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To determine which vectors are orthogonal to the plane described by the equation [tex]\(2(x-1) - (y+1) + z = 0\)[/tex], we first need to identify the normal vector to the plane. The normal vector to a plane given by the general equation [tex]\(Ax + By + Cz = D\)[/tex] is [tex]\(\langle A, B, C \rangle\)[/tex].
From the given equation [tex]\(2(x-1) - (y+1) + z = 0\)[/tex], we can rewrite it in the standard form:
[tex]\[2x - y + z - 3 = 0.\][/tex]
Here, [tex]\(A = 2\)[/tex], [tex]\(B = -1\)[/tex], and [tex]\(C = 1\)[/tex]. Therefore, the normal vector to the plane is [tex]\(\langle 2, -1, 1 \rangle\)[/tex].
For a vector to be orthogonal to the plane, it has to be perpendicular to the normal vector. Two vectors are perpendicular if and only if their dot product is zero. Hence, for each vector [tex]\(\vec{v} = \langle v_1, v_2, v_3 \rangle\)[/tex], we need to check if:
[tex]\[ \vec{v} \cdot \langle 2, -1, 1 \rangle = 0. \][/tex]
Let's compute the dot product for each given vector:
1. For [tex]\( \vec{v} = \langle 1, -1, 0 \rangle \)[/tex]:
[tex]\[ \langle 1, -1, 0 \rangle \cdot \langle 2, -1, 1 \rangle = 1 \cdot 2 + (-1) \cdot (-1) + 0 \cdot 1 = 2 + 1 + 0 = 3. \][/tex]
This dot product is [tex]\(3\)[/tex], not [tex]\(0\)[/tex], so [tex]\(\vec{v} = \langle 1, -1, 0 \rangle\)[/tex] is not orthogonal to the plane.
2. For [tex]\(\vec{v} = \langle -2, 2, 0 \rangle \)[/tex]:
[tex]\[ \langle -2, 2, 0 \rangle \cdot \langle 2, -1, 1 \rangle = -2 \cdot 2 + 2 \cdot (-1) + 0 \cdot 1 = -4 - 2 + 0 = -6. \][/tex]
This dot product is [tex]\(-6\)[/tex], not [tex]\(0\)[/tex], so [tex]\(\vec{v} = \langle -2, 2, 0 \rangle\)[/tex] is not orthogonal to the plane.
3. For [tex]\(\vec{v} = \langle -2, 1, -1 \rangle \)[/tex]:
[tex]\[ \langle -2, 1, -1 \rangle \cdot \langle 2, -1, 1 \rangle = -2 \cdot 2 + 1 \cdot (-1) + (-1) \cdot 1 = -4 - 1 - 1 = -6. \][/tex]
This dot product is [tex]\(-6\)[/tex], not [tex]\(0\)[/tex], so [tex]\(\vec{v} = \langle -2, 1, -1 \rangle\)[/tex] is not orthogonal to the plane.
4. For [tex]\(\vec{v} = \langle 4, 2, -2 \rangle \)[/tex]:
[tex]\[ \langle 4, 2, -2 \rangle \cdot \langle 2, -1, 1 \rangle = 4 \cdot 2 + 2 \cdot (-1) + (-2) \cdot 1 = 8 - 2 - 2 = 4. \][/tex]
This dot product is [tex]\(4\)[/tex], not [tex]\(0\)[/tex], so [tex]\(\vec{v} = \langle 4, 2, -2 \rangle\)[/tex] is not orthogonal to the plane.
5. For [tex]\(\vec{v} = \langle -1, 2, -3 \rangle \)[/tex]:
[tex]\[ \langle -1, 2, -3 \rangle \cdot \langle 2, -1, 1 \rangle = -1 \cdot 2 + 2 \cdot (-1) + (-3) \cdot 1 = -2 - 2 - 3 = -7. \][/tex]
This dot product is [tex]\(-7\)[/tex], not [tex]\(0\)[/tex], so [tex]\(\vec{v} = \langle -1, 2, -3 \rangle\)[/tex] is not orthogonal to the plane.
6. For [tex]\(\vec{v} = \langle 2, -1, 1 \rangle \)[/tex]:
[tex]\[ \langle 2, -1, 1 \rangle \cdot \langle 2, -1, 1 \rangle = 2 \cdot 2 + (-1) \cdot (-1) + 1 \cdot 1 = 4 + 1 + 1 = 6. \][/tex]
This dot product is [tex]\(6\)[/tex], not [tex]\(0\)[/tex], so [tex]\(\vec{v} = \langle 2, -1, 1 \rangle\)[/tex] is not orthogonal to the plane.
