Let L1 be the line passing through the points Q1=(−1, 5, 2) and Q2=(−2, 7, 3) and let L2 be the line passing through the point P1=(−8, 4, 9) with direction vector →d=[−6, 3, 6]T. Determine whether L1 and L2 intersect. If so, find the point of intersection Q.

yes they intersect

point of intersection is (-17.6,38.2,18.6)

Step-by-step explanation:

the direction vector for L2 is given but for L1 it is not given.

Let's first find the direction vector for the L1 as;

from points Q1 and Q2 we can find the direction vector by subtracting there corresponding coordinates (x,y,z) as;

-1-(-2), 5-7, 2-3

direction vector for L1= [1,-2,-1]k

Now the equation for L1 is;

x1=-2+k

y1=7-2k

z1=3-k

direction vector for L2 is =[−6, 3, 6]T

So the equation for the L2 is;

x2=-8-6T

y2=4+3T

z2=9+6T

If these points are intersecting then there x coordinates must be equal, y coordinates must be equal and z coordinates must also be equal as;

x1=x2; y1=y2; z1=z2;

-2+k=-8-6T(a)

7-2k=4+3T(b)

3-k=9+6T(c)

subtracting (a) and (c) we get

-2+k=-8-6T

-( 3-k=9+6T)

-5+2k=-17-12T(d)

adding (b) and (d) we get

7-2k=4+3T

-5+2k=-17-12T

-2=13-9T

T=1.6 putting in (a) we get k as;

-2+k=-8-6(1.6)

k=-15.6

putting T and k, if it satisfy the equation then they intersect

(c)=3-(-15.6)=9+6(1.6)

18.6=18.6 satisfied.

thus they intersect each other and their point of intersection is

x=-2-15.6

x=-17.6

y=7-2(-15.6)

y=38.2

z=3-(-15.6)

z=18.6

point of intersection is (-17.6,38.2,18.6)

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

the answer to this question is yes