Question

A vector field is specified as $\mathbf{G}=24xy{\mathbf{a}}_x+12(x^2+2){\mathbf{a}}_y+18z^2{\mathbf{a}}_z.$ Given two points, $P(1,2,-1)$ and $Q(-2,1,3)$, find $(a)\mathbf{G}$ at $P;(b)$ a unit vector in the direction of $\mathbf{G}$ at $Q;(c)$ a unit vector directed from $Q$ toward $P;(d)$ the equation of the surface on which $|\mathbf{G}|=60$.

Short answer

(b) Find a unit vector in the direction of the vector field $\mathbf{G}$ at point $Q(-2,1,3)$. (c) What is a unit vector directed from point $Q$ towards point $P$? (d) Determine the equation of the surface on which the magnitude of the vector field $\mathbf{G}$ is equal to 60.

Step-by-step solution

(a) Find $\mathbf{G}$ at point $P$(1,2,-1)

To find the value of $\mathbf{G}$ at point $P$, substitute the coordinates of point $P$ into the expressions for $\mathbf{G}$: $\mathbf{G}(x,y,z)=24xy{\mathbf{a}}_x+12(x^2+2){\mathbf{a}}_y+18z^2{\mathbf{a}}_z$. So, $\mathbf{G}(1,2,-1)=24(1)(2){\mathbf{a}}_x+12((1{)}^2+2){\mathbf{a}}_y+18(-1{)}^2{\mathbf{a}}_z=48{\mathbf{a}}_x+36{\mathbf{a}}_y+18{\mathbf{a}}_z$.

(b) Find a unit vector in the direction of $\mathbf{G}$ at point $Q$(-2,1,3)

First, we need to find the value of $\mathbf{G}$ at point $Q$. Substitute the coordinates of point $Q$ into the expressions for $\mathbf{G}$: $\mathbf{G}(-2,1,3)=24(-2)(1){\mathbf{a}}_x+12((-2{)}^2+2){\mathbf{a}}_y+18(3{)}^2{\mathbf{a}}_z=-48{\mathbf{a}}_x+60{\mathbf{a}}_y+162{\mathbf{a}}_z$. Now we will find the magnitude of $\mathbf{G}$ at point $Q$: $|\mathbf{G}|=\sqrt{(-48{)}^2+{60}^2+{162}^2}=6\sqrt{170}$. A unit vector in the direction of $\mathbf{G}$ at point $Q$: ${\hat{\mathbf{G}}}_Q=\frac{\mathbf{G}}{|\mathbf{G}|}=\frac{1}{6\sqrt{170}}(-48{\mathbf{a}}_x+60{\mathbf{a}}_y+162{\mathbf{a}}_z)$.

(c) Find a unit vector directed from $Q$ towards $P$

First, we need to find the vector from point $Q$ to point $P$: ${\mathbf{V}}_{QP}=(1-(-2)){\mathbf{a}}_x+(2-1){\mathbf{a}}_y+(-1-3){\mathbf{a}}_z=3{\mathbf{a}}_x+{\mathbf{a}}_y-4{\mathbf{a}}_z$. Now we will find the magnitude of this vector: $|{\mathbf{V}}_{QP}|=\sqrt{(3{)}^2+(1{)}^2+(-4{)}^2}=\sqrt{26}$. A unit vector directed from $Q$ towards $P$: ${\hat{\mathbf{V}}}_{QP}=\frac{{\mathbf{V}}_{QP}}{|{\mathbf{V}}_{QP}|}=\frac{1}{\sqrt{26}}(3{\mathbf{a}}_x+{\mathbf{a}}_y-4{\mathbf{a}}_z)$.

(d) Find the equation of the surface on which $|\mathbf{G}|=60$

First, we find the magnitude of $\mathbf{G}$ using the given expression for $\mathbf{G}$: $|\mathbf{G}|=\sqrt{(24xy{)}^2+(12(x^2+2){)}^2+(18z^2{)}^2}$. Now we will equate the magnitude of $\mathbf{G}$ to 60: $60=\sqrt{(24xy{)}^2+(12(x^2+2){)}^2+(18z^2{)}^2}$. Square the equation to get rid of the square root: $3600=(24xy{)}^2+(12(x^2+2){)}^2+(18z^2{)}^2$. This is the equation of the surface on which the magnitude of $\mathbf{G}$ is equal to 60.

Key concepts

Magnitude of a Vector

The magnitude of a vector is a measure of its length in space. It can be likened to the concept of distance. When you are given a vector, say $\mathbf{v}=a\mathbf{i}+b\mathbf{j}+c\mathbf{k}$, the magnitude is calculated using the formula:$$\Vert \mathbf{v}\Vert =\sqrt{a^2+b^2+c^2}$$This involves taking the square root of the sum of the squares of its components. In vector field analysis, calculating the magnitude helps us understand the vector's strength at a given point. It's like determining the overall force or speed of an entity at that point in space. By computing the magnitude $|\mathbf{G}|$ at different points, we can assess how the vector field influences those regions.

Unit Vector Calculation

A unit vector is a vector with a magnitude of one. It is often used to describe direction without concern for magnitude. To convert any vector into a unit vector, you divide it by its magnitude. For a vector $\mathbf{a}=xi+yj+zk$, the unit vector is calculated as follows:$$\hat{\mathbf{a}}=\frac{\mathbf{a}}{\Vert \mathbf{a}\Vert }=\frac{xi+yj+zk}{\sqrt{x^2+y^2+z^2}}$$

• Why Calculate a Unit Vector? - It simplifies the direction description of vectors. When considering direction separately from magnitude, unit vectors provide a standard method to represent directional components.
• Application: - In our exercise, unit vectors help us understand the direction of the vector field $\mathbf{G}$ at certain points $P$ and $Q$. It's crucial for understanding how these directions compare or interact in a given space.

Equation of a Surface

The equation of a surface describes the set of points that make up that surface. In vector field analysis, surfaces can often be defined by equating the magnitude of a vector to a constant value. When we have a vector $\mathbf{G}$ and we know $|\mathbf{G}|=C$, equating the vector's magnitude to a constant form can describe a surface.Consider our example with the vector $\mathbf{G}$, where we need to find the surface for which $|\mathbf{G}|=60$. This involves solving the expression for $|\mathbf{G}|$ and setting it equal to 60. It forms a quadratic equation of multiple variables ($x$, $y$, and $z$), which visually represents a 3D surface encompassing all points in space where this condition holds true. Solving this reveals not just the shape but also how the vector field value changes across different spatial coordinates.

Vector Direction

Determining vector direction is foundational in understanding how vectors interact and move within a space. It does not involve the magnitude but focuses purely on the path or axis along which a vector would extend.

• Relation to Unit Vectors: - Direction is often simplified to unit vectors since they ignore magnitude and stress purpose and orientation.
• Finding Direction: - In the exercise, finding a vector directed from one point ($Q$) towards another ($P$) involved calculating the difference between their coordinates. This resultant vector ${\mathbf{V}}_{QP}$ then needs to be converted into a unit vector to express its direction.

This helps, for example, when modeling physical phenomena like wind direction or water current in fluid dynamics, as understanding direction at different field points is crucial for predicting behavior and interactions.