Find a unit vector in the direction of r + 8s

Answer

To find the unit vector in the direction of $\vec{r} + 8\vec{s}$, we first compute the resultant vector and then normalize it by dividing by its magnitude. The resulting unit vector is $\begin{pmatrix} \frac{21}{\sqrt{457}} \\ 0 \end{pmatrix}$, which simplifies to $\begin{pmatrix} \frac{21}{\sqrt{457}} \\ 0 \end{pmatrix}$.

Explanation

1. Calculate the resultant vector $\vec{v} = \vec{r} + 8\vec{s}$ We perform scalar multiplication and vector addition component-wise: $\vec{v} = \begin{pmatrix} 5 \\ -8 \end{pmatrix} + 8 \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ -8 \end{pmatrix} + \begin{pmatrix} 16 \\ 8 \end{pmatrix} = \begin{pmatrix} 21 \\ 0 \end{pmatrix}$ This step combines the horizontal and vertical components of the two vectors to find the direction of the sum.

2. Calculate the magnitude $\|\vec{v}\|$ The magnitude of a vector $\vec{v} = \begin{pmatrix} x \\ y \end{pmatrix}$ is defined as $\|\vec{v}\| = \sqrt{x^2 + y^2}$. $\|\vec{v}\| = \sqrt{21^2 + 0^2} = \sqrt{441} = 21$ ⚠️ This step is required on exams because the unit vector formula relies explicitly on the length of the vector.

3. Determine the unit vector $\mathbf{\hat{u}}$ The unit vector $\mathbf{\hat{u}}$ is defined as the vector $\vec{v}$ divided by its magnitude: $\mathbf{\hat{u}} = \frac{\vec{v}}{\|\vec{v}\|} = \frac{1}{21} \begin{pmatrix} 21 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ This formula scales the vector to have a length of exactly $1$ while maintaining the original direction.

Final Answer

The unit vector in the same direction as $\vec{r} + 8\vec{s}$ is: $\boxed{\begin{pmatrix} 1 \\ 0 \end{pmatrix}}$

Common Mistakes

• Calculation Errors: Students often fail to distribute the scalar $8$ to both components of vector $\vec{s}$ before adding.
• Normalization Forgetfulness: Some students stop after finding $\vec{r} + 8\vec{s}$ and neglect the final step of dividing by the magnitude to make it a unit vector.