The vertices of a triangle are at A(1,3,2), B(4,1,3), C(0,1,-1).
(a) Find the perimeter of the triangle,
(b) Find the area of the triangle,
(c) Find the acute angle between the plane of this triangle and the xy-plane
(i.e. the plane z=0)
maths vertices of a triangle
2009-10-19 4:20 am
回答 (3)
2009-10-21 2:05 am
2009-10-19 6:24 am
cos@ = 〈1/sqrt3 (i + j - k) ‧1/sqrt2 (i + j)〉/│1/sqrt3 (i + j - k) │‧│1/sqrt2 (i + j)│ = 2/sqrt(6), right?
normal vector of xy-plane = k
required angle = angle between the 2 normals.
normal vector of xy-plane = k
required angle = angle between the 2 normals.
2009-10-19 5:11 am
a. Vector AB = (4,1,3) - (1,3,2) = (3,-2,1)
Vector BC = (0,1,-1) - (4,1,3) = (-4,0,-4)
Vector CA = (1,3,2) - (0,1,-1) = (1,2,3)
Perimeter of the triangle ABC
= │AB│+│BC│+│CA│
= sqrt[(3)^2 + (-2)^2 + (1)^2] + sqrt[(-4)^2 + (-4)^2] + sqrt[(1)^2 + (2)^2 + (3)^2]
= 2sqrt14 + 4sqrt2
= 13.14 units (2 d.p.)
b. Area of the triangle ABC
= 1/2 │AB X BC│ (Cross product of vectors)
│i j k│
= 1/2 │3 -2 1│
│-4 0 -4│
= 1/2 │8i + 8j - 8k│
= 1/2 sqrt(3 X 8^2)
= 4sqrt3 sq. units
c. The normal vector to the plane = 1/sqrt3 (i + j - k) (From part b)
The required angle is 90* - angle between the normal vector and the xy plane.
The unit vector in the xy plane = 1/sqrt2 (i + j)
cos@ = 〈1/sqrt3 (i + j - k) ‧1/sqrt2 (i + j)〉/│1/sqrt3 (i + j - k) │‧│1/sqrt2 (i + j)│
= 1/sqrt3
@ = 54.74*
So, the required angle = 90* - 54.74* = 35.26* (2 d.p.)
Vector BC = (0,1,-1) - (4,1,3) = (-4,0,-4)
Vector CA = (1,3,2) - (0,1,-1) = (1,2,3)
Perimeter of the triangle ABC
= │AB│+│BC│+│CA│
= sqrt[(3)^2 + (-2)^2 + (1)^2] + sqrt[(-4)^2 + (-4)^2] + sqrt[(1)^2 + (2)^2 + (3)^2]
= 2sqrt14 + 4sqrt2
= 13.14 units (2 d.p.)
b. Area of the triangle ABC
= 1/2 │AB X BC│ (Cross product of vectors)
│i j k│
= 1/2 │3 -2 1│
│-4 0 -4│
= 1/2 │8i + 8j - 8k│
= 1/2 sqrt(3 X 8^2)
= 4sqrt3 sq. units
c. The normal vector to the plane = 1/sqrt3 (i + j - k) (From part b)
The required angle is 90* - angle between the normal vector and the xy plane.
The unit vector in the xy plane = 1/sqrt2 (i + j)
cos@ = 〈1/sqrt3 (i + j - k) ‧1/sqrt2 (i + j)〉/│1/sqrt3 (i + j - k) │‧│1/sqrt2 (i + j)│
= 1/sqrt3
@ = 54.74*
So, the required angle = 90* - 54.74* = 35.26* (2 d.p.)
參考: Physics king
收錄日期: 2021-04-19 15:57:34
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