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Find the vector \(\mathbf{v}\) with the given magnitude and the same direction as \(\mathrm{u}\). $$ \|\mathbf{v}\|=4 \quad \mathbf{u}=\langle-1,1\rangle $$

Short Answer

Expert verified
The vector \(\mathbf{v}\) is \(\left\langle-2\sqrt{2},2\sqrt{2}\right\rangle\).

Step by step solution

01

Calculate the Magnitude of Vector \(\mathbf{u}\)

The first step is to calculate the magnitude or size of the vector \(\mathbf{u}\). For a vector \(\mathbf{u}=\langle u_1, u_2 \rangle\), the magnitude, denoted as \(\|\mathbf{u}\|\), can be calculated using the formula: \(\|\mathbf{u}\| = \sqrt{u_1^2 + u_2^2}\). So for \(\mathbf{u}=\langle-1,1\rangle\), \(\|\mathbf{u}\| = \sqrt{(-1)^2 + 1^2} = \sqrt{2}\).
02

Determine the Unit Vector in the Direction of Vector \(\mathbf{u}\)

Next, calculate the unit vector in the direction of \(\mathbf{u}\) by dividing each component in \(\mathbf{u}\) by the magnitude of \(\mathbf{u}\). The unit vector \(\mathbf{u}_{unit}\) can be calculated as \(\mathbf{u}_{unit} = \frac{\mathbf{u}}{\|\mathbf{u}\|}\). For our vector, \(\mathbf{u}_{unit} = \frac{\langle-1,1\rangle}{\sqrt{2}}= \left\langle-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right\rangle\).
03

Find the Required Vector \(\mathbf{v}\)

Lastly, to find the vector \(\mathbf{v}\) with the same direction as \(\mathbf{u}\) and the given magnitude 4 we multiply the unit vector by the given magnitude resulting in: \(\mathbf{v} = \|\mathbf{v}\|\cdot\mathbf{u}_{unit}\)Substituting the known values, \(\mathbf{v} = 4\cdot\left\langle-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right\rangle = \left\langle-2\sqrt{2},2\sqrt{2}\right\rangle\).

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Key Concepts

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

Vector Magnitude
In vector calculus, magnitude is a crucial concept. Think of it as the "length" or "size" of a vector. If you envision a vector in a 2-dimensional space as an arrow, the magnitude gives you the length of that arrow. For any vector \( \mathbf{u} = \langle u_1, u_2 \rangle \), the formula to calculate its magnitude is \( \|\mathbf{u}\| = \sqrt{u_1^2 + u_2^2} \).

Using this formula helps you to determine how long the vector is no matter its direction. For example, with the vector \( \mathbf{u} = \langle -1, 1 \rangle \), the magnitude is \( \sqrt{(-1)^2 + 1^2} = \sqrt{2} \).
Understanding the magnitude is important as it helps build the foundation for other vector operations like finding the unit vector or scaling vectors.
Unit Vector
A unit vector points in the same direction as a given vector but has a magnitude of 1. This standardized length makes unit vectors very useful for indicating direction without considering the vector's actual size.

To convert any vector \( \mathbf{u} \) into a unit vector \( \mathbf{u}_{unit} \), you divide each of its components by its magnitude: \( \mathbf{u}_{unit} = \frac{\mathbf{u}}{\|\mathbf{u}\|} \).
  • It keeps the direction unaltered.
  • It scales down the vector into a length of 1.
For instance, for the vector \( \mathbf{u} = \langle -1, 1 \rangle \), the unit vector would be \( \left\langle -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle \). This conversion is essential, especially when you need a directional component separate from magnitude.
Vector Direction
The direction of a vector is another fundamental aspect that defines its orientation in space. While magnitude tells us how long a vector is, the direction tells us where it points.

Direction is maintained when computing the unit vector of a given vector. Even after normalizing a vector to obtain its unit vector, the direction component remains unchanged.
For example, although \( \mathbf{u} = \langle -1, 1 \rangle \) becomes \( \mathbf{u}_{unit} = \left\langle -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle \), this unit vector still points in the same direction as \( \mathbf{u} \). This consistency is key when performing operations like scaling, which maintains the vector's unchanged direction.
Vector Scaling
Vector scaling refers to the process of adjusting the size or magnitude of a vector while keeping its direction consistent. It's like stretching or shrinking the vector's length without altering its path.

To scale a vector \( \mathbf{u} \), you multiply its unit vector \( \mathbf{u}_{unit} \) by the desired magnitude. This process ensures that the new vector \( \mathbf{v} \) maintains the same direction.
  • This operation is useful when you need vectors of a specific length, but their direction is predetermined by another vector.
  • The computation involves a multiplication step that rescales the vector components appropriately.
For instance, to find a vector \( \mathbf{v} \) with a magnitude of 4 in the same direction as \( \mathbf{u} \), we calculate \( \mathbf{v} = 4 \cdot \mathbf{u}_{unit} \). This gives you \( \left\langle -2\sqrt{2}, 2\sqrt{2} \right\rangle \), a scaled version of the unit vector, yet aligned in the same direction.

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