Q➊:
"Find the equation of the plane containing the lines 2x - y +z -3 =0 and 3x +y = z = 5 and at a distance of 1/ root6 from the point (2, 1, -1)."
There appears to be a typo in the Q ... "3x +y = z = 5" ?
Edit your Q and re-post; we'll answer then
Q➋:
"given three planes: P₁: x – y + z = 1 ; P₂: x + y – z = –1 and P₃: x – 3y + 3z = 2.
Let L₁ , L₂ , L₃ be the lines of intersection of the planes P₂ and P₃ , P₃ and P₁ , and P₁ and P₂ respectively.
i) Find the lines L₁ , L₂ , and L₃
ii) Check whether at least 2 lines are non-parallel or not and
whether the 3 planes have a common point of intersection or a common line of intersection"
AQ➋i):
P₁: x – y + z = 1
P₂: x + y – z = –1
adding: 2x = 0 ➞ P₁ ∩ P₂ = L₃: x = 0
P₂: x + y – z = –1
P₃: x – 3y + 3z = 2
adding P₂ + (–P₃) ➞ P₂ ∩ P₃ = L₁: y – z = ¾
P₁: x – y + z = 1
P₃: x – 3y + 3z = 2
adding P₁ + (–P₃) ➞ P₁ ∩ P₃ = L₂: y – z = ½
AQ➋ii):
Since L₁ ‖ L₂ ➞ the 3 planes can NOT have a common point of intersection
At least two lines are non-parallel, but the planes do NOT have a common line of intersection either.
There are two parallel lines, and a third line that is skew to the two parallel lines.
Q➌:
"If the plane x/2 + y/3 + z/6 =1 cuts the axes of co-ordinates at points A,B,C then the area of the triangle ABC is?
(in this sum i know the co-ordinates of A, B, and C are (2,0,0 ), ( 0,3,0 ) , (0,0,6), but i dont know how to find he area of this triangle.)"
AQ➌:
OPTION 1:
The area of a ∆ in 3-space can be found by determining ½ of the magnitude of the 'cross-product' of any two of the three vectors, V₁, V₂, V₃ representing the sides of the 3-D ∆
(I will use the symbol ↗ to denote a vector, if it is not clear from the context)
Area(3D∆) = ½ | V₁ X V₂ | ... OR ... = ½ | V₁ X V₃ | ... OR ... = ½ | V₂ X V₃ |,
where 'X' denotes 'cross product'
Any of the three vectors, say for example ↗V₁ = ↗AB, representing one of the sides of the 3-D ∆, can be determined by the vector subtraction of the two Origin-Vertex vectors, ↗OA – ↗OB, which can be easily determined by the subtraction of the corresponding pairs of co-ordinates.
the co-ordinates of A, B, and C are A(2, 0, 0), B(0, 3, 0), C(0, 0, 6)
V₁ = ↗AB = ↗OA – ↗OB = V₁[(0 – 2), (3 – 0), (0 – 0)] = V₁[–2, 3, 0]
V₂ = ↗AC = ↗OA – ↗OC = V₂[(0 – 2), (0 – 0), (6 – 0)] = V₂[–2, 0, 6]
Area(∆ABC) = ½ | ↗AB X ↗AC | = ½ | V₁ X V₂ |
= ½ | [–2, 3, 0] X [–2, 0, 6] |
= ½ | [(3)(6) – 0(0), –2(6) –(–2)(0), –2(0) –(–2)(3)] |
= ½ | [18, –12, –6] |
= ½ √{18² +(–12)² + (–6)²}
= ½ √{324 + 144 + 36}
= ½(√504)
= ½(22.44994...)
= 11.2249...
= 11.2 units² (1 dec pl)
OPTION 2:
You could also use Heron's formula: A = √[s(s – a)(s – b)(s – c)] where s = (a + b + c)/2
Using the 'distance between points' formula, determine the length of each side a, b, c.
Then calculate 's', 's – a', 's – b', 's – c'
Finally, plug results into Heron's formula.
Hope this is helpful! Cheers! ☺
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