To solve this problem, let’s go through it step-by-step.

1. Finding the Length of Vector a

The vector a is given as: $\mathbf{a} = \begin{bmatrix} 6 \\ 3 \\ 1 \end{bmatrix}$ The length (or magnitude) of a vector a with components $a_x$, $a_y$, and $a_z$ is calculated using the formula: $\|\mathbf{a}\| = \sqrt{a_x^2 + a_y^2 + a_z^2}$ For this vector: $\|\mathbf{a}\| = \sqrt{6^2 + 3^2 + 1^2} = \sqrt{36 + 9 + 1} = \sqrt{46}$ Thus, the length of a is: $\|\mathbf{a}\| = \sqrt{46} \approx 6.78$

2. Finding the Unit Vector in the Same Direction as a

The unit vector in the direction of a is obtained by dividing each component of a by the magnitude $\|\mathbf{a}\|$. The unit vector u in the direction of a is given by: $\mathbf{u} = \frac{1}{\|\mathbf{a}\|} \mathbf{a} = \frac{1}{\sqrt{46}} \begin{bmatrix} 6 \\ 3 \\ 1 \end{bmatrix}$ Breaking this down, each component of the unit vector is: $u_x = \frac{6}{\sqrt{46}}, \quad u_y = \frac{3}{\sqrt{46}}, \quad u_z = \frac{1}{\sqrt{46}}$ If we approximate these values: $u_x \approx 0.884, \quad u_y \approx 0.442, \quad u_z \approx 0.147$ Thus, the unit vector u is: $\mathbf{u} \approx \begin{bmatrix} 0.884 \\ 0.442 \\ 0.147 \end{bmatrix}$

Summary of Answers

• Length of $\mathbf{a}$: $\sqrt{46} \approx 6.78$
• Unit Vector in the Direction of $\mathbf{a}$: $\begin{bmatrix} 0.884 \\ 0.442 \\ 0.147 \end{bmatrix}$

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Related Questions

1. How do you find the magnitude of a 2-dimensional vector?
2. What is the purpose of finding a unit vector?
3. Can a unit vector have a negative component?
4. How is vector addition performed?
5. What are the applications of vectors in physics?

Tip

Always double-check your square root calculations, as rounding errors can impact the precision of your answers, especially when finding unit vectors.