Type the correct answer in each box. Round the vector's magnitude to the nearest tenth.

Vector [tex]$u$[/tex] has its initial point at [tex]$(14, -6)$[/tex] and its terminal point at [tex][tex]$(-4, 7)$[/tex][/tex]. Write the component form of [tex]$u$[/tex] and find its magnitude.

[tex]$u = \langle \square, \square \rangle$[/tex], and [tex][tex]$\|u\| \approx \square$[/tex][/tex] units.



Answer :

To find the component form and magnitude of vector [tex]\( u \)[/tex] with its initial point at [tex]\((14, -6)\)[/tex] and terminal point at [tex]\((-4, 7)\)[/tex], we need to perform the following steps:

1. Component Form of Vector [tex]\( u \)[/tex]:

The component form of vector [tex]\( u \)[/tex] can be determined by subtracting the coordinates of the initial point from the coordinates of the terminal point. For a vector with initial point [tex]\((x_1, y_1)\)[/tex] and terminal point [tex]\((x_2, y_2)\)[/tex], the component form [tex]\( u \)[/tex] is given by:
[tex]\[ u = \][/tex]

Here, the initial point is [tex]\((14, -6)\)[/tex] and the terminal point is [tex]\((-4, 7)\)[/tex]. Thus:
[tex]\[ u_x = -4 - 14 = -18 \][/tex]
[tex]\[ u_y = 7 - (-6) = 7 + 6 = 13 \][/tex]

So, the component form of vector [tex]\( u \)[/tex] is:
[tex]\[ u = <-18, 13> \][/tex]

2. Magnitude of Vector [tex]\( u \)[/tex]:

The magnitude (or length) of vector [tex]\( u \)[/tex], denoted by [tex]\( \|u\| \)[/tex], can be calculated using the Pythagorean theorem. For a vector [tex]\( u = \)[/tex], the magnitude is given by:
[tex]\[ \|u\| = \sqrt{u_x^2 + u_y^2} \][/tex]

Using the components [tex]\( u_x = -18 \)[/tex] and [tex]\( u_y = 13 \)[/tex]:
[tex]\[ \|u\| = \sqrt{(-18)^2 + 13^2} = \sqrt{324 + 169} = \sqrt{493} \approx 22.2 \text{ units} \][/tex]

So, the answers are:
[tex]\[ u = <-18, 13>, \quad \text{and} \quad \|u\| \approx 22.2 \text{ units} \][/tex]