Answer :
Let's find the unit vector in the direction of the vector n. Given a vector n, the unit vector in the direction of n is a vector that has the same direction as n but a magnitude of 1.
Here's a step-by-step solution:
### Step 1: Represent the Vector
The vector given is [tex]\( n = 21 + 3 \)[/tex]. When expressed in component form, it can be written as:
[tex]\[ \mathbf{n} = \begin{bmatrix} 21 \\ 3 \end{bmatrix} \][/tex]
### Step 2: Calculate Magnitude of the Vector
The magnitude [tex]\(\|\mathbf{n}\|\)[/tex] of the vector [tex]\(\mathbf{n}\)[/tex] can be calculated using the Euclidean norm formula:
[tex]\[ \|\mathbf{n}\| = \sqrt{x^2 + y^2} \][/tex]
where [tex]\(x = 21\)[/tex] and [tex]\(y = 3\)[/tex].
[tex]\[ \|\mathbf{n}\| = \sqrt{21^2 + 3^2} = \sqrt{441 + 9} = \sqrt{450} \][/tex]
[tex]\[ \|\mathbf{n}\| = \sqrt{450} \approx 21.213 \][/tex]
### Step 3: Calculate the Unit Vector
The unit vector [tex]\(\hat{\mathbf{n}}\)[/tex] in the direction of [tex]\(\mathbf{n}\)[/tex] is obtained by dividing each component of [tex]\(\mathbf{n}\)[/tex] by the magnitude [tex]\(\|\mathbf{n}\|\)[/tex]:
[tex]\[ \hat{\mathbf{n}} = \frac{\mathbf{n}}{\|\mathbf{n}\|} = \begin{bmatrix} \frac{21}{\sqrt{450}} \\ \frac{3}{\sqrt{450}} \end{bmatrix} \][/tex]
This simplifies to:
[tex]\[ \hat{\mathbf{n}} = \begin{bmatrix} \frac{21}{21.213} \\ \frac{3}{21.213} \end{bmatrix} \approx \begin{bmatrix} 0.990 \\ 0.141 \end{bmatrix} \][/tex]
So, the unit vector in the direction of the vector [tex]\(\mathbf{n} = 21 + 3\)[/tex] is approximately:
[tex]\[ \hat{\mathbf{n}} \approx \begin{bmatrix} 0.990 \\ 0.141 \end{bmatrix} \][/tex]
### Summary
- The magnitude of the vector [tex]\(\mathbf{n}\)[/tex] is approximately 21.213.
- The unit vector in the direction of [tex]\(\mathbf{n}\)[/tex] is approximately [tex]\(\begin{bmatrix} 0.990 \\ 0.141 \end{bmatrix}\)[/tex].
Here's a step-by-step solution:
### Step 1: Represent the Vector
The vector given is [tex]\( n = 21 + 3 \)[/tex]. When expressed in component form, it can be written as:
[tex]\[ \mathbf{n} = \begin{bmatrix} 21 \\ 3 \end{bmatrix} \][/tex]
### Step 2: Calculate Magnitude of the Vector
The magnitude [tex]\(\|\mathbf{n}\|\)[/tex] of the vector [tex]\(\mathbf{n}\)[/tex] can be calculated using the Euclidean norm formula:
[tex]\[ \|\mathbf{n}\| = \sqrt{x^2 + y^2} \][/tex]
where [tex]\(x = 21\)[/tex] and [tex]\(y = 3\)[/tex].
[tex]\[ \|\mathbf{n}\| = \sqrt{21^2 + 3^2} = \sqrt{441 + 9} = \sqrt{450} \][/tex]
[tex]\[ \|\mathbf{n}\| = \sqrt{450} \approx 21.213 \][/tex]
### Step 3: Calculate the Unit Vector
The unit vector [tex]\(\hat{\mathbf{n}}\)[/tex] in the direction of [tex]\(\mathbf{n}\)[/tex] is obtained by dividing each component of [tex]\(\mathbf{n}\)[/tex] by the magnitude [tex]\(\|\mathbf{n}\|\)[/tex]:
[tex]\[ \hat{\mathbf{n}} = \frac{\mathbf{n}}{\|\mathbf{n}\|} = \begin{bmatrix} \frac{21}{\sqrt{450}} \\ \frac{3}{\sqrt{450}} \end{bmatrix} \][/tex]
This simplifies to:
[tex]\[ \hat{\mathbf{n}} = \begin{bmatrix} \frac{21}{21.213} \\ \frac{3}{21.213} \end{bmatrix} \approx \begin{bmatrix} 0.990 \\ 0.141 \end{bmatrix} \][/tex]
So, the unit vector in the direction of the vector [tex]\(\mathbf{n} = 21 + 3\)[/tex] is approximately:
[tex]\[ \hat{\mathbf{n}} \approx \begin{bmatrix} 0.990 \\ 0.141 \end{bmatrix} \][/tex]
### Summary
- The magnitude of the vector [tex]\(\mathbf{n}\)[/tex] is approximately 21.213.
- The unit vector in the direction of [tex]\(\mathbf{n}\)[/tex] is approximately [tex]\(\begin{bmatrix} 0.990 \\ 0.141 \end{bmatrix}\)[/tex].
