Answer :
To solve this problem, we need to write the vector [tex]\( u = \langle -8, -8 \rangle \)[/tex] as a sum of two orthogonal vectors, one of which is the projection of [tex]\( u \)[/tex] onto [tex]\( v \)[/tex], denoted as [tex]\( \operatorname{proj}_v u \)[/tex].
Given the vectors [tex]\( u = \langle -8, -8 \rangle \)[/tex] and [tex]\( v = \langle -1, 2 \rangle \)[/tex]:
Step 1: Calculate the dot product of [tex]\( u \)[/tex] and [tex]\( v \)[/tex].
The dot product [tex]\( u \cdot v \)[/tex] is calculated as follows:
[tex]\[ u \cdot v = (-8)(-1) + (-8)(2) = 8 - 16 = -8 \][/tex]
Step 2: Calculate the dot product of [tex]\( v \)[/tex] with itself.
The dot product [tex]\( v \cdot v \)[/tex] is calculated as follows:
[tex]\[ v \cdot v = (-1)^2 + (2)^2 = 1 + 4 = 5 \][/tex]
Step 3: Calculate the projection of [tex]\( u \)[/tex] onto [tex]\( v \)[/tex].
The projection formula is given by:
[tex]\[ \operatorname{proj}_v u = \left( \frac{u \cdot v}{v \cdot v} \right) v = \left( \frac{-8}{5} \right) \langle -1, 2 \rangle \][/tex]
So,
[tex]\[ \operatorname{proj}_v u = \left( \frac{-8}{5} \right) \times \langle -1, 2 \rangle = \langle 1.6, -3.2 \rangle \][/tex]
Step 4: Calculate the orthogonal component.
The orthogonal component is found by subtracting the projection from [tex]\( u \)[/tex]:
[tex]\[ \text{Orthogonal component} = u - \operatorname{proj}_v u = \langle -8, -8 \rangle - \langle 1.6, -3.2 \rangle = \langle -8 - 1.6, -8 + 3.2 \rangle \][/tex]
So,
[tex]\[ \text{Orthogonal component} = \langle -9.6, -4.8 \rangle \][/tex]
Therefore, [tex]\( u \)[/tex] can be expressed as:
[tex]\[ u = \operatorname{proj}_v u + \text{Orthogonal component} = \langle 1.6, -3.2 \rangle + \langle -9.6, -4.8 \rangle \][/tex]
From the given options:
- a. [tex]\( \langle -0.5, 1 \rangle + \langle -7.5, -9 \rangle \)[/tex]
- b. [tex]\( \langle -1, 2 \rangle + \langle -7, -10 \rangle \)[/tex]
- c. [tex]\( \langle -1, 2 \rangle + \langle -2, -1 \rangle \)[/tex]
- d. [tex]\( \langle 1.6, -3.2 \rangle + \langle -9.6, -4.8 \rangle \)[/tex]
The correct choice is:
d. [tex]\( \langle 1.6, -3.2 \rangle + \langle -9.6, -4.8 \rangle \)[/tex]
So, the best answer from the choices provided is:
D
Given the vectors [tex]\( u = \langle -8, -8 \rangle \)[/tex] and [tex]\( v = \langle -1, 2 \rangle \)[/tex]:
Step 1: Calculate the dot product of [tex]\( u \)[/tex] and [tex]\( v \)[/tex].
The dot product [tex]\( u \cdot v \)[/tex] is calculated as follows:
[tex]\[ u \cdot v = (-8)(-1) + (-8)(2) = 8 - 16 = -8 \][/tex]
Step 2: Calculate the dot product of [tex]\( v \)[/tex] with itself.
The dot product [tex]\( v \cdot v \)[/tex] is calculated as follows:
[tex]\[ v \cdot v = (-1)^2 + (2)^2 = 1 + 4 = 5 \][/tex]
Step 3: Calculate the projection of [tex]\( u \)[/tex] onto [tex]\( v \)[/tex].
The projection formula is given by:
[tex]\[ \operatorname{proj}_v u = \left( \frac{u \cdot v}{v \cdot v} \right) v = \left( \frac{-8}{5} \right) \langle -1, 2 \rangle \][/tex]
So,
[tex]\[ \operatorname{proj}_v u = \left( \frac{-8}{5} \right) \times \langle -1, 2 \rangle = \langle 1.6, -3.2 \rangle \][/tex]
Step 4: Calculate the orthogonal component.
The orthogonal component is found by subtracting the projection from [tex]\( u \)[/tex]:
[tex]\[ \text{Orthogonal component} = u - \operatorname{proj}_v u = \langle -8, -8 \rangle - \langle 1.6, -3.2 \rangle = \langle -8 - 1.6, -8 + 3.2 \rangle \][/tex]
So,
[tex]\[ \text{Orthogonal component} = \langle -9.6, -4.8 \rangle \][/tex]
Therefore, [tex]\( u \)[/tex] can be expressed as:
[tex]\[ u = \operatorname{proj}_v u + \text{Orthogonal component} = \langle 1.6, -3.2 \rangle + \langle -9.6, -4.8 \rangle \][/tex]
From the given options:
- a. [tex]\( \langle -0.5, 1 \rangle + \langle -7.5, -9 \rangle \)[/tex]
- b. [tex]\( \langle -1, 2 \rangle + \langle -7, -10 \rangle \)[/tex]
- c. [tex]\( \langle -1, 2 \rangle + \langle -2, -1 \rangle \)[/tex]
- d. [tex]\( \langle 1.6, -3.2 \rangle + \langle -9.6, -4.8 \rangle \)[/tex]
The correct choice is:
d. [tex]\( \langle 1.6, -3.2 \rangle + \langle -9.6, -4.8 \rangle \)[/tex]
So, the best answer from the choices provided is:
D