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Math  /  Geometry

QuestionQ6 (6 points) Find the general equation of the plane containing the origin and points P(1,2,3)P(1,2,3) and Q(1,−1,1)Q(1,-1,1).

Studdy Solution

STEP 1

1. A plane can be defined using a point and a normal vector.
2. The origin and points P(1,2,3) P(1,2,3) and Q(1,−1,1) Q(1,-1,1) lie on the plane.
3. The normal vector to the plane can be found using the cross product of two vectors lying on the plane.

STEP 2

1. Find two vectors on the plane.
2. Compute the normal vector to the plane using the cross product.
3. Use the normal vector and a point on the plane to find the general equation of the plane.

STEP 3

Find two vectors on the plane. Use the origin O(0,0,0) O(0,0,0) , point P(1,2,3) P(1,2,3) , and point Q(1,−1,1) Q(1,-1,1) .
Vector OP⃗ \vec{OP} from the origin to point P P : OP⃗=⟨1−0,2−0,3−0⟩=⟨1,2,3⟩ \vec{OP} = \langle 1 - 0, 2 - 0, 3 - 0 \rangle = \langle 1, 2, 3 \rangle
Vector OQ⃗ \vec{OQ} from the origin to point Q Q : OQ⃗=⟨1−0,−1−0,1−0⟩=⟨1,−1,1⟩ \vec{OQ} = \langle 1 - 0, -1 - 0, 1 - 0 \rangle = \langle 1, -1, 1 \rangle

STEP 4

Compute the normal vector to the plane using the cross product of OP⃗ \vec{OP} and OQ⃗ \vec{OQ} .
n⃗=OP⃗×OQ⃗ \vec{n} = \vec{OP} \times \vec{OQ}
Calculate the cross product: n⃗=∣ijk1231−11∣ \vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 1 & -1 & 1 \end{vmatrix}
=i(2⋅1−3⋅(−1))−j(1⋅1−3⋅1)+k(1⋅(−1)−2⋅1) = \mathbf{i}(2 \cdot 1 - 3 \cdot (-1)) - \mathbf{j}(1 \cdot 1 - 3 \cdot 1) + \mathbf{k}(1 \cdot (-1) - 2 \cdot 1)
=i(2+3)−j(1−3)+k(−1−2) = \mathbf{i}(2 + 3) - \mathbf{j}(1 - 3) + \mathbf{k}(-1 - 2)
=5i+2j−3k = 5\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}
Thus, the normal vector is n⃗=⟨5,2,−3⟩ \vec{n} = \langle 5, 2, -3 \rangle .

STEP 5

Use the normal vector and a point on the plane (e.g., the origin) to find the general equation of the plane.
The general equation of a plane is: ax+by+cz=d ax + by + cz = d
Substitute the normal vector ⟨5,2,−3⟩ \langle 5, 2, -3 \rangle and the origin (0,0,0) (0,0,0) : 5x+2y−3z=0 5x + 2y - 3z = 0
Since the plane passes through the origin, d=0 d = 0 .
The general equation of the plane is: 5x+2y−3z=0 \boxed{5x + 2y - 3z = 0}

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