Write the scalar equation of the line given the normal vector n and Point P.
n = [3,1], P = (2,4).
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To write the scalar equation of the line, we need the direction vector of the line. The direction vector can be obtained by taking any two non-parallel vectors that are perpendicular to the normal vector.
Let's find the direction vector:
Since n = [3, 1], we can find a vector that is perpendicular to n by swapping the components and negating one of them. Let's choose [-1, 3]:
Direction vector, d = [-1, 3]
Now, we can use the point-slope form of a linear equation:
(x - x₀)/a = (y - y₀)/b
Where (x₀, y₀) is a point on the line, and (a, b) is the direction vector.
Substituting the values:
(x - 2)/(-1) = (y - 4)/3
Multiplying through by -1:
(x - 2) = -3(y - 4)
Expanding:
x - 2 = -3y + 12
Rearranging:
x + 3y = 14
So, the scalar equation of the line is x + 3y = 14.
To write the scalar equation of a line given the normal vector and a point on the line, we can use the following steps:
Step 1: Write the equation in the form of n · (x - P) = 0, where n is the normal vector and P is the given point.
Step 2: Substitute the values of n and P into the equation.
For the given normal vector n = [3,1] and point P = (2,4), the scalar equation of the line will be:
(3,1) · (x - (2,4)) = 0
Step 3: Simplify the equation.
Using the dot product formula, we have:
3(x - 2) + 1(x - 4) = 0
Step 4: Expand and simplify the equation.
3x - 6 + x - 4 = 0
4x - 10 = 0
Step 5: Rearrange the equation to the standard form.
4x = 10
x = 10/4
x = 5/2
Therefore, the scalar equation of the line given the normal vector n = [3,1] and point P = (2,4) is 4x - 10 = 0.