Nice problem Another method (geometrical approach) First equation represents a plane whose perpendicular distance from origin is 15/√3 or 5√3 Second equation represents a sphere with centre at origin and radius 5√3 Obviously given plane is tangent plane Suppose that point of contact is (u,v,w) then plane will be ux + vy + wz = 75 Comparing it with x+y+z = 15 u/1=v/1=w/1= 75/15 Hence (u,v,w) =(5,5,5) is solution of problem.
Your problem can also be solved fairly easily using the Cauchy-Swartz inequality, which says that for vectors a and b in ℝⁿ (n-tuples of real numbers), a·b ≤ |a|·|b| with equality happening if and only if a = k·b or b = k·a for some real number k. (a·b is the dot product, also called the inner product, of vectors a and b.) Let a = (x, y, z) and b = (1, 1, 1). Then a·b = x + y + z = 15, |a| = sqrt(x² + y² + z²) = sqrt(75), and |b| = sqrt(3). Substituting that into Cauchy-Swartz, 15 ≤ sqrt(75) · sqrt(3) which is actually an equality. According to Cauchy-Swartz we must have (x, y, z) = k · (1, 1, 1) = (k, k, k). We immediately get x = y = z = 5.
Beautiful Algebra ❤
Nice problem
Another method (geometrical approach)
First equation represents a plane
whose perpendicular distance from origin is 15/√3 or 5√3
Second equation represents a sphere with centre at origin and radius 5√3
Obviously given plane is tangent plane
Suppose that point of contact is (u,v,w) then plane will be
ux + vy + wz = 75
Comparing it with x+y+z = 15
u/1=v/1=w/1= 75/15
Hence (u,v,w) =(5,5,5) is solution of problem.
Sir!! You second solution is so cool. I love it❤ thank youu
Your problem can also be solved fairly easily using the Cauchy-Swartz inequality, which says that for vectors a and b in ℝⁿ (n-tuples of real numbers),
a·b ≤ |a|·|b| with equality happening if and only if a = k·b or b = k·a for some real number k. (a·b is the dot product, also called the inner product, of vectors a and b.)
Let a = (x, y, z) and b = (1, 1, 1). Then a·b = x + y + z = 15, |a| = sqrt(x² + y² + z²) = sqrt(75), and |b| = sqrt(3). Substituting that into Cauchy-Swartz,
15 ≤ sqrt(75) · sqrt(3)
which is actually an equality. According to Cauchy-Swartz we must have (x, y, z) = k · (1, 1, 1) = (k, k, k). We immediately get x = y = z = 5.
Other way:
x²+y²+z²-10(x+y+z)+3(25)=75-15*10+75=0
=> (x-5)²+(y-5)²+(z-5)²=0
So x=y=z=5.
Can be hard to follow when you go fast and writing gets sloppy.
X^2+y2+z^2= xy + yz + zx= 75
Its clear by anyways
X= y = z= 5
No more fuss r method
Do not write X and x. Those are two different variables.
"It's clear by anyways"
Ahh, is "anyways" a new axiom we can use for proofs?