Let L1 be the line passing through the points Q1=(−2, 4, −2) and Q2=(−3, 5, 1) and let L2 be the line passing through the point P1=(−2, 10, −5) with direction vector →d=[−2, 6, 4]T. Determine whether L1 and L2 intersect. If so, find the point of intersection Q.

The answers to the questions are;

Yes, the lines  L1 and L2 intersect.

The point of intersection Q of the lines L1 and L2 is (1, 1, -11).

Step-by-step explanation:

The direction vector for L1 is (1, -1, -3)

and an equation is

x = -3 + t

y =  5 - t

z = 1 - 3·t

The possible equation for L2 is

x = -2 - 2·k

y = 10 + 6·k

z = -5 + 4·k

For intersection, we have

-3 + t = -2 - 2·k AND 5 - t = 10 + 6·k AND 1 - 3·t = -5 + 4 k

From the first equation, we have

t = 1 - 2·k and substituting in the second equation, we get

5 - (1 - 2·k) = 10 + 6·k

4 + 2·k = 10 + 6·k

6 = -4·k and k = -3/2

From which t = 1 - 2·k = 1 -2×(-3/2) = 1 + 3 = 4

t = 4

We now check if the results for t and k satisfies the third equation as follows

Left Hand Side is  1 - 3·t = 1 - 3×4 = -11

Right Hand Side is  -5 + 4 k = -5 + (4×(-3/2)) = -5 -6 = -11

Therefore the results from the first two equations satisfy the third equation and we now look for the intersection point as

x = -3 + t  ⇒ -3 + 4 = 1

y =  5 - t ⇒    5 - 4 = 1

z = 1 - 3·t ⇒    1 - 3×4 = -11

Giving the intersection point Q as (1, 1, -11).

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb