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2 years ago

Vector v is shown in the graph.

Mathematics

1 answer:

ololo11 [35]2 years ago

8 0

The graph of vector v with an initial point at (2, 6) and terminal point at

(5, -3). Thus, v = 3i - 9j represents the linear form of v.

<h3>How to find the vector component?</h3>

If we have given a vector v of initial point A and terminal point B

v = ai + bj

then the components form as;

AB = xi + yj

Here, xi and yj are the components of the vector.

Given;

The graph of vector v with an initial point at (2, 6) and terminal point at

(5, -3).

v = (5 - 2)i + (-3 - 6)j

v = (3)i + (-9)j

v = 3i - 9j

Thus, v = 3i - 9j represents the linear form of v.

Learn more about vectors;

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How do you write 3,500 in scientific notation

kolbaska11 [484]

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3 years ago

What is the equation of the line perpendicular to 3x+y= -8that passes through -3,1? Write your answer in slope-intercept form. S

Gekata [30.6K]

Slope intercept form of a line perpendicular to 3x + y = -8, and passing through (-3,1) is $y=\frac{1}{3} x+2$

<u>Solution:</u>

Need to write equation of line perpendicular to 3x+y = -8 and passes through the point (-3,1).

Generic slope intercept form of a line is given by y = mx + c

where m = slope of the line.

Let's first find slope intercept form of 3x + y = -8

3x + y = -8

=> y = -3x - 8

On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c , we can say that for line 3x + y = -8 , slope m = -3

And as the line passing through (-3,1) and is perpendicular to 3x + y = -8, product of slopes of two line will be -1 as lies are perpendicular.

Let required slope = x

$\begin{array}{l}{=x \times-3=-1} \\\\ {=>x=\frac{-1}{-3}=\frac{1}{3}}\end{array}$

So we need to find the equation of a line whose slope is $\frac{1}{3}$ and passing through (-3,1)

Equation of line passing through $(x_1 , y_1)$ and having lope of m is given by

$\left(y-y_{1}\right)=\mathrm{m}\left(x-x_{1}\right)$

$\text { In our case } x_{1}=-3 \text { and } y_{1}=1 \text { and } \mathrm{m}=\frac{1}{3}$

Substituting the values we get,

$\begin{array}{l}{(\mathrm{y}-1)=\frac{1}{3}(\mathrm{x}-(-3))} \\\\ {=>\mathrm{y}-1=\frac{1}{3} \mathrm{x}+1} \\\\ {=>\mathrm{y}=\frac{1}{3} \mathrm{x}+2}\end{array}$