Question

True or False? In Exercises 115-118, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. $$ \begin{aligned} &\text { If } \mathbf{v}=a_{1} \mathbf{i}+b_{1} \mathbf{j}+c_{1} \mathbf{k} \text { is any vector in the plane given by }\\\ &a_{2} x+b_{2} y+c_{2} z+d_{2}=0, \text { then } a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \end{aligned} $$

Short answer

True. The dot product of any vector in the plane and the normal to the plane equals zero. Therefore, the provided statement is true.

Step-by-step solution

Understanding the given Information

The given statement implies that any vector in the plane is orthogonal to the normal vector of this plane. The equation of the plane \(a_{2} x+b_{2} y+c_{2} z+d_{2}=0\) suggests that the normal vector to this plane is \(N=a_{2} \mathbf{i}+b_{2} \mathbf{j}+c_{2} \mathbf{k}\) . A vector in the plane would be orthogonal to the normal vector.

Evaluating the Statement

The vector \(\mathbf{v}=a_{1} \mathbf{i}+b_{1} \mathbf{j}+c_{1} \mathbf{k}\) is in the plane provided by \(a_{2} x+b_{2} y+c_{2} z+d_{2}=0\). This requires the dot product of vector \(\mathbf{v}\) and normal vector \(N\) to be zero. Hence \(a_{1} a_{2}+b_{1} b_{2}+c_{1}c_{2}=0\). This is because, if the dot product of the vectors is zero, this indicates that the vectors are orthogonal to each other.

Conclusion

Since the dot product of any vector in the plane and the normal to the plane equals zero, the statement \(a_{1} a_{2}+b_{1} b_{2}+c_{1}c_{2}=0\) is true, if \(\mathbf{v}=a_{1} \mathbf{i}+b_{1} \mathbf{j}+c_{1} \mathbf{k}\) is any vector in the plane defined by \(a_{2} x+b_{2} y+c_{2} z+d_{2}=0\). Therefore, the provided assertion is correct.

Key concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector calculus. It describes the product of two vectors and results in a scalar value, not a vector. The dot product of two vectors \(\textbf{u} = u_1 \textbf{i} + u_2 \textbf{j} + u_3 \textbf{k}\) and \(\textbf{v} = v_1 \textbf{i} + v_2 \textbf{j} + v_3 \textbf{k}\) is calculated as \( \textbf{u} \. \textbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3\).

To visualize the practical application of the dot product, imagine you are pushing a cart along a path. The dot product can help quantify the amount of push you exert in the direction of the path. A larger dot product value indicates a stronger push in the direction of movement, while a value of zero suggests no force exerted in that direction.

In the context of the exercise, the dot product is used to judge the orthogonality of vectors. Since a zero dot product implies orthogonality, it serves as an essential tool in analyzing the geometrical relationship between vectors in a plane.

Orthogonal Vectors

Orthogonal vectors are two or more vectors that meet at a right angle (90 degrees). In mathematical terms, their dot product is equal to zero. This concept extends to any number of dimensions, not just the familiar three-dimensional space.

Orthogonality is a critical concept in various fields, such as computer graphics, engineering, and physics. For example, navigational systems use orthogonal vectors to ensure accurate positioning, as they help to define directions that do not interfere with one another.

In the solution provided for the exercise, the fact that any vector in the plane is orthogonal to the plane's normal vector is key. The normal vector is perpendicular to the plane, and therefore, its dot product with any vector lying entirely within that plane will be zero. This property is what makes the initial statement true, confirming the relationship between the components of the vectors and the necessity for their dot product to equal zero.

Plane Equation

A plane in three-dimensional space can be defined by a linear equation. The general form of the equation for a plane is \(a x + b y + c z + d = 0\), where \((a, b, c)\) are the coefficients that define the plane's normal vector and \(d\) is the scalar representing the plane's distance from the origin along the normal vector.

The normal vector to a plane is crucial for understanding its orientation in space. It is perpendicular to every vector that lies on the plane and is often denoted by \(\textbf{n} = a \textbf{i} + b \textbf{j} + c \textbf{k}\). Any vector lying on the plane will therefore have a zero dot product with the normal vector, indicating that it is orthogonal to the plane.

Through this exercise, the concept of a plane equation becomes tangible. The relationship between a plane's definition and vectors in three-dimensional space, such as the connection to orthogonality through the dot product, reveals the deep interconnection between geometry and algebra in vector calculus.