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Find the vector pointing in the direction opposite that of \langle 6,-8\rangle with length 20

Short Answer

Expert verified
Answer: The vector pointing in the direction opposite that of the given vector ⟨6, -8⟩, with a length of 20 is ⟨-12, 16⟩.

Step by step solution

01

Find the negative of the given vector

The given vector is \langle 6, -8 \rangle. To find the vector pointing in the opposite direction, we simply change the sign of each component: \langle -6, 8 \rangle.
02

Find the magnitude of the new vector

Now, we need to find the magnitude (length) of the new vector \langle -6, 8 \rangle. We use the formula for the magnitude of a vector: Magnitude = \sqrt{x^2 + y^2} In our case, x = -6 and y = 8, so the magnitude of the vector is: Magnitude = \sqrt{(-6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10
03

Calculate the scalar to scale the vector to length 20

We want the final vector to have a length of 20. We will achieve this by multiplying the current vector (which has a length of 10) by a scalar. To find the scalar, we simply divide the desired length (20) by the current length (10): Scalar = Desired length / Current length = 20 / 10 = 2
04

Scale the vector to length 20

Now that we have calculated the scalar (2), we can multiply the current vector \langle -6, 8 \rangle by this scalar to get the final vector with length 20: Final vector = Scalar × Current vector = 2 × \langle -6, 8 \rangle = \langle -12, 16 \rangle The vector pointing in the direction opposite that of \langle 6, -8 \rangle with length 20 is \langle -12, 16 \rangle.

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Key Concepts

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

Negative Vector
A negative vector is simply the vector that has the same magnitude as the original but points in the exact opposite direction. Imagine an arrow flying north; its negative would travel south with the same speed and straightness. In our original problem, the negative of the vector \(\langle 6, -8 \rangle\) is calculated by reversing the sign of each component, resulting in \(\langle -6, 8 \rangle\). The concept of negative vectors is vital in physics and engineering to denote forces or movements in opposite directions,

When dealing with negative vectors, a common mistake is to misunderstand their representation. They are not 'less' or 'without value'; they have a direction and magnitude, just like positive vectors – only in the reverse direction. It's useful to think of these vectors as arrows on a graph; flipping an arrow to point in the opposite direction creates the 'negative' version of that arrow. This operation is essential in vector subtraction where to subtract is to add the negative.
Magnitude of a Vector
The magnitude of a vector represents its length, or how far it stretches from its origin to its endpoint. We often refer to this concept as the vector's 'size' or 'strength'. For the vector \(\langle -6, 8 \rangle\), the magnitude is calculated using the formula \( \sqrt{x^2 + y^2} \), where \(x\) and \(y\) are the vector's horizontal and vertical components, respectively.

In our example, we find the magnitude to be \( \sqrt{(-6)^2 + (8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \). Understanding magnitude is fundamental as it helps measure quantities like velocity, force, and displacement in a scalar (non-directional) sense. Without knowing a vector's magnitude, we cannot fully understand its potential impact or use it accurately in calculations.
Scalar Multiplication
Scalar multiplication involves scaling a vector by a real number, called a scalar. This process changes the length of the vector but not its direction. To scale our vector \(\langle -6, 8 \rangle\) to a new length of 20, we multiply each component by a scalar. If the original length is 10, as we have calculated, then the required scalar is the ratio of the new length to the original length, which is \(20 / 10 = 2\).

  • To stretch the vector, we use a scalar greater than one.
  • To shrink it, we use a scalar between zero and one.
  • Applying a negative scalar flips the vector's direction as well as scales it.
For our vector, the scalar operation is \(2 \times \langle -6, 8 \rangle = \langle -12, 16 \rangle\), giving us a new vector with the same direction as \(\langle -6, 8 \rangle\), but twice the magnitude. Scalar multiplication is widely used to adjust forces, velocities, or other vector quantities in physics and engineering.
Vector Components
Vector components are the building blocks of a vector, representing its effect along the horizontal (x) and vertical (y) axes. A vector in two dimensions can be thought of as an instruction to move right or left a certain amount (the x-component) and up or down a certain amount (the y-component). For instance, the original vector in our problem, \(\langle 6, -8 \rangle\), suggests moving 6 units to the right and 8 units down.

Understanding vector components is critical as they let us analyze and manipulate vectors in terms of familiar axes, often simplifying complex movement or force problems into manageable parts. In many applications, breaking down vectors into their components is the first step in solving the problem at hand.

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Most popular questions from this chapter

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. José travels from point \(A\) to point \(B\) in the plane by following vector \(\mathbf{u},\) then vector \(\mathbf{v},\) and then vector \(\mathbf{w} .\) If he starts at \(A\) and follows \(\mathbf{w},\) then \(\mathbf{v},\) and then \(\mathbf{u},\) he still arrives at \(B\) b. Maria travels from \(A\) to \(B\) in the plane by following the vector u. By following \(-\mathbf{u},\) she returns from \(B\) to \(A\) c. \(|\mathbf{u}+\mathbf{v}| \geq|\mathbf{u}|,\) for all vectors \(\mathbf{u}\) and \(\mathbf{v}\) d. \(|\mathbf{u}+\mathbf{v}| \geq|\mathbf{u}|+|\mathbf{v}|,\) for all vectors \(\mathbf{u}\) and \(\mathbf{v}\) e. Parallel vectors have the same length. f. If \( {A B}= {C D},\) then \(A=C\) and \(B=D\) g. If \(\mathbf{u}\) and \(\mathbf{v}\) are perpendicular, then \(|\mathbf{u}+\mathbf{v}|=|\mathbf{u}|+|\mathbf{v}|\) h. If \(\mathbf{u}\) and \(\mathbf{v}\) are parallel and have the same direction, then \(|\mathbf{u}+\mathbf{v}|=|\mathbf{u}|+|\mathbf{v}|\)

Airplanes and crosswinds Assume each plane flies horizontally in a crosswind that blows horizontally. An airplane flies east to west at 320 mi/hr relative to the air in a crosswind that blows at 40 mi/hr toward the southwest (45° south of west). a. Find the velocity of the plane relative to the air \(v_{a},\) the velocity of the crosswind \(\mathbf{w},\) and the velocity of the plane relative to the ground \(\mathbf{v}_{s}\) b. Find the ground speed and heading of the plane relative to the ground.

Prove that \(|c \mathbf{v}|=|c||\mathbf{v}|,\) where \(c\) is a scalar and \(\mathbf{v}\) is a vector.

Parallel vectors of varying lengths Find vectors parallel to \(\mathbf{v}\) of the given length. $$\mathbf{v}=P Q \text { with } P(3,4,0) \text { and } Q(2,3,1) ; \text { length }=3$$

Linear combinations A sum of scalar multiples of two or more vectors (such as \(c_{1} \mathbf{u}+c_{2} \mathbf{v}+c_{3} \mathbf{w},\) where \(c_{i}\) are scalars) is called a linear combination of the vectors. Let \(\mathbf{i}=\langle 1,0\rangle, \mathbf{j}=\langle 0,1\rangle\) \(\mathbf{u}=\langle 1,1\rangle,\) and \(\mathbf{v}=\langle-1,1\rangle\) For arbitrary real numbers \(a\) and \(b\), express \(\langle a, b\rangle\) as a linear combination of \(\mathbf{u}\) and \(\mathbf{v}\)

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