1 i - 2 j slope = -2
our slope = -1/-2 = 1/2
vector direction = 2 i + 1 j
so
r = (1i +0j) + (2 i + j) t
our slope = -1/-2 = 1/2
vector direction = 2 i + 1 j
so
r = (1i +0j) + (2 i + j) t
V1 = 1 i - 2 j
V2 = a i + b j
V1 dot V2 = a -2 b = 0
so no matter what t is, a = 2 b
V2 = 2 i + 1 j or 4 i + 2 j
or in general
V2 = (2 i + 1 j) t
1. Find the direction vector of the given line.
2. Determine the direction vector of the perpendicular line.
3. Use the point-slope form to write the equation of the perpendicular line.
Let's begin step by step.
Step 1: Find the direction vector of the given line.
The given line is represented in vector form as:
r = (2i + 4j) + (i - 2j)t
The coefficient of 't' gives us the direction vector of the line. So, the direction vector of the given line is (1i - 2j).
Step 2: Determine the direction vector of the perpendicular line.
For two lines to be perpendicular, their direction vectors must be perpendicular to each other.
If we take the dot product of the direction vector of the given line with the direction vector of the perpendicular line, the result should be zero. Therefore, the dot product of these two vectors should be zero.
Let's choose a direction vector for the perpendicular line as (a i + b j):
(1i - 2j) • (a i + b j) = 0
(a - 2b) = 0 (using dot product definition)
From this equation, we can see that a = 2b.
Step 3: Use the point-slope form to write the equation of the perpendicular line.
Now we have the direction vector of the perpendicular line, which is (2i + b j). We also have a point through which the line passes, (1,0).
Using the point-slope form, the equation of the perpendicular line is:
r = (1i + 0j) + (2i + b j)t
= (i + 2t i + b t j)
So, the parametric form of the equation for the line perpendicular to r=(2i + 4j) + (i - 2j)t and passes through (1,0) is:
x = 1 + 2t
y = 0 + bt
where 't' is a parameter that can take any real value, and 'b' can be any constant where a = 2b.