The two lines with the equation r=(1,2,-44)+ t(k+1, 3k+1,k-3) and x=2-3s, y=1-10s, z=3-5s, are given.

a) determine the vaulue for K when the lines are paralel
b) determine the value of K when the lines are perpendictular

Vector 1 = (K+1) i +(3K+1)j + (K-3) k

Vector 2 = -3 i -10 j -5 k

If those are parallel the cross product is zero
i j k
K+1 3K+1 K-3
-3 -10 -5

i term -5(3K+1) + 10(K-3) = 0
K = -7
same for j and k directions, K = -7

If they are perpendicular the dot product is 0

(K+1)(-3) +(3K+1)(-10) + (K-3)(-5) = 0
solve for K

To determine the values of K when the lines are parallel or perpendicular, we need to compare their direction vectors. The direction vector of a line is the coefficient vector (k+1, 3k+1, k-3).

a) When the lines are parallel, their direction vectors are proportional, which means they differ only by a scalar multiple. So, we can set up the following equation:

(k+1, 3k+1, k-3) = (a(k'+1), a(3k'+1), a(k'-3))

where a is a non-zero scalar and (k'+1, 3k'+1, k'-3) is the direction vector of the second line.

By comparing the corresponding components, we get:

k+1 = a(k'+1)
3k+1 = a(3k'+1)
k-3 = a(k'-3)

We can solve this system of equations to find the value of K when the lines are parallel.

b) When the lines are perpendicular, their direction vectors are orthogonal, which means their dot product is zero. So, we can set up the following equation:

(k+1, 3k+1, k-3) β‹… (k', 3k', k'-3) = 0

Using the dot product formula: (a, b, c) β‹… (d, e, f) = ad + be + cf, we can rewrite the equation as:

(k+1)(k') + (3k+1)(3k') + (k-3)(k'-3) = 0

We can solve this equation to find the value of K when the lines are perpendicular.

To determine the value of k when the two lines are parallel, we need to compare the direction vectors of the two lines.

For the first line, the direction vector is given as (k+1, 3k+1, k-3).

For the second line, the direction vector is (-3, -10, -5).

Two vectors are parallel if they are scalar multiples of each other. In other words, the first vector can be written as a constant multiple of the second vector.

For example, (k+1, 3k+1, k-3) = c(-3, -10, -5), where c is a constant.

We can equate corresponding components to find the value of k.

k+1 = -3c
3k+1 = -10c
k-3 = -5c

Now we will solve these three equations simultaneously to find the values of k and c.

Adding the first and third equations, we get:
k + (k-3) = -3c - 5c
2k -3 = -8c

So, k = (-8c + 3)/2.

Substituting this value of k in the second equation, we get:

3((-8c + 3)/2) + 1 = -10c
(-24c + 9)/2 + 1 = -10c
-24c + 9 + 2 = -20c
-24c + 11 = -20c
4c = 11
c = 11/4

Substituting the value of c back into the equation for k, we get:

k = (-8(11/4) + 3)/2
k = (-22 + 12)/2
k = -5/2

Therefore, the value of k when the two lines are parallel is k = -5/2.

To determine the value of k when the two lines are perpendicular, we need to check if the dot product of the direction vectors is zero.

The dot product of the direction vectors (k+1, 3k+1, k-3) and (-3, -10, -5) is given by:

(k+1)(-3) + (3k+1)(-10) + (k-3)(-5) = 0

Simplifying this equation gives:

-3k - 3 - 30k - 10 + 5k - 15 = 0
-28k - 28 = 0
k = -1

Therefore, the value of k when the two lines are perpendicular is k = -1.