Determine the value of k that makes the line (x+2)/4=(y+1)/5=(z-3)/3 and r=(1,3,6 +s(-2k,2,k), SeR, perpendicular.
To determine the value of k that makes the line perpendicular to the given line, we need to find the direction vector of the given line and equate it to the direction vector of the perpendicular line.
Given line: (x+2)/4 = (y+1)/5 = (z-3)/3
The direction vector of the given line can be found by taking the coefficients of x, y, and z in the equation and setting them as the components of the direction vector.
Direction vector of given line, say d₁: [4, 5, 3]
Now, we want to find the direction vector of the perpendicular line, say d₂. In order for the two lines to be perpendicular, their direction vectors must be orthogonal, meaning their dot product must be zero.
The dot product of d₁ and d₂ should be zero:
d₁ · d₂ = 4d₂₁ + 5d₂₂ + 3d₂₃ = 0
We can substitute the components of the direction vector d₂ as [a, b, c] and solve for them.
4a + 5b + 3c = 0
To find a particular solution for d₂, we set one of the variables (a, b, or c) as a constant. Let's choose a = 1.
4(1) + 5b + 3c = 0
4 + 5b + 3c = 0
Now, we can solve the system of equations by choosing a value for b and subsequently finding c.
Let's choose b = 0:
4 + 5(0) + 3c = 0
4 + 3c = 0
3c = -4
c = -4/3
So, a particular solution for d₂ is [1, 0, -4/3].
Now, we have the direction vector of the perpendicular line, which is [1, 0, -4/3]. We can represent this as (x, y, z) = (1, 3, 6) + t[1, 0, -4/3], where t is a scalar.
To find k, we can equate the corresponding components of the direction vector of the given line and the perpendicular line:
-2k = 1
2 = 0
k = -1/2
Therefore, the value of k that makes the line (x+2)/4=(y+1)/5=(z-3)/3 and r=(1,3,6 +s(-2k,2,k) perpendicular is k = -1/2.