The general coordinates of any point lying on the two lines may be expressed as:
View attachment 59766
If these two lines are to join, these coordinates need to become equal to each other at some value of λ and μ.
So:
- 3 - λ = 4 + aμ
- -2 + 2λ = 4 + bμ
- 1+λ = 2 - μ
You'll notice the last equation doesn't contain a or b. It can be used to eliminate either λ or μ from the first two equation:
1+ λ = 2 - μ
λ= 1 - μ
So equation 1 becomes (by substituting λ = 1 - μ into it):
3 - 1 + μ = 4 + aμ
μ = 2 + aμ
(1-a)μ = 2
μ = 2/(1-a)
The reason we made μ the subject here will become clear later.
Equation 2 becomes (again after substituting λ = 1 - μ into it):
-2 + 2(1 - μ) = 4 + bμ
-2 + 2 - 2μ = 4 + bμ
(-b-2)μ = 4
Now we can substitute μ from above into here:
(-b-2)*( 2/(1-a) ) = 4
-2b - 4 = 4 - 4a
4a - 2b = 8
2a - b = 4
Part (ii):
If two lines are perpendicular, the dot product of their direction vectors = 0
So:
[-1 2 1] ⋅ [a b -1] = 0
-a + 2b -1 = 0
-a + 2b = 1
Solve this simultaneously with 2a- b = 4 and you'll get:
a = 3, b = 2
Now for the location of their intersection, you just have to substitute these values and do it the usual way:
View attachment 59767
Solve for λ and μ from any two of these three components (check for consistency by substituting into the third component), then obtain the coordinate by substituting either λ or μ into either of the line equations.
You'll get λ = 2, and μ = -1
So the coordinates will be (1,2,3)
Thank You so much!Sorry forgot your last question:
For these types of question just apply log to both sides so that they become like linear equations:
5^(2x-1) = 2(3^x)
ln(5^(2x-1)) = ln(2(3^x))
(2x-1)*ln5 = ln2 + x*ln3
(2ln5)x - ln5 = ln2 + (ln3)x
(2ln5 - ln3)x = ln2 + ln5
Do the rest using calculator:
x = 1.085990045
