Find the area of triangle PQR, P(0,-2,0), Q(4,1,-2), R(5,3,1)

signokodo7h

signokodo7h

Answered question

2021-11-15

Find the area of triangle PQR, P(0,-2,0), Q(4,1,-2), R(5,3,1)

Answer & Explanation

May Dunn

May Dunn

Beginner2021-11-16Added 12 answers

Area of a vector with vertices at P,Q and R is
Area=12|PQ×PR|
PQ=<4,2,0<0,0,3<4,2,3>
PR=<3,3,1<0,0,3<3,3,4>
Area=12|<4,2,3>×<3,3,4>|=12|ijk423334|
=|i(89)j(169)+k(126)|2
=i7j+6k2
=12+72+622
=1+49+362=862
madeleinejames20

madeleinejames20

Skilled2023-05-25Added 165 answers

Step 1:
First, let's find the vectors PQ and PR. The vector PQ is given by subtracting the coordinates of point P from those of point Q:
PQ=[401(2)20]=[432]
Similarly, the vector PR is given by subtracting the coordinates of point P from those of point R:
PR=[503(2)10]=[551]
Step 2:
Next, we calculate the cross product of PQ and PR. The cross product is obtained by taking the determinants of the following matrix:
PQ×PR=|𝐢𝐣𝐤432551|
Expanding this determinant, we have:
PQ×PR=(3·1(2)·5)𝐢(4·1(2)·5)𝐣+(4·53·5)𝐤
Simplifying the expression, we get:
PQ×PR=[13145]
Step 3:
Now, we can calculate the magnitude of this cross product vector:
|PQ×PR|=(13)2+142+52
Simplifying further, we obtain:
|PQ×PR|=169+196+25=390
Finally, we can find the area of triangle PQR by substituting this value into the formula:
{Area}=12(390)=3902
Hence, the area of triangle PQR is 3902.
Eliza Beth13

Eliza Beth13

Skilled2023-05-25Added 130 answers

Answer:
12390
Explanation:
First, we find the vectors 𝐯1 and 𝐯2 that are formed by the edges PQ and PR, respectively:
𝐯1=𝐐𝐏=(4,1,2)(0,2,0)=(4,3,2)
𝐯2=𝐑𝐏=(5,3,1)(0,2,0)=(5,5,1)
Next, we calculate the cross product of 𝐯1 and 𝐯2:
𝐧=𝐯1×𝐯2
Using the determinant method, we have:
𝐧=|𝐢𝐣𝐤432551|
Expanding the determinant, we get:
𝐧=𝐢|3251|𝐣|4251|+𝐤|4355|
Calculating the determinants, we find:
𝐧=𝐢((3×1)(2×5))𝐣((4×1)(2×5))+𝐤((4×5)(3×5))
𝐧=𝐢(3+10)𝐣(4+10)+𝐤(2015)
𝐧=𝐢(13)𝐣(14)+𝐤(5)
Therefore, the normal vector to the triangle is 𝐧=13𝐢14𝐣+5𝐤.
Now, we can calculate the magnitude of the normal vector:
|𝐧|=132+(14)2+52=169+196+25=390
Finally, we can find the area of triangle PQR using the formula:
A=12|𝐧|
Substituting the values, we have:
A=12390
Therefore, the area of triangle PQR is 12390 square units.
Mr Solver

Mr Solver

Skilled2023-05-25Added 147 answers

To find the area of triangle PQR with vertices P(0,-2,0), Q(4,1,-2), and R(5,3,1), we can use the formula for the area of a triangle in 3D space. Let's denote the position vectors of the vertices as 𝐏, 𝐐, and 𝐑, respectively. The area of the triangle is given by:
Area=12(𝐐𝐏)×(𝐑𝐏)
where × represents the cross product and · represents the magnitude of a vector.
First, let's calculate the position vectors of 𝐐𝐏 and 𝐑𝐏:
𝐐𝐏=[401(2)(2)0]=[432]
𝐑𝐏=[503(2)10]=[551]
Next, let's calculate the cross product of 𝐐𝐏 and 𝐑𝐏:
(𝐐𝐏)×(𝐑𝐏)=[432]×[551]=[(2×5)(3×1)(4×1)(5×2)(4×5)(3×5)]=[13145]
Finally, let's calculate the magnitude of (𝐐𝐏)×(𝐑𝐏) and divide it by 2 to obtain the area:
(𝐐𝐏)×(𝐑𝐏)=(13)2+142+52=169+196+25=390=310
Therefore, the area of triangle PQR is 12×310=3210.

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