Find a vector perpendicular ( = normal) to the line 2y = 5-3x in the xy-plane.
the line has slope -3/2
So, what is a vector with slope 2/3 ?
2i-3j
Did you divded by 2?-->(5-3x)/2 -->y=5/2-3x/2-->m=-3/2 right? How does the slope -3/2 become 2/3?
if two lines are perpendicular, their slopes are negative reciprocals.
To find a vector perpendicular to a given line in the xy-plane, we need to determine the slope of the given line and then obtain the negative reciprocal of that slope.
Let's start by rearranging the equation of the given line 2y = 5 - 3x to the standard form: 3x + 2y = 5.
The standard form of a line equation is Ax + By = C, where A, B, and C are constants.
Now, let's examine the coefficient of x and y in the standard form equation: A = 3 and B = 2.
The slope of the line can be calculated using the formula: slope = -A / B.
In this case, the slope of the line is: slope = -3 / 2.
To find a vector perpendicular to this line, we need to determine the negative reciprocal of the slope: -1 / slope.
Substituting the slope value, we get: perpendicular slope = -1 / (-3 / 2).
Simplifying, we find: perpendicular slope = 2 / 3.
This means that the slope of the vector perpendicular to the given line is 2/3.
Since the line is in the xy-plane, the vector perpendicular to the line will be in the form of <a, b, 0>, where a and b are constants.
Now, we can express the perpendicular vector as <2, 3, 0> (where a = 2 and b = 3) since the slope of the vector perpendicular is 2/3.
Hence, the vector <2, 3, 0> is perpendicular to the line 2y = 5 - 3x in the xy-plane.