Find the value of k so that the lines 3x + ky + z - 6 = 0 and 6x + (1-k)y + 2z - 9 = 0 are perpendicular.
Two lines are perpendicular if the dot product of their direction vectors is equal to zero.
The direction vector of the line 3x + ky + z - 6 = 0 is <3, k, 1>.
The direction vector of the line 6x + (1-k)y + 2z - 9 = 0 is <6, 1-k, 2>.
Taking the dot product of these two vectors, we have:
(3)(6) + (k)(1-k) + (1)(2) = 0
18 + k - k^2 + 2 = 0
k^2 - k - 20 = 0
(k - 5)(k + 4) = 0
Solving for k, we have k = 5 or k = -4.
Therefore, the values of k that make the lines perpendicular are k = 5 and k = -4.
To find the value of k so that the lines are perpendicular, we need to compare the directional vectors of the lines. The directional vector of a line can be obtained by taking the coefficients of x, y, and z in the equation of the line.
For the first line, the directional vector is <3, k, 1>.
For the second line, the directional vector is <6, 1-k, 2>.
For two lines to be perpendicular, the dot product of their directional vectors should be zero. So, we can find the dot product of the directional vectors and set it equal to zero:
(3)(6) + (k)(1-k) + (1)(2) = 0
18 + k - k^2 + 2 = 0
-k^2 + k + 20 = 0
Next, we can solve this quadratic equation for k. Let's use the quadratic formula:
k = (-1 ± √(1^2 - 4(-1)(20))) / (2*(-1))
k = (-1 ± √(1 + 80)) / (-2)
k = (-1 ± √81) / (-2)
k = (-1 ± 9) / (-2)
Simplifying further gives two possible values for k:
k = (8) / (-2) = -4
k = (-10) / (-2) = 5
Therefore, the two possible values of k that make the lines perpendicular are -4 and 5.
To check if two lines are perpendicular, we need to find the dot product of their direction vectors and check if it is equal to zero.
First, we need to rewrite the equations of the lines in vector form.
For the first line, 3x + ky + z - 6 = 0, we can write it as:
A: (3, k, 1) · (x, y, z) = 6
Similarly, for the second line, 6x + (1 - k)y + 2z - 9 = 0, we can write it as:
B: (6, 1 - k, 2) · (x, y, z) = 9
Now, let's find the direction vectors of the lines.
The direction vector of line A will be the coefficients of x, y, and z:
u = (3, k, 1)
The direction vector of line B will be the coefficients of x, y, and z:
v = (6, 1 - k, 2)
Now, we can find the dot product of u and v:
u · v = (3, k, 1) · (6, 1 - k, 2)
The dot product of two vectors can be found by multiplying corresponding components and adding them up:
u · v = (3)(6) + (k)(1 - k) + (1)(2) = 18 + k - k^2 + 2 = 20 + k - k^2
To find the value of k that makes the lines perpendicular, we need to set the dot product equal to zero:
20 + k - k^2 = 0
Now we have a quadratic equation. Let's solve it:
k^2 - k - 20 = 0
We can factor this quadratic equation:
(k - 5)(k + 4) = 0
Setting each factor equal to zero:
k - 5 = 0 or k + 4 = 0
Solving these equations for k:
k = 5 or k = -4
So, the values of k that make the lines perpendicular are k = 5 or k = -4.