/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 58 Prove that the vectors \((3,2,-1... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 58

Prove that the vectors \((3,2,-1),(5,-7,3)\) and \((11,-3,1)\) are coplanar.

Short Answer

Expert verified
The vectors are coplanar as their scalar triple product is zero.

Step by step solution

01

Understanding Coplanarity

Three vectors are coplanar if they lie within the same plane. Mathematically, vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) are coplanar if the scalar triple product is zero. The scalar triple product is calculated as \(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})\).
02

Define the vectors

Let \(\mathbf{a} = (3, 2, -1)\), \(\mathbf{b} = (5, -7, 3)\), and \(\mathbf{c} = (11, -3, 1)\). We need to calculate the scalar triple product \(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})\).
03

Calculate the cross product \(\mathbf{b} \times \mathbf{c}\)

To find \(\mathbf{b} \times \mathbf{c}\), use the determinant of the following matrix:\[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \5 & -7 & 3 \11 & -3 & 1 \ \end{vmatrix}\]Calculating the determinant gives:\[\mathbf{b} \times \mathbf{c} = \mathbf{i}(({-7})(1) - (3)(-3)) - \mathbf{j}((5)(1) - (3)(11)) + \mathbf{k}((5)(-3) - (-7)(11))\]\[= \mathbf{i}(-7 + 9) - \mathbf{j}(5 - 33) + \mathbf{k}(-15 + 77)\]\[= 2\mathbf{i} + 28\mathbf{j} + 62\mathbf{k}\]So, \(\mathbf{b} \times \mathbf{c} = (2, 28, 62)\).
04

Calculate the dot product \(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})\)

Compute the dot product of \((3, 2, -1)\) with \((2, 28, 62)\):\[\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 3 \cdot 2 + 2 \cdot 28 + (-1) \cdot 62\]\[= 6 + 56 - 62\]\[= 0\]
05

Conclusion

Since the scalar triple product is zero, the vectors \(\mathbf{a}, \mathbf{b}, \) and \(\mathbf{c}\) are coplanar.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Scalar Triple Product
The scalar triple product is a mathematical operation involving three vectors. It helps us determine if these vectors are coplanar, meaning they lie on the same plane.
To compute it, you need two key operations: the cross product and the dot product.
  • You first find the cross product of two vectors, which results in a new vector perpendicular to both.
  • Then, take the dot product of that resulting vector with the third vector.
The formula for the scalar triple product is \[\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}).\]In this problem, this computation results in zero, confirming the vectors are coplanar.
Cross Product
The cross product is a binary operation on two vectors in three-dimensional space. It produces a third vector perpendicular to both original vectors.
This perpendicular vector's direction is determined using the right-hand rule, and its magnitude corresponds to the area of the parallelogram formed by the original vectors.
  • The cross product of vectors \(\mathbf{b}\) and \(\mathbf{c}\) is calculated using the determinant of a matrix:
\[\begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \5 & -7 & 3 \11 & -3 & 1 \\end{vmatrix}\]. This results in a new vector, which is used in the scalar triple product computation.
Determinants
Determinants are mathematical numbers that are associated with a square matrix. They are extremely useful in various calculations, including cross products.
For a 3x3 matrix, the determinant can be thought of as a value that describes certain properties about the matrix and the linear system it represents.
  • In this exercise, determinants solve the cross product efficiently using the formula:
\[\det \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\mathbf{b}_x & \mathbf{b}_y & \mathbf{b}_z \\mathbf{c}_x & \mathbf{c}_y & \mathbf{c}_z \\end{vmatrix}.\] This calculation gives us the vector \((2, 28, 62)\), aiding in proving vector coplanarity.
Vector Dot Product
The dot product is a basic operation involving two vectors, and it results in a scalar (a real number). This is the product of the magnitudes of the two vectors and the cosine of the angle between them.
It measures the extent to which one vector goes in the direction of another.
  • In this context, the dot product is performed between vector \(\mathbf{a}\) and the result from the cross product \(\mathbf{b} \times \mathbf{c}\):
\[\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = 3 \cdot 2 + 2 \cdot 28 + (-1) \cdot 62.\] This operation leads to the scalar zero, verifying that the original vectors are indeed coplanar.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The line of intersection of two planes \(\boldsymbol{r} \cdot \boldsymbol{n}_{1}=p_{1}\) and \(\boldsymbol{r} \cdot \boldsymbol{n}_{2}=p_{2}\) lies in both planes. It is therefore perpendicular to both \(\boldsymbol{n}_{1}\) and \(\boldsymbol{n}_{2}\). Give an expression for this direction, and so show that the equation of the line of intersection may be written as \(\boldsymbol{r}=\boldsymbol{r}_{0}+t\left(\boldsymbol{n}_{1} \times \boldsymbol{n}_{2}\right)\), where \(\boldsymbol{r}_{0}\) is any vector satisfying \(\boldsymbol{r}_{0} \cdot \boldsymbol{n}_{1}=p_{1}\) and \(\boldsymbol{r}_{0} \cdot \boldsymbol{n}_{2}=p_{2}\). Hence find the line of intersection of the planes \(\boldsymbol{r} \cdot(1,1,1)=5\) and \(\boldsymbol{r} \cdot(4,1,2)=15\).

An aeroplane flies \(100 \mathrm{~km}\) in a NE direction, then \(120 \mathrm{~km}\) in a ESE direction and finally \(\mathrm{S}\) for a further \(50 \mathrm{~km}\). Sketch the vectors representing this flight path. What is the distance from start to finish and also the length of the flight path?

The points A, B and C have coordinates \((1,2,2)\), \((7,2,1)\) and \((2,4,1)\) relative to rectangular coordinate axes. Find: (a) the vectors \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{AC}}\) (b) \(|\overrightarrow{\mathrm{AB}}-3 \overrightarrow{\mathrm{AC}}|\) (c) the unit vector in the direction of \(\overrightarrow{\mathrm{AB}}-3 \overrightarrow{\mathrm{AC}}\) (d) the lengths of the vectors \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{AC}}\) (e) the vector \(\overrightarrow{\mathrm{AM}}\) where \(\mathrm{M}\) is the midpoint of \(\mathrm{BC}\).

Find the vector equation of the plane that passes through the points \((1,2,3),(2,4,5)\) and \((4,5,6)\) What is its cartesian equation?

Find the value of \(t\) that makes the angle between the two vectors \(\boldsymbol{a}=(3,1,0)\) and \(\boldsymbol{b}=(t, 0,1)\) equal to \(45^{\circ}\).
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Translational Dynamics

Read Explanation

Quantum Physics

Read Explanation

Oscillations

Read Explanation

Electromagnetism

Read Explanation

Electricity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.