Problem 50 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find a unit vector in the direction of the given vector. Verify that the result has a magnitude of 1. $$\mathbf{v}=\langle-2,2\rangle$$

Step-by-Step Solution

Answer

The unit vector in the direction of the vector \(\mathbf{v} = < -2, 2 >\) is \(\mathbf{u} = < -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} >\).

Step 1: Finding the Magnitude of the given vector

We'll begin by finding the magnitude of the given vector \(\mathbf{v} = <-2, 2 >\). The magnitude or norm of a vector \(\mathbf{v} = \) can be found using the formula \(\|\mathbf{v}\| = \sqrt{a^2 + b^2}\). Here, \(a = -2\) and \(b = 2\). So, the magnitude would be \(\sqrt{(-2)^2+(2)^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}\).

Step 2: Finding the Unit Vector

The next step is to find the unit vector in the same direction as the given vector \(\mathbf{v}\). We can achieve this by dividing each component of the vector by its magnitude. That is, the unit vector \(\mathbf{u}\) is given by \(\mathbf{u} = \frac{\mathbf{v}}{\|\mathbf{v}\|}\). Thus, \(\mathbf{u}=<\frac{-2}{2\sqrt{2}},\frac{2}{2\sqrt{2}}>= <-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}>\).

Step 3: Verifying the magnitude of the Unit Vector

The final step is to verify that the magnitude of the unit vector is 1. To do this, we must find the magnitude of vector \(\mathbf{u}\) using the formula given in Step 1. Therefore, the norm of \(\mathbf{u}\) can be calculated as \(\|\mathbf{u}\| = \sqrt{(-1/\sqrt{2})^2 + (1/\sqrt{2})^2}=\sqrt{1/2 + 1/2}= \sqrt{1}=1\). This verifies that the obtained vector is indeed a unit vector.