Answered: find (a) the area of the parallelogram…

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN:9781305658004

Author:Ron Larson

Publisher:Ron Larson

Chapter5: Inner Product Spaces

Section5.CR: Review Exercises

Problem 73CR

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Question

find (a) the area of the parallelogram determined
by vectors u and v and (b) the volume of the parallelepiped
determined by the vectors u, v, and w. u = i + j - k, v = 2i + j + k, w = -i - 2j + 3k

Expert Solution

Step 1: Given

The vectors u, v, and w.

$u = i + j - k, v = 2 i + j + k, w = - i - 2 j + 3 k .$

Step 2: (a)

To determine the area of the parallelogram determined by vectors u and v.

The area of the parallelogram determined by vectors u and v is given by the cross product of u and v.

That is,

$\begin{array}{rcl} u \times v & = & |\begin{matrix} i & j & k \\ 1 & 1 & - 1 \\ 2 & 1 & 1 \end{matrix}| \\ = & i |\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix}| - j |\begin{matrix} 1 & - 1 \\ 2 & 1 \end{matrix}| + k |\begin{matrix} 1 & 1 \\ 2 & 1 \end{matrix}| \\ = & i (1 + 1) - j (1 + 2) + k (1 - 2) \\ = & 2 i - 3 j - k \\ = & <2, - 3, - 1> \end{array}$

And Area of the parallelogram is given by,

$A = |u \times v| = \sqrt{{(2)}^{2} + {(- 3)}^{2} + {(- 1)}^{2}} = \sqrt{4 + 9 + 1} = \sqrt{14}$ .

Therefore, the area of the parallelogram determined by vectors u and v is $\sqrt{14}$ .

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