Question

a. Does the line through \(P(1,2,-3)\) with direction vector \(\mathbf{d}=\left[\begin{array}{r}1 \\ 2 \\ -3\end{array}\right]\) lie in the plane \(2 x-y-z=3 ?\) Explain. b. Does the plane through \(P(4,0,5), Q(2,2,1),\) and \(R(1,-1,2)\) pass through the origin? Explain.

Short answer

a. No, the line does not lie entirely in the plane. b. Yes, the plane passes through the origin.

Step-by-step solution

Verify if the point lies in the line

To determine if the line through point \(P(1,2,-3)\) lies in the plane given by the equation \(2x - y - z = 3\), substitute the coordinates of point \(P\) into the plane's equation:\[2(1) - (2) - (-3) = 3.\]Calculate:\[2 - 2 + 3 = 3.\]This equation holds true. Thus, point \(P\) does lie in the plane.

Check direction vector in plane

Next, we must determine if the line's direction vector is parallel to the plane. The plane's normal vector is \(\mathbf{n} = \left[\begin{array}{r}2 \ -1 \ -1\end{array}\right]\). Calculate the dot product of \(\mathbf{d}\) and \(\mathbf{n}\):\[\mathbf{d} \cdot \mathbf{n} = \left[\begin{array}{r}1 \ 2 \ -3\end{array}\right] \cdot \left[\begin{array}{r}2 \ -1 \ -1\end{array}\right] = 1 \cdot 2 + 2 \cdot (-1) + (-3) \cdot (-1).\]Calculate:\[2 - 2 + 3 = 3.\]The dot product is not zero, which means the line is not parallel to the plane. Therefore, the line does not lie entirely within the plane.

Determine the equation of the plane through P, Q, R

Find the normal vector to the plane that passes through points \(P(4,0,5)\), \(Q(2,2,1)\), and \(R(1,-1,2)\). First, calculate vectors \(PQ\) and \(PR\):\[\mathbf{PQ} = \left[\begin{array}{r}2 \ 2 \ 1\end{array}\right] - \left[\begin{array}{r}4 \ 0 \ 5\end{array}\right] = \left[\begin{array}{r}-2 \ 2 \ -4\end{array}\right],\]\[\mathbf{PR} = \left[\begin{array}{r}1 \ -1 \ 2\end{array}\right] - \left[\begin{array}{r}4 \ 0 \ 5\end{array}\right] = \left[\begin{array}{r}-3 \ -1 \ -3\end{array}\right].\]

Calculate the normal vector

Compute the cross product \(\mathbf{n} = \mathbf{PQ} \times \mathbf{PR}\) to find the normal vector:\[\mathbf{n} = \left[\begin{array}{c}2 \times (-3) - (-4) \times (-1) \ -((-2) \times (-3) - (-4) \times (-3)) \ (-2) \times (-1) - 2 \times (-3)\end{array}\right] = \left[\begin{array}{r}-6 - 4 \ -(6 + 12) \ 2 + 6\end{array}\right] = \left[\begin{array}{r}-10 \ -18 \ 8\end{array}\right].\]

Determine if the plane passes through the origin

Use point \(P(4,0,5)\) to find the equation of the plane: \(-10(x-4) - 18(y-0) + 8(z-5) = 0\), simplifying to:\[-10x - 18y + 8z = -10\times 4 + 8 \times 5.\]Which simplifies further:\[-10x - 18y + 8z = -40 + 40 = 0.\]The origin \((0,0,0)\) satisfies this equation \(-10(0) - 18(0) + 8(0) = 0.\), indicating that the plane does pass through the origin.

Key concepts

Direction Vector

In analytic geometry, a direction vector is a vector that indicates the direction of a line or a path in space. It can tell you how the line is oriented. For this exercise, the direction vector of the line through point \(P(1,2,-3)\) is represented as \(\mathbf{d} = \begin{bmatrix} 1 \ 2 \ -3 \end{bmatrix}\). This means that for every one unit the line moves along the x-axis, it moves two units along the y-axis and negative three units along the z-axis.

• The vector helps in describing the line's direction but does not give its position on its own.
• Direction vectors are key in determining if lines and planes are parallel, by checking if multiples of their direction vectors match.

To see if a line lies on a plane, it's not enough to know the line's direction. You must also confirm that every point along the line satisfies the plane's equation. This can involve verifying if the dot product of the direction vector with the plane's normal vector is zero, indicating they are perpendicular.

Plane Equation

A plane in 3D space can be defined using a plane equation of the form \(Ax + By + Cz = D\), where \(A, B,\) and \(C\) are the components of the normal vector to the plane, and \(D\) is a constant. For the plane described in the exercise \(2x - y - z = 3\), the normal vector \(\mathbf{n}\) is \(\begin{bmatrix} 2 \ -1 \ -1 \end{bmatrix}\).

• The normal vector is perpendicular to the plane, providing a reference for its orientation in space.
• To determine if a point lies on the plane, substitute the point's coordinates into the plane's equation to check if the equation holds.

In this concept, understanding that the plane is infinitely flat and extends in all directions is vital. So if a single point \(P\) lies on the plane, one must consider the entirety of the line through \(P\) in verifying full containment within the plane, which often involves verifying parallel and perpendicular relationships between vectors.

Cross Product

The cross product is a fundamental operation in vector mathematics and analytic geometry. It is used to find a vector that is perpendicular to two given vectors. The ability to calculate a cross product allows us to find a normal vector to a plane when given positions of three points on that plane.
The solution demonstrates this by calculating \(\mathbf{PQ} \times \mathbf{PR}\) to find the normal vector \(\mathbf{n}\) of the plane through points \(P, Q, R\), where \(\mathbf{PQ} = \begin{bmatrix} -2 \ 2 \ -4 \end{bmatrix}\) and \(\mathbf{PR} = \begin{bmatrix} -3 \ -1 \ -3 \end{bmatrix}\).

• The resulting vector \(\mathbf{n}\) from the cross product gives the plane's normal direction.
• This is essential for constructing the plane equation and checking if the plane passes through specific points, such as the origin.

The cross product itself is defined mathematically as \(\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\), making use of the determinant of a matrix. This operation is crucial since it helps in establishing planes and understanding their orientations relative to other geometric entities in space.