Problem 18 - Algebra and Trigonometry

Question

Find a vector that is perpendicular to the plane passing through the three given points. $$ P(3,4,5), Q(1,2,3), R(4,7,6) $$

Step-by-Step Solution

Answer

A vector perpendicular to the plane is (4, 0, -4).

Step 1: Understand the Problem

We need to find a vector that is perpendicular to the plane defined by three points: \( P(3,4,5), Q(1,2,3), R(4,7,6) \). This means we are looking for a normal vector to the plane. This can be found using the cross product of two vectors that lie on the plane.

Step 2: Determine Two Vectors on the Plane

To find vectors on the plane, we compute two distinct vectors using the given points. One vector can be \( \vec{PQ} = Q - P \) and another can be \( \vec{PR} = R - P \). Calculate these vectors:- \( \vec{PQ} = (1-3, 2-4, 3-5) = (-2, -2, -2) \)- \( \vec{PR} = (4-3, 7-4, 6-5) = (1, 3, 1) \)

Step 3: Calculate the Cross Product

The normal vector can be found by calculating the cross product \( \vec{PQ} \times \vec{PR} \). The cross product is computed as follows:\[\vec{PQ} \times \vec{PR} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \-2 & -2 & -2 \1 & 3 & 1 \end{vmatrix}\]Compute the determinant:\(\vec{PQ} \times \vec{PR} = \mathbf{i}((-2)(1) - (-2)(3)) - \mathbf{j}((-2)(1) - (-2)(1)) + \mathbf{k}((-2)(3) - (-2)(1))\)\(= \mathbf{i}(4) - \mathbf{j}(0) + \mathbf{k}(-4)\)\(= (4, 0, -4)\)

Step 4: Verify Perpendicularity

The result from Step 3, \( (4, 0, -4) \), is our candidate for the normal vector. Check if it is perpendicular by confirming that the dot product of this vector with either \( \vec{PQ} \) or \( \vec{PR} \) is zero:- \( (4, 0, -4) \cdot (-2, -2, -2) = 4(-2) + 0(-2) + (-4)(-2) = -8 + 0 + 8 = 0 \)Thus, \( (4, 0, -4) \) is indeed perpendicular to the plane.

Key Concepts

cross product

The cross product is a fundamental operation when working with vectors, especially in three-dimensional space. It is used to determine a vector that is perpendicular to two given vectors in space. This concept is particularly useful for finding the normal vector of a plane, as it allows us to establish perpendicularity.

- To compute the cross product of two vectors, say \( \vec{A} = (a_1, a_2, a_3) \) and \( \vec{B} = (b_1, b_2, b_3) \), we use the following determinant-based formula:
\[\vec{A} \times \vec{B} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \a_1 & a_2 & a_3 \b_1 & b_2 & b_3\end{vmatrix}\]Where \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are the unit vectors along the x, y, and z-axes.

- The result of this operation gives us a new vector, which in our example is \( (4, 0, -4) \). This vector is orthogonal to both \( \vec{A} \) and \( \vec{B} \).

Knowing how to compute and utilize the cross product is crucial for understanding geometric relations in vector calculus.

normal vector

A normal vector is a vector that is perpendicular to a surface or a curve at a given point. In the context of a plane, the normal vector gives directional information about the plane's orientation in space.

- In our exercise, the plane is defined by three points, \( P(3,4,5) \), \( Q(1,2,3) \), and \( R(4,7,6) \). To find the normal vector, we use the cross product of two vectors lying on the plane.

- These vectors, \( \vec{PQ} \) and \( \vec{PR} \), are derived by subtracting coordinates of given points:

• \( \vec{PQ} = (-2, -2, -2) \)
• \( \vec{PR} = (1, 3, 1) \)

- The cross product \( \vec{PQ} \times \vec{PR} \) yields the normal vector \( (4, 0, -4) \).

Identifying the normal vector is essential for numerous applications, including finding angles, calculating areas, and determining spatial relations.

perpendicular vectors

Perpendicular vectors are two vectors that intersect at a right angle (90 degrees). The fundamental property of perpendicular vectors is that their dot product is always zero.

- When working in three-dimensional space, finding perpendicular vectors, such as normal vectors to planes, is a frequent requirement in geometry and physics.

- In our task, the vector \( (4, 0, -4) \) serves as a normal vector to the plane formed by points \( P, Q, R \). We confirmed its perpendicularity by checking that the dot product of \( (4, 0, -4) \) with \( \vec{PQ} = (-2, -2, -2) \) is zero:
\[ (4, 0, -4) \cdot (-2, -2, -2) = 0 \]

Understanding how to determine and verify perpendicular vectors is vital for solving problems involving spatial relations and constructing orthogonal projections.

dot product

The dot product, also known as the scalar product, is a fundamental operation in vector mathematics. It is used to measure the extent to which two vectors are pointing in the same direction.

- The dot product between two vectors \( \vec{A} = (a_1, a_2, a_3) \) and \( \vec{B} = (b_1, b_2, b_3) \) is calculated as:
\[ \vec{A} \cdot \vec{B} = a_1b_1 + a_2b_2 + a_3b_3 \]

- A key property of the dot product is that it equals zero when the vectors are perpendicular to each other.

- In this exercise, we use this property to validate our normal vector. The dot product \( (4, 0, -4) \cdot (-2, -2, -2) = 0 \) confirms perpendicularity, indicating that \( (4, 0, -4) \) is indeed a normal vector to the plane.

Mastering the concept of the dot product is essential, as it is widely applicable in physics, engineering, and computer graphics.