Question

Write the name of a vector with a magnitude of 13 in algebraic form that is perpendicular to the line 7x + 2y =6. Please explain step by step.

Answer 1

Answer:

The vector is < 91 / sqrt(53),26 / sqrt(53)>  or <-91 / sqrt(53),-26 / sqrt(53)>

Step-by-step explanation:

7x + 2y =6

We need to find the slope so we solve for y

Subtract 7x from each side

7x -7x+ 2y =-7x+6

2y = -7x +6

Divide by 2

2y/2 = -7x /2 + 6/2

y = -7/2 x +3

The slope is -7/2

We want it perpendicular so the slope must be the negative reciprocal

m = -(-2/7)

   = 2/7

Vectors are written in the form  

v = <  x, y>

slope = change in y over change in x

So y/x = 2/7  or 7y =2x

Now we worry about the magnitude

Magnitude is sqrt(x^2 + y^2) and it equals 13

sqrt(x^2 + y^2)= 13

square both sides

x^2 + y^2 = 169

We have 2 equations and 2 unknowns

y = (2/7x)

x^2 +(2/7x) ^2 = 169

x^2 +4/49 x^2 =169

49/49 x^2 +4/49 x^2 = 169

53/49  x^2 = 169

x^2 = 169 *49/53

Take the square root of each side

x = ±91 / sqrt(53)

y =  (2/7x)

   = 2/7 * ±91 / sqrt(53)

   = ±26 / sqrt(53)

The vector is < 91 / sqrt(53),26 / sqrt(53)>  or <-91 / sqrt(53),-26 / sqrt(53)>