \]
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle 2, -6, -1 \right\rangle,\quad \mathbf r(t) = \left\langle 2 t + 1, - 6 t - 1, 2 - t \right\rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
∇F at the point is normal to the level surface.✓ Proved
\[ \left[\begin{matrix}\left. 2 t + 1 \right|_{\substack{ t=0 }}\\\left. 2 t + 1 \right|_{\substack{ t=0 }}\\\left. 1 - t \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}1\\1\\1\end{matrix}\right] \]
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle 2, 2, -1 \right\rangle,\quad \mathbf r(t) = \left\langle 2 t + 1, 2 t + 1, 1 - t \right\rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
∇F at the point is normal to the level surface.✓ Proved
\[ \left[\begin{matrix}\left. 2 t + 1 \right|_{\substack{ t=0 }}\\\left. - 2 t - 1 \right|_{\substack{ t=0 }}\\\left. 2 - t \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}1\\-1\\2\end{matrix}\right] \]
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle 2, -2, -1 \right\rangle,\quad \mathbf r(t) = \left\langle 2 t + 1, - 2 t - 1, 2 - t \right\rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
∇F at the point is normal to the level surface.✓ Proved
\[ \left[\begin{matrix}\left. 2 t + 1 \right|_{\substack{ t=0 }}\\\left. 2 - 4 t \right|_{\substack{ t=0 }}\\\left. 2 t + 1 \right|_{\substack{ t=0 }}\end{matrix}\right] = \left[\begin{matrix}1\\2\\1\end{matrix}\right] \]
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle 2, -4, 2 \right\rangle,\quad \mathbf r(t) = \left\langle 2 t + 1, 2 - 4 t, 2 t + 1 \right\rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
Find a normal vector and the normal line to the surface \( \displaystyle x y + x z + y z = -1 \) at \( \displaystyle (1, -1, 2) \).
\[ \left. x y + x z + y z \right|_{\substack{ x=1\\ y=-1\\ z=2 }} = -1 \]
The point is on the surface.✓ Proved
\[ \left[\begin{matrix}\left. \frac{\partial}{\partial x} \left(x y + x z + y z\right) \right|_{\substack{ x=1\\ y=-1\\ z=2 }}\\\left. \frac{\partial}{\partial y} \left(x y + x z + y z\right) \right|_{\substack{ x=1\\ y=-1\\ z=2 }}\\\left. \frac{\partial}{\partial z} \left(x y + x z + y z\right) \right|_{\substack{ x=1\\ y=-1\\ z=2 }}\end{matrix}\right] = \left[\begin{matrix}1\\3\\0\end{matrix}\right] \]
∇F at the point is normal to the level surface.✓ Proved
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle 1, 3, 0 \right\rangle,\quad \mathbf r(t) = \left\langle t + 1, 3 t - 1, 2 \right\rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
The line r(t) = P + t∇F passes through P at t = 0.✓ Proved
Answer \( \mathbf n = \left\langle 0, 2, -1 \right\rangle,\quad \mathbf r(t) = \left\langle 0, 2 t + 1, 2 - t \right\rangle \)
✓ Nihil obstat Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
More normal lines and normal vectors to surfaces practice