the acute angle between the two planes x+y+2z=3 and 3x-2y+2z=7 is.__
Answers
Given The acute angle between the two planes x+y+2z=3 and 3x-2y+2z=7 is.
We know that cos θ = mod n1 n2 / mod n1 n2
We have n1 = i + j + 2k and n2 = 3i - 2j + 2k
mod n1 = √1^2 + 1^2 + 2^2
= √6
mod n2 = √3^2 + (-2)^2 + 2^2
= √17
cosθ = mod (i + j + k). (3i - 2j + 2k) / √6.√17
cosθ = mod (1 x 3) + (1 x -2) + (2 x 2) / √102
= mod 3 - 2 + 4/√102
cosθ = mod 5/√102
θ = cos⁻¹ 5/√102
θ is approximately equal to 60 degree
For finding the angle between two planes you need to find the angle between two vectors that are normal to planes.
Let
A: x + y + 2z = 3 ---> for this plane appropriate normal vector is a = (1, 1, 2)
and
B: 3x - 2y + 2z = 7 ---> for this plane appropriate normal vector is b = (3, -2, 2)
Okay, so now we have two vectors a and b. The angle between planes A and B equals the angle between vectors a and b.
To find this angle we use formula for multiplying vectors:
a · b = ║a║ · ║b║ · cos(a,b)
cos∡(a,b) = (a · b)/ (║a║ · ║b║ )
∡(a,b) = arccos ( (a · b) / (║a║ · ║b║) )
Okay, now when calculating magnitude of a vector formula: a = (1,1,2) ---> ║a║ = √(1² + 1²+ 2²) = √6
Using same formula we get that ║b║ = √17
When calculating a·b where a = (a1, a2, a3) and b = (b1, b2 ,b3 ) we say that
a · b = a1·b1 + a2·b2 + a3·b3
So for our vectors
a = (1,1,2) and b = (3,-2,2) we have that a · b = 1·3 + 1·(-2) + 2·2 = 3 - 2 + 4 = 5
Now we have everything we need to calculate the angle :
∡(a,b) = arccos ( (a · b) / (║a║ · ║b║) )
║a║ = √6
║b║ = √17
a · b = 5 5 = √25
---> ∡(a,b) = arccos ( √25/(√6 · √17))
= arccos ( √(25/102) )
≈ arccos ( 0.4949... )
≈ 60°
So the angle between planes A and B ∡(A,B) ≈ 60°
Note: If by using this method by any chance you get obtuse angle . Use this formula
+
= 180° And from this equation you can calculate the acute angle.