Given following vectors;
\(u\quad =\quad \left[ \begin{matrix} 0.5 \\ 0.5 \end{matrix} \right]\) and \(v\quad =\quad \left[ \begin{matrix} 3 \\ 4 \end{matrix} \right]\)
Perform the following:
1. Calculate the dot product \(u\cdot v\)
2. The lengths of u and v respectively
3. The linear Combination; \(3u-2v\)
4. Angle between u and v
Given \(u\quad =\quad \left[ \begin{matrix} { u }_{ 1 } \\ { u }_{ 2 } \end{matrix} \right]\) and \(v\quad =\quad \left[ \begin{matrix} { v }_{ 1 } \\ { v }_{ 2 } \end{matrix} \right]\)
The dot product \(u\cdot v\quad =\quad { u }_{ 1 }\cdot { v }_{ 1 }\quad +\quad { u }_{ 2 }\cdot { v }_{ 2 }\)
Hence, in this case; \(u\cdot v\quad =\quad { 0.5\times }3\quad +\quad 0.5\times \left( -4 \right) \quad =\quad -0.5\)
The legnths of a vector u in the mathematical sense is calculated by the following; \(\left\| u \right\| \quad =\quad \sqrt { u\cdot u }\)
In Computer Science, the length of a vector refers to the mathematical notion of dimension of vector.
\(\left\| u \right\| \quad =\quad \sqrt { u\cdot u } \quad =\quad \sqrt { { 0.5 }^{ 2 }\quad +\quad { 0.5 }^{ 2 } } \quad =\quad \sqrt { 0.5 } =\quad 0.71\)
\(\left\| v \right\| \quad =\quad \sqrt { v\cdot v } \quad =\quad \sqrt { { 3 }^{ 2 }\quad +\quad { \left( -4 \right) }^{ 2 } } \quad =\quad \sqrt { 9\quad +\quad 16 } =\quad \sqrt { 25 } \quad =\quad 5\)
The linear combination \(3u-2v\) is a vector calculated as follows;
\(3u\quad -2v\quad =\quad 3\left[ \begin{matrix} 0.5 \\ 0.5 \end{matrix} \right] \quad -2\left[ \begin{matrix} 3 \\ -4 \end{matrix} \right] \quad =\quad \left[ \begin{matrix} 3\times 0.5\quad -2\times 3 \\ 3\times 0.5\quad -2\times \left( -4 \right) \end{matrix} \right] \quad =\quad \left[ \begin{matrix} 1.5\quad -\quad 6 \\ 1.5\quad +\quad 8 \end{matrix} \right] \quad =\quad \left[ \begin{matrix} -4.5 \\ 9.5 \end{matrix} \right]\)
If \(\theta\) is the angle between vector u and v, then if u and v are non-zero vectors which is the case, then
\(\cos { \theta } \quad =\quad \frac { u\cdot v }{ \left\| u \right\| \cdot \left\| v \right\| } \quad =\quad \frac { -0.5 }{ 0.71\times 5 } \quad 0.14\)
\(\theta \quad =\quad 1.43\quad radiant\)
# Helper Function to swap rows of matrix
swap_rows <-function(A, m, n){
temp <- A[m,]
A[m,] <- A[n,]
A[n,] <- temp
return(A)
}
# Helper Function to swap elements of vector
swap_elements <- function(v, m, n){
temp_v <- v[m]
v[m] <- v[n]
v[n] <- temp_v
return(v)
}
# Helper function to find pivot
find_pivot <- function (A, r, c){
# Initialize return variables and processing variables
swap <- FALSE
p <- 0
swap_r <- 0
loop <- TRUE
i <- r
#loop through each of matrix and find non-zero pivot point for column c
while(loop){
if(A[i,c] != 0){
loop <- FALSE
p<-A[i,c]
if(i>r){
swap_r <- i
swap <- TRUE
} # end of inner if statement
} # end of outer if statement
i <- i + 1 #increment row
if (i > nrow(A)){
loop <- FALSE
}
} # end of while loop
return(list(p, swap, swap_r))
} # end of function
# Main function to Solve equation (3x3)
solve_equation <- function (A, b){
# Validate input parameters, A, b, and whether row(A)==length(b)
if(missing(A)){
stop("Need to specify matrix A in equation Ax = b.")
}
if(missing(b)){
stop("Needs to specify vector b in equation Ax =b.")
}
if(nrow(A) != ncol(A)){
stop("Needs A to be square matrix.")
}
if(nrow(A) != length(b)){
stop("Needs to have number of rows of A same as dimension of b.")
}
# store nrow and ncol for matrix A for reference
num_rows <- nrow(A)
num_col <- ncol(A)
# store Matrix and Vector in temporary variables
U <- A
w <- b
#-----------------#
# Build Upper-Triangular Matrix
# Initialize colum cursor
c <- 1
while (c <= num_col){
# Initialize row cursor
r <- c
# find pivot point
l_result <- find_pivot (U, r, c)
p <- unlist(l_result[1])
swap_indicator <- unlist(l_result[2])
swap_row <- unlist(l_result[3])
# Validate that pivot point is found
if (p==0){
text <- "No pivot point found in column"
txt_col <- as.character(c)
msg <- paste(text,txt_col)
stop(msg)
}
# if pivot point found, check for required swap
if(swap_indicator){
U <- swap_rows(U,r,swap_row)
w <- swap_elements (w, r, swap_row)
}
# Proceed with elimination for each rows
i <- r + 1
while (i <= num_rows){
# Determine multiplier
m = U[i,c]/p
U[i, ] <- U[i, ] - m*U[r, ]
w[i] <- w[i] - m*w[r]
i <- i + 1
} # end of for while for rows
c <- c + 1
} # end of while loop for columns
#------------#
# Solve equation from last row up
v_equation <- vector()
i <- num_col
v_equation[i] <- w[i]/U[i,i]
z<- w[3]/U[3,3]
y<- (w[2] - U[2,3]*z)/U[2,2]
x<- (w[1] - U[1,2]*y - U[1,3]*z)/U[1,1]
result <- c(round(x, digits = 2), round(y, digits = 2), round(z, digits = 2))
return(result)
}
# -------- Invoquing function ----------#
M = matrix(c(1, 2, -1, 1, -1, -2, 3, 5, 4), nrow=3, ncol=3)
v = c(1, 2, 6)
result_equation <- solve_equation(M,v)
result_equation## [1] -1.55 -0.32 0.95