From our calculations, we can see that _none_ of the given vectors are orthogonal to the plane described by the equation [tex]\(2(x-1) - (y+1) + z = 0\)[/tex].
From the given equation [tex]\(2(x-1) - (y+1) + z = 0\)[/tex], we can rewrite it in the standard form:
[tex]\[2x - y + z - 3 = 0.\][/tex]
Here, [tex]\(A = 2\)[/tex], [tex]\(B = -1\)[/tex], and [tex]\(C = 1\)[/tex]. Therefore, the normal vector to the plane is [tex]\(\langle 2, -1, 1 \rangle\)[/tex].
For a vector to be orthogonal to the plane, it has to be perpendicular to the normal vector. Two vectors are perpendicular if and only if their dot product is zero. Hence, for each vector [tex]\(\vec{v} = \langle v_1, v_2, v_3 \rangle\)[/tex], we need to check if:
[tex]\[ \vec{v} \cdot \langle 2, -1, 1 \rangle = 0. \][/tex]
Let's compute the dot product for each given vector:
1. For [tex]\( \vec{v} = \langle 1, -1, 0 \rangle \)[/tex]:
[tex]\[ \langle 1, -1, 0 \rangle \cdot \langle 2, -1, 1 \rangle = 1 \cdot 2 + (-1) \cdot (-1) + 0 \cdot 1 = 2 + 1 + 0 = 3. \][/tex]
This dot product is [tex]\(3\)[/tex], not [tex]\(0\)[/tex], so [tex]\(\vec{v} = \langle 1, -1, 0 \rangle\)[/tex] is not orthogonal to the plane.
2. For [tex]\(\vec{v} = \langle -2, 2, 0 \rangle \)[/tex]:
[tex]\[ \langle -2, 2, 0 \rangle \cdot \langle 2, -1, 1 \rangle = -2 \cdot 2 + 2 \cdot (-1) + 0 \cdot 1 = -4 - 2 + 0 = -6. \][/tex]
This dot product is [tex]\(-6\)[/tex], not [tex]\(0\)[/tex], so [tex]\(\vec{v} = \langle -2, 2, 0 \rangle\)[/tex] is not orthogonal to the plane.
3. For [tex]\(\vec{v} = \langle -2, 1, -1 \rangle \)[/tex]:
[tex]\[ \langle -2, 1, -1 \rangle \cdot \langle 2, -1, 1 \rangle = -2 \cdot 2 + 1 \cdot (-1) + (-1) \cdot 1 = -4 - 1 - 1 = -6. \][/tex]
This dot product is [tex]\(-6\)[/tex], not [tex]\(0\)[/tex], so [tex]\(\vec{v} = \langle -2, 1, -1 \rangle\)[/tex] is not orthogonal to the plane.
4. For [tex]\(\vec{v} = \langle 4, 2, -2 \rangle \)[/tex]:
[tex]\[ \langle 4, 2, -2 \rangle \cdot \langle 2, -1, 1 \rangle = 4 \cdot 2 + 2 \cdot (-1) + (-2) \cdot 1 = 8 - 2 - 2 = 4. \][/tex]
This dot product is [tex]\(4\)[/tex], not [tex]\(0\)[/tex], so [tex]\(\vec{v} = \langle 4, 2, -2 \rangle\)[/tex] is not orthogonal to the plane.
5. For [tex]\(\vec{v} = \langle -1, 2, -3 \rangle \)[/tex]:
[tex]\[ \langle -1, 2, -3 \rangle \cdot \langle 2, -1, 1 \rangle = -1 \cdot 2 + 2 \cdot (-1) + (-3) \cdot 1 = -2 - 2 - 3 = -7. \][/tex]
This dot product is [tex]\(-7\)[/tex], not [tex]\(0\)[/tex], so [tex]\(\vec{v} = \langle -1, 2, -3 \rangle\)[/tex] is not orthogonal to the plane.
6. For [tex]\(\vec{v} = \langle 2, -1, 1 \rangle \)[/tex]:
[tex]\[ \langle 2, -1, 1 \rangle \cdot \langle 2, -1, 1 \rangle = 2 \cdot 2 + (-1) \cdot (-1) + 1 \cdot 1 = 4 + 1 + 1 = 6. \][/tex]
This dot product is [tex]\(6\)[/tex], not [tex]\(0\)[/tex], so [tex]\(\vec{v} = \langle 2, -1, 1 \rangle\)[/tex] is not orthogonal to the plane.
From our calculations, we can see that _none_ of the given vectors are orthogonal to the plane described by the equation [tex]\(2(x-1) - (y+1) + z = 0\)[/tex].
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