Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Explore our Q&A platform to find in-depth answers from a wide range of experts in different fields. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
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].
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
