Please use Theorems at Geometry of Vectors of Linear Algebra to prove that inequ
ID: 3169481 • Letter: P
Question
Please use Theorems at Geometry of Vectors of Linear Algebra to prove that inequaity of Topic 18.
{Theorem 1: Law of Cosines
Consider a triangle with side lengths a, b, and c with angle opposite to side with length c. Then
Theorem 2
If v and w are two vectors in 2 then
v . w = ||v|| ||w|| cos
where is the angle between the vectors v and w.Theorem 3 The Dot Product in 2
Two vectors v, w 2 are orthogonal if and only if v . w = 0.
Notice that this definition shows the zero vector 0 is trivially orthogonal to every vector.
Theorem 3
Two vectors v, w 2 are orthogonal if and only if v . w = 0.
Notice that this definition shows the zero vector 0 is trivially orthogonal to every vector.
Theorem 4: Cauchy-Schwarz inequality
For all vectors v, w n we have the following inequality holds
|v . w| ||v|| ||w||
Equality holds when and only when v and w are scalar multiples of each other.
Definition 5: Orthogonal and Orthonormal Sets
S = {v1, ..., vk} n is said to be an orthogonal set of vectors in n if vi . vj = 0 whenever i j. The set S is said to be an orthonormal set of vectors if it’s orthogonal and all vectors vi have length 1, i.e. ||vi|| = 1.
We use the Cauchy-Schwarz inequality to prove the next geometric fact about n which is illustrated in the following figure
Theorem 6: Triangle Inequality
For all v, w n we have that ||v + w|| ||v|| + ||w||
Theorem 7: Triangle Inequality (Second Version)
For all vectors u, v, w in n we have that
d(v, w) d(v, u) + d(u, w)}
Definition 8: Vector Projection
Let v and d 0 n. The projection of v onto d is given by
and is the unique vector parallel to d such that d and v - projdv are orthogonal. We call the scalar
the component of v along d.
Theorem 9
Let v and d 0 n. The vector projection projdv minimizes the distance from v to the line parallel to d in the sense that
holds for all
Explanation / Answer
a1^2/b1^2+a2^2/b2+ .............................>=(a1+a2...............an)^2/b1+b2..........bn
(a1/b1+a2/b2+ ............................)^2-2(a1/b1*a2/b2+a2/b2*a3/b3+......................) so its only possible when
inequality holds when and only when a1/b1=a2/b2..........................................
We can simply write out the terms of a squared multinomial by writing out the squared terms and then, for each letter, add to the answer twice the product of this letter and the remaining ones to the right.
Check the sum of the coefficients is equal to (1+1+1+1)2=16.
a1^2/b1^2+a2^2/b2+ .............................>=(a1+a2...............an)^2/b1+b2..........bn
(a1/b1+a2/b2+ ............................)^2-2(a1/b1*a2/b2+a2/b2*a3/b3+......................) so its only possible when
inequality holds when and only when a1/b1=a2/b2..........................................
We can simply write out the terms of a squared multinomial by writing out the squared terms and then, for each letter, add to the answer twice the product of this letter and the remaining ones to the right.
(a+b+c+d)2 a2+b2+c2+d2 Write down the squares 2ab+2ac+2ac Then, starting with a, write down twice a times the other letters: 2a(b+c+d) 2bc+2bd Similarly for b, with the remaining letters: 2b(c+d) 2cd Finally with c. ("Finally" because no letters follow d).2c(d)
Check the sum of the coefficients is equal to (1+1+1+1)2=16.