Answer :
To find the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] such that the vector [tex]\( \vec{a} = \begin{pmatrix} a \\ 2b \\ 8 \end{pmatrix} \)[/tex] is orthogonal to both [tex]\( \vec{v} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \)[/tex] and [tex]\( \vec{\omega} = \begin{pmatrix} 4 \\ 2 \\ -4 \end{pmatrix} \)[/tex], we need to ensure that the dot products of [tex]\( \vec{a} \)[/tex] with each of these vectors are zero. Let's set up the equations step by step.
### Step 1: Ensure [tex]\( \vec{a} \)[/tex] is orthogonal to [tex]\( \vec{v} \)[/tex]
The dot product of [tex]\( \vec{a} \)[/tex] and [tex]\( \vec{v} \)[/tex] is given by:
[tex]\[ \vec{a} \cdot \vec{v} = a \cdot 2 + 2b \cdot (-1) + 8 \cdot 1 = 0 \][/tex]
Simplifying the expression:
[tex]\[ 2a - 2b + 8 = 0 \][/tex]
Divide the entire equation by 2:
[tex]\[ a - b + 4 = 0 \quad \implies \quad a - b = -4 \tag{1} \][/tex]
### Step 2: Ensure [tex]\( \vec{a} \)[/tex] is orthogonal to [tex]\( \vec{\omega} \)[/tex]
The dot product of [tex]\( \vec{a} \)[/tex] and [tex]\( \vec{\omega} \)[/tex] is given by:
[tex]\[ \vec{a} \cdot \vec{\omega} = a \cdot 4 + 2b \cdot 2 + 8 \cdot (-4) = 0 \][/tex]
Simplifying the expression:
[tex]\[ 4a + 4b - 32 = 0 \][/tex]
Divide the entire equation by 4:
[tex]\[ a + b - 8 = 0 \tag{2} \][/tex]
### Step 3: Solve the system of linear equations
Now we solve the system of equations obtained from the dot products:
1. [tex]\( a - b = -4 \)[/tex]
2. [tex]\( a + b = -8 \)[/tex]
Add the two equations to eliminate [tex]\( b \)[/tex]:
[tex]\[ (a - b) + (a + b) = -4 - 8 \][/tex]
This simplifies to:
[tex]\[ 2a = -12 \quad \implies \quad a = -6 \][/tex]
Substitute [tex]\( a = -6 \)[/tex] back into equation (1):
[tex]\[ -6 - b = -4 \quad \implies \quad -b = 2 \quad \implies \quad b = -2 \][/tex]
Therefore, the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are:
[tex]\[ a = -6 \quad \text{and} \quad b = -2 \][/tex]
Therefore, the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] such that [tex]\( \vec{a} \)[/tex] is orthogonal to both [tex]\( \vec{v} \)[/tex] and [tex]\( \vec{\omega} \)[/tex] are [tex]\( a = -6 \)[/tex] and [tex]\( b = -2 \)[/tex].
### Step 1: Ensure [tex]\( \vec{a} \)[/tex] is orthogonal to [tex]\( \vec{v} \)[/tex]
The dot product of [tex]\( \vec{a} \)[/tex] and [tex]\( \vec{v} \)[/tex] is given by:
[tex]\[ \vec{a} \cdot \vec{v} = a \cdot 2 + 2b \cdot (-1) + 8 \cdot 1 = 0 \][/tex]
Simplifying the expression:
[tex]\[ 2a - 2b + 8 = 0 \][/tex]
Divide the entire equation by 2:
[tex]\[ a - b + 4 = 0 \quad \implies \quad a - b = -4 \tag{1} \][/tex]
### Step 2: Ensure [tex]\( \vec{a} \)[/tex] is orthogonal to [tex]\( \vec{\omega} \)[/tex]
The dot product of [tex]\( \vec{a} \)[/tex] and [tex]\( \vec{\omega} \)[/tex] is given by:
[tex]\[ \vec{a} \cdot \vec{\omega} = a \cdot 4 + 2b \cdot 2 + 8 \cdot (-4) = 0 \][/tex]
Simplifying the expression:
[tex]\[ 4a + 4b - 32 = 0 \][/tex]
Divide the entire equation by 4:
[tex]\[ a + b - 8 = 0 \tag{2} \][/tex]
### Step 3: Solve the system of linear equations
Now we solve the system of equations obtained from the dot products:
1. [tex]\( a - b = -4 \)[/tex]
2. [tex]\( a + b = -8 \)[/tex]
Add the two equations to eliminate [tex]\( b \)[/tex]:
[tex]\[ (a - b) + (a + b) = -4 - 8 \][/tex]
This simplifies to:
[tex]\[ 2a = -12 \quad \implies \quad a = -6 \][/tex]
Substitute [tex]\( a = -6 \)[/tex] back into equation (1):
[tex]\[ -6 - b = -4 \quad \implies \quad -b = 2 \quad \implies \quad b = -2 \][/tex]
Therefore, the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are:
[tex]\[ a = -6 \quad \text{and} \quad b = -2 \][/tex]
Therefore, the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex] such that [tex]\( \vec{a} \)[/tex] is orthogonal to both [tex]\( \vec{v} \)[/tex] and [tex]\( \vec{\omega} \)[/tex] are [tex]\( a = -6 \)[/tex] and [tex]\( b = -2 \)[/tex].