Tue orth J-J, 2025. Mic isth Meh, 2487.
9) Given that [tex]$z=\binom{4}{4}, \underset{\sim}{b}=\binom{5}{12}, c\binom{-4}{-2}$[/tex] and [tex]$d=\left(-\frac{12}{4}\right.$[/tex] Worker x in column form each of the following
a) [tex]$\underline{x}+\underline{d}=\underline{g}$[/tex]
b) [tex]$c-b=a-x$[/tex]
c) [tex]$2 b=3 c-x$[/tex]
d) [tex]$2 x+q=x+b$[/tex]
e) [tex]$3 x-a=x+d$[/tex]
b
[tex]$\begin{array}{rlrl}
2(x)+1 & 3 x-1 & =x+d \\
4 & =\{x+-12 \\
4 & 3 x-1 & =x+12 \\
2 x-12 & 3 x-1 & =x-12
\end{array}$[/tex]
Answer (2)
Solve for x in each vector equation by rearranging the terms and performing vector addition and subtraction.
a) x + d = a ⟹ x = a − d = ( 0 16 )
b) c − b = a − x ⟹ x = a + b − c = ( 18 13 )
c) 2 b = 3 c − x ⟹ x = 3 c − 2 b = ( − 30 − 22 )
d) 2 x + a = x + b ⟹ x = b − a = ( 8 1 )
e) 3 x − a = x + d ⟹ x = 2 1 ( a + d ) = ( 4 − 4 ) Therefore, the solutions are: a = ( 0 16 ) , b = ( 18 13 ) , c = ( − 30 − 22 ) , d = ( 8 1 ) , e = ( 4 − 4 )
Explanation
Understanding the Problem We are given four vectors: a = ( 4 4 ) , b = ( 12 5 ) , c = ( − 2 − 4 ) , and d = ( 4 − 12 ) . Our objective is to find the vector x in each of the given equations.
Solving for x in equation a a) We have the equation x + d = a . To solve for x , we subtract d from both sides: x = a − d . Substituting the given vectors, we get x = ( 4 4 ) − ( 4 − 12 ) = ( 4 − 4 4 − ( − 12 ) ) = ( 0 16 ) So, x = ( 0 16 ) .
Solving for x in equation b b) We have the equation c − b = a − x . To solve for x , we rearrange the equation: x = a + b − c . Substituting the given vectors, we get x = ( 4 4 ) + ( 12 5 ) − ( − 2 − 4 ) = ( 4 + 12 − ( − 2 ) 4 + 5 − ( − 4 ) ) = ( 18 13 ) So, x = ( 18 13 ) .
Solving for x in equation c c) We have the equation 2 b = 3 c − x . To solve for x , we rearrange the equation: x = 3 c − 2 b . Substituting the given vectors, we get x = 3 ( − 2 − 4 ) − 2 ( 12 5 ) = ( − 6 − 12 ) − ( 24 10 ) = ( − 6 − 24 − 12 − 10 ) = ( − 30 − 22 ) So, x = ( − 30 − 22 ) .
Solving for x in equation d d) We have the equation 2 x + a = x + b . To solve for x , we subtract x and a from both sides: x = b − a . Substituting the given vectors, we get x = ( 12 5 ) − ( 4 4 ) = ( 12 − 4 5 − 4 ) = ( 8 1 ) So, x = ( 8 1 ) .
Solving for x in equation e e) We have the equation 3 x − a = x + d . To solve for x , we rearrange the equation: 2 x = a + d , so x = 2 1 ( a + d ) . Substituting the given vectors, we get x = 2 1 ( ( 4 4 ) + ( 4 − 12 ) ) = 2 1 ( 4 + 4 4 − 12 ) = 2 1 ( 8 − 8 ) = ( 4 − 4 ) So, x = ( 4 − 4 ) .
Final Answer Therefore, the solutions for x in each equation are: a) x = ( 0 16 ) b) x = ( 18 13 ) c) x = ( − 30 − 22 ) d) x = ( 8 1 ) e) x = ( 4 − 4 )
Examples
Vector equations are used in physics to describe forces and motion. For example, if you have two forces acting on an object, you can represent them as vectors and add them to find the net force. Similarly, in computer graphics, vectors are used to represent the position and orientation of objects in 3D space. Solving vector equations allows you to determine how these objects move and interact with each other.
The solutions for x in the vector equations are: x = ( 0 16 ) , x = ( 18 13 ) , x = ( − 30 − 22 ) , x = ( 8 1 ) , and x = ( 4 − 4 ) .
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