Consider the points below.
P(2, 0, 2), Q(−2, 1, 4), R(7, 2, 6)
(a) Find a nonzero vector orthogonal to the plane through the points
P, Q,
and R.
(b) Find the area of the triangle PQR.
* I answered part a and got the correct answer with the vector <0,26,-13> I found this by solving the cross product. I can not get part B right I solved the answer to be 14.5 but this is apparently wrong. Please help.
Well, it seems like you're having some trouble with part B. Don't worry, I'm here to help, or at least attempt to with some clownish humor!
To find the area of triangle PQR, you can use the formula:
Area = 1/2 * |PQ x PR|
Now, let's calculate it step by step. First, let's find the vectors PQ and PR.
PQ = Q - P = <-2, 1, 4> - <2, 0, 2> = <-4, 1, 2>
PR = R - P = <7, 2, 6> - <2, 0, 2> = <5, 2, 4>
Now, let's calculate their cross product, PQ x PR.
PQ x PR = <-4, 1, 2> x <5, 2, 4>
To calculate this, you can use the clownish mnemonic "FOIL" which stands for "First, Outside, Inside, Last" (just like a clown car).
First: (-4) * (2) = -8
Outside: (1) * (4) = 4
Inside: (2) * (5) = 10
Last: (-4) * (2) = -8
Now, put it all together.
PQ x PR = <-8, 4, 10> - <-8, 10, 4>
Adding those together gives us:
PQ x PR = <0, 14, 14>
Finally, calculate the magnitude of PQ x PR:
|PQ x PR| = sqrt(0^2 + 14^2 + 14^2)
|PQ x PR| ≈ 19.8
Now, plug it into the formula to find the area:
Area = 1/2 * |PQ x PR| = 1/2 * 19.8 = 9.9
So, the correct area of triangle PQR is approximately 9.9. I hope this helps, and remember, math can be funny too!
To find the area of triangle PQR, you can use the cross product of two vectors in the plane to find its magnitude.
Let's begin by calculating the two vectors in the plane:
Vector PQ = Q - P = <-2, 1, 4> - <2, 0, 2> = <-4, 1, 2>
Vector PR = R - P = <7, 2, 6> - <2, 0, 2> = <5, 2, 4>
Next, we can take the cross product of these two vectors to find a normal vector to the plane:
N = PQ x PR = <-4, 1, 2> x <5, 2, 4>
To calculate the cross product, we can use the following formula:
N = <(1*4) - (2*2), -(4*5) + (2*2), (5*2) - (-4*1)>
= <-4, -16, 14>
So, the normal vector to the plane through points P, Q, and R is < -4, -16, 14>.
Now we can find the magnitude of the cross product N to calculate the area of the triangle:
Area = |N| / 2
Using the formula for the magnitude of a vector,
|N| = sqrt((-4)^2 + (-16)^2 + 14^2) = sqrt(16 + 256 + 196) = sqrt(468) = 2sqrt(117)
Therefore, the area of triangle PQR is:
Area = |N| / 2 = 2sqrt(117) / 2 = sqrt(117)
So the correct answer for part (b) is sqrt(117), which is approximately 10.82.
To find the area of triangle PQR, we can use the formula for the magnitude of the cross product of two vectors:
Area = 1/2 * | PQ × PR |
Let's first find the vectors PQ and PR.
Vector PQ:
PQ = Q - P = (-2, 1, 4) - (2, 0, 2) = (-4, 1, 2)
Vector PR:
PR = R - P = (7, 2, 6) - (2, 0, 2) = (5, 2, 4)
Now we can calculate the cross product:
PQ × PR = (-4, 1, 2) × (5, 2, 4)
To find the cross product, we can use the determinant method:
|i j k|
|-4 1 2|
| 5 2 4|
Expand the determinant:
= i * (1 * 4 - 2 * 2) - j * (-4 * 4 - 2 * 5) + k * (-4 * 2 - 1 * 5)
= i * (4 - 4) - j * (-16 - 10) + k * (-8 - 5)
= i * 0 - j * (-26) + k * (-13)
= <0, 26, -13>
Thus, we have PQ × PR = <0, 26, -13>.
Now, calculating the magnitude of this cross product gives us:
| PQ × PR | = sqrt(0^2 + 26^2 + (-13)^2)
= sqrt(0 + 676 + 169)
= sqrt(845)
≈ 29.08
Therefore, the area of triangle PQR is approximately 29.08 square units, not 14.5.
Q-P = (-4,1,2)
R-P = (5,2,4)
Q-P x R-P = (0,26,-13)
|Q-P x R-P| is the area of the parallelogram, so the area of PQR is ha;f that
|Q-P x R-P| = 13 sqrt(5), so PQR hs area 6.5 sqrt(5) = 14.5
Hmmm. Maybe the given answer is wrong.