Question

The vector from the origin to point $A$ is given as $(6,-2,-4)$, and the unit vector directed from the origin toward point $B$ is $(2,-2,1)/3$. If points $A$ and $B$ are ten units apart, find the coordinates of point $B$.

Short answer

Answer: The coordinates of point B are ($\frac{8}{5},-\frac{8}{5},\frac{4}{5}$).

Step-by-step solution

Derive the vector from A to O.

The coordinates of point A are (6, -2, -4) and it is mentioned that A is at the origin (0, 0, 0). Therefore, Vector OA is: $OA=(-6,2,4)$

Compute the vector along the line AB.

The unit vector directed from O to B is given as (2, -2, 1) / 3. Therefore, the vector OB can be expressed as: $OB=(2t,-2t,t)$ Where $t$ is the scalar multiple. We know that the distance between A and B is 10 units. Let's represent the vector from A to B as Vector AB: $AB=OB-OA=(2t-(-6),-2t-2,t-4)$ Since we know the distance between A and B which is 10 units, we can apply the distance formula on the AB vector: $\sqrt{(2t+6{)}^2+(-2t-2{)}^2+(t-4{)}^2}=10$ Now, we need to solve the equation to find the value of $t$.

Determine the coordinates of point B.

To find the coordinates of point B, we need to solve the distance equation to find the value of $t$: $(2t+6{)}^2+(-2t-2{)}^2+(t-4{)}^2=100$ Expanding and simplifying the equation, we get: $10t^2-8t-48=0$ Factoring: $(5t-4)(2t+3)=0$ So, t can have two possible values: - $t_1=\frac{4}{5}$ - $t_2=\frac{-3}{2}$ However, considering the given unit vector directed from the origin to point B is $(2,-2,1)/3$, we know that the magnitude of t should be positive. So, we should consider only the positive value of t, $t_1=\frac{4}{5}$. Now, we can find the coordinates of point B by substituting the value of $t_1$ in OB: $OB=(2(\frac{4}{5}),-2(\frac{4}{5}),\frac{4}{5})=(\frac{8}{5},-\frac{8}{5},\frac{4}{5})$ Therefore, the coordinates of point B are $(\frac{8}{5},-\frac{8}{5},\frac{4}{5})$.

Key concepts

Vector Coordinates

Vector coordinates are a way to specify the position of a point in space relative to an origin. In this exercise, we are dealing with three-dimensional vectors. Each vector is described by three numerical coordinates, which can be thought of as steps along the x, y, and z axes respectively. For example, the vector from the origin to point A is given as $(6,-2,-4)$. This tells us that point A is 6 units along the x-axis, -2 units along the y-axis, and -4 units along the z-axis from the origin.

Understanding vector coordinates is essential as they allow us to easily visualize and calculate distances and directions in space, which forms the foundation for solving vector calculus problems.

Unit Vector

A unit vector is a vector that has a magnitude of one. It is used to specify direction without regard to magnitude. In vector calculus, unit vectors are often employed to express directions in a standardized way. In this problem, the unit vector directed from the origin to point B is given as $(\frac{2}{3},-\frac{2}{3},\frac{1}{3})$.

• It helps define the direction from the origin toward point B.
  
• It's calculated by dividing each component of the vector by the vector's magnitude, ensuring the vector’s length remains one.
  

Using unit vectors helps simplify the process of scaling vectors when calculating positions, as it reduces complexity by removing length constraints.

Distance Formula

The distance formula is pivotal in finding the distance between two points in space based on their coordinates. It is derived from the Pythagorean theorem and, in three dimensions, it is expressed as:
$$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$$

This exercise uses the distance formula to express the known distance of 10 units between points A and B. By substituting the components of vector AB into the formula, we establish a relationship that leads to finding t, the scalar multiplier for the vector OB. Once we know the distance and direction, we can find unknown coordinates, like those of point B in this problem.

Scalar Multiplication

Scalar multiplication involves multiplying a vector by a scalar (which is a real number) to adjust its magnitude without changing its direction. In this exercise, scalar multiplication is used to scale the unit vector OB by a scalar 't', which helps in reaching the proper length or magnitude that matches the distance between points A and B.

• When the unit vector $(\frac{2}{3},-\frac{2}{3},\frac{1}{3})$ is multiplied by $t$, each component of the vector is multiplied by $t$.
  
• This changes the vector's magnitude from one to $t$, defining the position of point B at 10 units from point A.
  
• Effectively, scalar multiplication allows utilization of the unit vector's pure directionality to calculate the specific point in space.
  

This technique makes it seamless to adjust vectors for practical physical applications, such as defining paths or distances in space.