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Q. 36

Expert-verified

Calculus

Found in: Page 824

Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

In Exercises 36–41 use the given sets of points to find:
(a) A nonzero vector N perpendicular to the plane determined by the points.
(b) Two unit vectors perpendicular to the plane determined by the points.
(c) The area of the triangle determined by the points.
$P (1, 4, 6), Q (- 3, 5, 0), R (3, 2, - 1)$

(a) A nonzero vector N perpendicular to the plane determined by the points are $(- 13, - 16, 6)$ .

(b) Two unit vectors perpendicular to the plane determined by the points are $\pm \frac{1}{\sqrt{461}} (- 13, - 16, 6)$ .

(c) The area of the triangle determined by the points is $\frac{\sqrt{461}}{2}$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1. Given Information

In the given exercises use the given sets of points to find:

(a) A nonzero vector N perpendicular to the plane determined by the points.

(b) Two unit vectors perpendicular to the plane determined by the points.

(c) The area of the triangle determined by the points.

The given points are $P (1, 4, 6), Q (- 3, 5, 0), R (3, 2, - 1)$

Part (a) Step 1. firstly finding a nonzero vector N perpendicular to the plane determined by the points.

We have $P (1, 4, 6), Q (- 3, 5, 0), R (3, 2, - 1)$

Now

role="math" localid="1649674392445" $\vec{P Q} = (- 3 - 1, 5 - 4, 0 - 6) = (- 4, 1, - 6) \vec{P R} = (3 - 1, 2 - 4, (- 1) - 6) = (2, - 2, - 1)$

Part (a) Step 2. Now finding PQ→×PR→

$\vec{P Q} \times \vec{P R} = |\begin{matrix} i & j & k \\ - 4 & 1 & - 6 \\ 2 & - 2 & - 1 \end{matrix}| \vec{P Q} \times \vec{P R} = i |\begin{matrix} 1 & - 6 \\ - 2 & - 1 \end{matrix}| - j |\begin{matrix} - 4 & - 6 \\ 2 & - 1 \end{matrix}| + k |\begin{matrix} - 4 & 1 \\ 2 & - 2 \end{matrix}| \vec{P Q} \times \vec{P R} = i {1 \times (- 1) - (- 6) \times (- 2)} - j {(- 4) \times (- 1) - (- 6) \times 2} + k {(- 4) \times (- 2) - 1 \times 2} \vec{P Q} \times \vec{P R} = i (- 1 - 12) - j (4 + 12) + k (8 - 2) \vec{P Q} \times \vec{P R} = - 13 i - 16 j + 6 k$

The points are $(- 13, - 16, 6)$ .

Part (b) Step 1. Now finding two unit vectors perpendicular to the plane determined by the points.

So,

$||\vec{P Q} \times \vec{P R}|| = \sqrt{{(- 13)}^{2} + {(- 16)}^{2} + {(6)}^{2}} ||\vec{P Q} \times \vec{P R}|| = \sqrt{169 + 256 + 36} ||\vec{P Q} \times \vec{P R}|| = \pm \sqrt{461}$

Required vector

$\frac{\vec{P Q} \times \vec{P R}}{||\vec{P Q} \times \vec{P R}||} = \frac{(- 13, - 16, 6)}{\pm \sqrt{461}} \frac{\vec{P Q} \times \vec{P R}}{||\vec{P Q} \times \vec{P R}||} = \pm \frac{1}{\sqrt{461}} (- 13, - 16, 6)$

Part (c) Step 1. Now finding the area of the triangle determined by the points.

$Area ∆ A B C = \frac{1}{2} ||\vec{P Q} \times \vec{P R}|| Area ∆ A B C = \frac{1}{2} ||\sqrt{461}|| Area ∆ A B C = \frac{\sqrt{461}}{2}$

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