/*! 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 20 Find a vector that is perpendicu... [FREE SOLUTION] | 91Ó°ÊÓ

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Find a vector that is perpendicular to the plane passing through the three given points. $$P(3,0,0), Q(0,2,-5), R(-2,0,6)$$

Short Answer

Expert verified
The vector \( \langle 12, 43, 10 \rangle \) is perpendicular to the plane.

Step by step solution

01

Understand the Problem

We need to find a vector that is perpendicular to the plane. The plane is defined by three points: \( P(3,0,0), Q(0,2,-5), \text{ and } R(-2,0,6) \). A vector that is perpendicular to the plane is called the normal vector of the plane.
02

Find Two Vectors in the Plane

To find two vectors in the plane, we subtract the coordinates of the given points. Using points \( P \) and \( Q \), we find the vector \( \overrightarrow{PQ} = \langle 0 - 3, 2 - 0, -5 - 0 \rangle = \langle -3, 2, -5 \rangle \). Similarly, using points \( P \) and \( R \), we find the vector \( \overrightarrow{PR} = \langle -2 - 3, 0 - 0, 6 - 0 \rangle = \langle -5, 0, 6 \rangle \).
03

Calculate the Cross Product

The cross product of two vectors in the plane gives us a vector that is perpendicular to both, and thus, to the plane itself. For vectors \( \overrightarrow{PQ} = \langle -3, 2, -5 \rangle \) and \( \overrightarrow{PR} = \langle -5, 0, 6 \rangle \), the cross product is calculated as follows:\[\overrightarrow{PQ} \times \overrightarrow{PR} = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -3 & 2 & -5 \ -5 & 0 & 6 \end{array} \right|\]Evaluating this determinant:\[= \mathbf{i}(2 \cdot 6 - (-5) \cdot 0) - \mathbf{j}(-3 \cdot 6 - (-5) \cdot -5) + \mathbf{k}(-3 \cdot 0 - 2 \cdot -5)\]\[= \mathbf{i}(12) - \mathbf{j}(-18 - 25) + \mathbf{k}(10) \]\[= \langle 12, 43, 10 \rangle \]
04

Verify the Perpendicular Vector

To confirm that the calculated cross product \( \langle 12, 43, 10 \rangle \) is a normal vector to the plane, check if the dot product of the normal vector with any of the plane vectors (e.g., \( \overrightarrow{PQ} \)) results in zero.\[\langle 12, 43, 10 \rangle \cdot \langle -3, 2, -5 \rangle = (12 \cdot -3) + (43 \cdot 2) + (10 \cdot -5) = -36 + 86 - 50 = 0\]Since the dot product is zero, the vector is indeed perpendicular to the vectors in the plane.

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Key Concepts

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

Normal Vector
A normal vector is a vector that is perpendicular to a given surface or plane. In simpler terms, it points directly outwards from the plane and forms a 90-degree angle with any vector lying on the plane. This property makes the normal vector crucial for various applications, including computer graphics, physics simulations, and geometric problem-solving.
When dealing with a plane defined by three points, finding a normal vector helps in understanding the plane's orientation in space. This is often done by calculating the cross product of two vectors that reside on the plane (formed by the given points). For instance, in our example, by determining the vectors \( \overrightarrow{PQ} = \langle -3, 2, -5 \rangle \) and \( \overrightarrow{PR} = \langle -5, 0, 6 \rangle \) and finding their cross product, we derive a vector perpendicular to the plane.
Thus, the normal vector serves as a key mathematical tool for defining the spatial properties and orientation of planes.
Cross Product
The cross product is a mathematical operation used to find a vector that is perpendicular to two given vectors in three-dimensional space. It is denoted by \( \times \) and is an essential concept in vector calculus and physics for determining directions perpendicular to two interacting forces or directions.
In our exercise example, we calculated the cross product of \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \) to find a normal vector to the plane. The cross product is computed using a determinant formed from a 3x3 matrix composed of the unit vectors and the components of the two vectors involved:
\[ \overrightarrow{PQ} \times \overrightarrow{PR} = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -3 & 2 & -5 \ -5 & 0 & 6 \end{array} \right| \]
The evaluated result gives us the vector \( \langle 12, 43, 10 \rangle \). This vector is perpendicular to both \( \overrightarrow{PQ} \) and \( \overrightarrow{PR} \), thus serving as the normal vector for the plane.
Understanding how to compute and apply the cross product is vital for solving geometric problems where perpendicular vectors are required.
Dot Product
The dot product is another fundamental operation involving two vectors. It results in a scalar value and is calculated by multiplying corresponding components of the vectors and then summing these products. The dot product has several uses, including determining the angle between vectors and checking perpendicularity.
In the context of confirming whether a vector is perpendicular to a plane, the dot product provides a simple verification method. If the dot product between a normal vector and any vector in the plane is zero, then these vectors are perpendicular by definition.
For example, in the final step of our problem, we find the dot product of the calculated normal vector \( \langle 12, 43, 10 \rangle \) with \( \overrightarrow{PQ} = \langle -3, 2, -5 \rangle \):
\[ \langle 12, 43, 10 \rangle \cdot \langle -3, 2, -5 \rangle = (12 \cdot -3) + (43 \cdot 2) + (10 \cdot -5) = -36 + 86 - 50 = 0 \]
Since the result is zero, this confirms the perpendicularity and correctness of our calculated normal vector. Understanding and using the dot product is essential for verifying vector directions in three-dimensional space.

