Problem 49 - Precalculus with Limits

Question

In Exercises 49-52, find the vector \(\mathbf{v}\) with the given magnitude and the same direction as \(\mathbf{u}\). Magnitude - ||\(\mathbf{v}\)|| \(= 10\) Direction - \(\mathbf{u} = \langle -3, 4 \rangle\)

Step-by-Step Solution

Answer

The vector \(\mathbf{v}\) with magnitude 10 and the same direction as \(\mathbf{u} = \langle -3, 4 \rangle\) is \(\mathbf{v} = \langle -6, 8 \rangle\).

Step 1: Understanding the exercise and finding the magnitude of the given vector \(\mathbf{u}\)

The given vector \(\mathbf{u}\) is \(\langle -3,4 \rangle\). To find the magnitude of vector \(\mathbf{u}\), the formula used is \(\|\mathbf{u}\|=\sqrt{u_{1}^{2}+u_{2}^{2}}\). Substituting the values, we get \(\|\mathbf{u}\|=\sqrt{(-3)^2+4^2} = 5\).

Step 2: Finding the direction of given vector \(\mathbf{u}\)

Direction of a vector is \(\mathbf{a}=\mathbf{u}/\|\mathbf{u}\|\). So substituting the values, we get \(\mathbf{a}=\langle -3/5, 4/5 \rangle\). This is the unit vector in the direction of \(\mathbf{u}\).

Step 3: Finding vector \(\mathbf{v}\)

Now that we have found the direction vector (\(\mathbf{a}\)), we can find the vector \(\mathbf{v}\) which has the same direction as \(\mathbf{u}\) and magnitude = 10. This can be found by multiplying the direction vector by the magnitude, i.e. \(\mathbf{v} = \)Magnitude\(\times \mathbf{a}\). So \(\mathbf{v} = 10 \times \langle -3/5, 4/5 \rangle = \langle -6, 8 \rangle.\)

Key Concepts

Magnitude of a Vector

The magnitude of a vector can be thought of as its length. This is similar to the length of a hypotenuse in a right triangle. For a vector \(\mathbf{u} = \langle x, y \rangle\), the magnitude is calculated with the formula: \(|\mathbf{u}| = \sqrt{x^2 + y^2}\). This formula is derived from the Pythagorean theorem. Let's consider our vector \(\mathbf{u} = \langle -3, 4 \rangle\). By substituting these values into the formula, we obtain: \(|\mathbf{u}| = \sqrt{(-3)^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\). Thus, the magnitude of vector \(\mathbf{u}\) is 5.

• Vector components are used to determine direction and magnitude.
• A vector’s magnitude is always non-negative.
• Knowing the magnitude helps to scale vectors accurately.

Direction of a Vector

The direction of a vector expresses where the vector points, similar to a compass pointing north. To find the direction, we often use a unit vector. This is a vector that has the same direction as the original vector but a magnitude of 1. For any vector \(\mathbf{u}\), its direction can be expressed as: \(\mathbf{a} = \frac{\mathbf{u}}{|\mathbf{u}|}\).In our exercise, once the magnitude of \(\mathbf{u} = \langle -3, 4 \rangle\) is found to be 5, the direction becomes: \(\mathbf{a} = \left\langle \frac{-3}{5}, \frac{4}{5} \right\rangle\). This vector shows the direction \(\mathbf{u}\) points in.

• Direction vectors give orientation but no size.
• They are essential in defining the path or alignment of vectors.
• The unit vector ensures consistency in direction calculation.

Unit Vector

A unit vector is a vector that has a length of 1. It points in the same direction as the original vector. Unit vectors are important because they ensure the direction remains unchanged while standardizing the length.To convert a vector \(\mathbf{u}\) into a unit vector, you divide each component by its magnitude. If \(\mathbf{u} = \langle x, y \rangle\), then the unit vector \(\mathbf{a} = \left\langle \frac{x}{|\mathbf{u}|}, \frac{y}{|\mathbf{u}|} \right\rangle\).From our prior calculations with \(\mathbf{u} = \langle -3, 4 \rangle\) and its magnitude of 5, the unit vector is calculated as: \(\mathbf{a} = \left\langle \frac{-3}{5}, \frac{4}{5} \right\rangle\).

• Unit vectors are an essential part of vector normalization.
• They simplify various operations like dot products and rotations.
• They maintain direction while standardizing vector calculations.