Calc Help !?

Question

Let P=(3,2,-1) and Q=(-2,1,c) and R=(c,1,0) be points in R3 (3 spaces). For what values of c is PQR a right triangle?

I got 34/8 for one value of c, but I'm not sure if I'm right or not. Can anyone help?

Answer 1

let's consider linear algebra and do it by vectors and dot product. (i don't know the geometric solution, if there is any.)

we have three vectors which are (P-Q), (P-R) AND (Q-R).

(P-Q)=( 3 - -2, 2-1, -1-c) = (5, 1, -1-c)

(P-R) = (3-c, 1, -1)

(Q-R)= (-2-c, 0, c)

(P-Q) AND (P-R) are prependicular to each other when:
(P-Q).(P-R)=0 ==> (5, 1, -1-c) . (3-c, 1, -1) = (15 - 5c) + 1 + (1+c) = -4c + 17= 0 ==> c=17/4

u got the same answer.

next step:

(P-Q).(Q-R) = 0 ==> (5, 1, -1-c) . (-2-c, 0, c) = 0

u will get a square function and probably two answers for C.

the last step is:

(Q-R).(P-R) = 0 and then solve for x.

if it still does not make sense, email me.

Answer 2

This can be solved by using the Pythagorean theorem and the 3d distance equation. d^2 = (delta x)^2 + (delta y)^2 + (delta z)^2.

Distance between P & Q
sqrt((3+2)^2 + (2-1)^2 + (-1-c)^2)
sqrt(c^2 + 2c + 27)

Distance between Q & R
sqrt(-2-c)^2 + (1-1)^2 + (c-0)^2)
sqrt(2c^2 + 4c + 4)

Distance between P & R
sqrt((3-c)^2 + (2-1)^2 + (-1-0)^2)
sqrt(c^2 + 6c + 11)

Solution 1
sqrt(c^2 + 2c + 27)^2 + sqrt(2c^2 + 4c + 4)^2 = sqrt(c^2 + 6c + 11)^2
c^2 + 2c + 27 + 2c^2 + 4c + 4 = c^2 + 6c + 11
2c^2 + 20 = 0
c^2 + 10 = 0
No real solution

Solution 2
sqrt(c^2 + 2c + 27)^2 + sqrt(c^2 + 6c + 11)^2 = sqrt(2c^2 + 4c + 4)^2
c^2 + 2c + 27 + c^2 + 6c + 11 = 2c^2 + 4c + 4
4c + 34 = 0
c = -17/2

Solution 3
sqrt(2c^2 + 4c + 4)^2 + sqrt(c^2 + 6c + 11)^2 = sqrt(c^2 + 2c + 27)^2
2c^2 + 4c + 4 + c^2 + 6c + 11 = c^2 + 2c + 27
2c^2 + 8c - 12 = 0
c^2 + 4c - 6 = 0
c = (-4 +/-sqrt(16 - (4)(1)(-6))) / (2)
c = -2 +/- sqrt(10)

Answer 3

I like your answer. Using the Pythagorean Theorem, the distance from P to Q squared and P to R squared is equal to Q to R squared. If I replace c with 34/8 and just work the distances with a calculator, your answer is correct.

I would guess that you might have come up with that answer by working the algebra the same way.