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Most popular questions from this chapter

Let a \(=\langle 2,2,2\rangle\) \(\mathbf{b}=\langle- 2,-2,0\rangle,\) and \(\mathbf{r}=\langle x, y, z\rangle\) (a) Show that the vector equation \((\mathbf{r}-\mathbf{a}) \cdot(\mathbf{r}-\mathbf{b})=0\) represents a sphere, by expanding the dot product and simplifying the resulting algebraic equation. (b) Find the center and radius of the sphere. (c) Interpret the result of part (a) geometrically, using the fact that the dot product of two vectors is 0 only if the vectors are perpendicular. endpoints of the vectors \(\mathbf{a}, \mathbf{b},\) and \(\mathbf{r},\) noting that the end. points of a and b are the endpoints of a diameter and the endpoint of \(\mathbf{r} \text { is an arbitrary point on the sphere. }]\) (d) Using your observations from part (a), find a vector equation for the sphere in which the points \((0,1,3)\) and \((2,-1,4)\) form the endpoints of a diameter. Simplify the vector equation to obtain an algebraic equation for the sphere. What are its center and radius?

Let \(\mathbf{u}, \mathbf{v},\) and \(\mathbf{w}\) be vectors, and let \(a\) be a scalar. Prove the given property. $$(\mathbf{u}-\mathbf{v}) \cdot(\mathbf{u}+\mathbf{v})=|\mathbf{u}|^{2}-|\mathbf{v}|^{2}$$

Find the horizontal and vertical components of the vector with given length and direction, and write the vector in terms of the vectors i and j. $$|\mathbf{v}|=1, \quad \theta=225^{\circ}$$

Find \(|\mathbf{u}|,|\mathbf{v}|,|2 \mathbf{u}|,\left|\frac{1}{2} \mathbf{v}\right|,|\mathbf{u}+\mathbf{v}|,|\mathbf{u}-\mathbf{v}|,\) and \(|\mathbf{u}|-|\mathbf{v}|\) $$\mathbf{u}=2 \mathbf{i}+\mathbf{j}, \quad \mathbf{v}=3 \mathbf{i}-2 \mathbf{j}$$

Every line can be described by infinitely many different sets of parametric equations, since any point on the line and any vector parallel to the line can be used to construct the equations. But how can we tell whether two sets of parametric equations represent the same line? Consider the following two sets of parametric equations: Line \(1: \quad x=1-t, \quad y=3 t, \quad z=-6+5 t\) Line \(2: \quad x=-1+2 t, \quad y=6-6 t, \quad z=4-10 t\) (a) Find two points that lie on Line 1 by setting \(t=0\) and \(t=1\) in its parametric equations. Then show that these points also lie on Line 2 by finding two values of the parameter that give these points when substituted into the parametric equations for Line 2 . (b) Show that the following two lines are not the same by finding a point on Line 3 and then showing that it does not lie on Line 4 Line \(3: \quad x=4 t, \quad y=3-6 t, \quad z=-5+2 t\) Line \(4: \quad x=8-2 t, \quad y=-9+3 t, \quad z=6-t\)

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