Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Get detailed answers to your questions from a community of experts dedicated to providing accurate information. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
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].
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
