Chapter 9: Q. 52 (page 627)

find the area of the parallelogram with vertices P1 , P2 , P3 , and P4 .
$P_{1} = (2, 1, 1) P_{2} = (2, 3, 1) P_{3} = (- 2, 4, 1) P_{4} = (- 2, 6, 1)$

Short Answer

Expert verified

$A r e a = 5 s q u n i t s$

Step by step solution

Step 1. Area of parallelogram

We know that area of parallelogram is given as :

$\vec{u} \times \vec{v}$

where ;

u and v being the vector of adjacent sides of parallelogram

Step 2. Solving for vectors u→ & v→

$\vec{u} = P_{1} P_{2} \vec{u} = (2, 3, 1) - (2, 1, 1) \vec{u} = (0, 2, 0) \vec{v} = \vec{P_{1} P_{4}} \vec{v} = (- 2, 4, 1) - (2, 1, 1) \vec{v} = (- 4, 3, 0)$

We can write these vectors as:

$\vec{u} = 0 \hat{i} + 2 \hat{j} + 0 \hat{k} \vec{v} = - 4 \hat{i} + 3 \hat{j} + 0 \hat{k}$

Step 3. Area is equals to

$A r e a = | \vec{u} \times \vec{v} | \vec{u} \times \vec{v} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & 0 \\ - 4 & 3 & 0 \end{matrix}| \vec{u} \times \vec{v} = \hat{i} (0) - \hat{j} (0) + \hat{k} (8) \vec{u} \times \vec{v} = 8 \hat{k} \Rightarrow A r e a = | \vec{u} \times \vec{v} | = 8 s q u n i t s$

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find the area of the parallelogram with vertices P1 , P2 , P3 , and P4 .
$P_{1} = (2, 1, 1) P_{2} = (2, 3, 1) P_{3} = (- 2, 4, 1) P_{4} = (- 2, 6, 1)$

Short Answer

Step by step solution

Step 1. Area of parallelogram

Step 2. Solving for vectors u→ & v→

Step 3. Area is equals to

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find the area of the parallelogram with vertices P1 , P2 , P3 , and P4 .P1=(2,1,1)P2=(2,3,1)P3=(-2,4,1)P4=(-2,6,1)

Short Answer

Step by step solution

Step 1. Area of parallelogram

Step 2. Solving for vectors u→ & v→

Step 3. Area is equals to

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Geometry

Logic and Functions

Applied Mathematics

Mechanics Maths

Pure Maths

Study anywhere. Anytime. Across all devices.

find the area of the parallelogram with vertices P1 , P2 , P3 , and P4 .
$P_{1} = (2, 1, 1) P_{2} = (2, 3, 1) P_{3} = (- 2, 4, 1) P_{4} = (- 2, 6, 1)$