Question

\(\triangle \mathrm{ABC}\) side \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\) position vectors are \(\underline{\mathrm{a}}, \underline{\mathrm{b}}\) and \(\underline{\mathrm{c}}\) find the length of line segment from \(\mathrm{A}\) to \(\underline{\mathrm{BC}}\). (a) \([(\underline{\mathrm{b}} \times \underline{\mathrm{c}}) /(\underline{\mathrm{b}}-\underline{\mathrm{c}})]\) (b) \([(|\underline{\mathbf{c}} \times \underline{\mathbf{a}}|) /(|\underline{\mathbf{c}}-\underline{\mathrm{a}}|)]\) (c) \([(|\underline{a} \times \underline{b}+\underline{b} \times \underline{c}+\underline{c} \times \underline{a}|) /(|\underline{b}-\underline{c}|)]\) (d) \([(\mid \underline{\mathrm{a}} \times \underline{\mathrm{b}}+\underline{\mathrm{b}} \times \underline{\mathrm{c}}+\underline{\mathrm{c}} \times \underline{\mathrm{a}}) /(|\underline{\mathrm{b}}+\underline{\mathrm{c}}|)]\)

Short answer

The short answer is: \[ \mathrm{Length} = \frac{|\underline{\mathbf{c}} \times \underline{\mathbf{a}}|}{|\underline{\mathbf{c}}-\underline{\mathrm{a}}|} \]

Step-by-step solution

Write down the known facts

: We know that, the position vectors of points A, B, and C are given as: \[ A: \underline{\mathrm{a}} \] \[ B: \underline{\mathrm{b}} \] \[ C: \underline{\mathrm{c}} \]

Compute vectors BC and BA

: To find the length of the line segment from point A to line BC, we need the vectors \(\underline{\mathrm{BC}}\) and \(\underline{\mathrm{BA}}\), which can be found by calculating: \[ \underline{\mathrm{BC}} = \underline{\mathrm{c}} - \underline{\mathrm{b}} \] \[ \underline{\mathrm{BA}} = \underline{\mathrm{a}} - \underline{\mathrm{b}} \]

Calculate the cross product of the vectors BC and BA

: Let's calculate the cross product of the two vectors \(\underline{\mathrm{BC}}\) and \(\underline{\mathrm{BA}}\): \[ \underline{\mathrm{n}} = \underline{\mathrm{BC}} \times \underline{\mathrm{BA}} \]

Find the length of the cross product

: Now, we'll find the length or magnitude of the cross product: \[ |\underline{\mathrm{n}}| = |\underline{\mathrm{BC}} \times \underline{\mathrm{BA}}| \]

Compute the length of vector BC

: Find the length, or magnitude, of vector \(\underline{\mathrm{BC}}\): \[ |\underline{\mathrm{BC}}| = |\underline{\mathrm{c}} - \underline{\mathrm{b}}| \]

Find the length of line segment from A to line BC

: The length of the line segment from point A to line BC can be given by the formula: \[ \mathrm{Length} = \frac{|\underline{\mathrm{n}}|}{|\underline{\mathrm{BC}}|} \] \[ = \frac{|\underline{\mathrm{BC}} \times \underline{\mathrm{BA}}|}{|\underline{\mathrm{c}} - \underline{\mathrm{b}}|} \] Now, comparing and matching the derived formula with options given: The correct option is (b). \[ \mathrm{Length} = \frac{|\underline{\mathbf{c}} \times \underline{\mathbf{a}}|}{|\underline{\mathbf{c}}-\underline{\mathrm{a}}|} \]

Key concepts

Vector Cross Product

The vector cross product, also known as the vector product, is a mathematical operation in three-dimensional space that results in a new vector perpendicular to the plane spanned by the original two vectors. It can be thought of as a way to find a vector that is not in the same direction as either of the two vectors being multiplied.

The formula for calculating the cross product of two vectors \( \underline{\mathbf{u}} \) and \( \underline{\mathbf{v}} \) is given by:
\[ \underline{\mathbf{u}} \times \underline{\mathbf{v}} \]
It's important to note that the order of multiplication matters—crossing \( \underline{\mathbf{v}} \) with \( \underline{\mathbf{u}} \) would give a vector pointing in the opposite direction. To solve problems involving vector cross products, we typically use the right-hand rule to determine the direction of the resulting vector.

The magnitude of this vector tells us about the area of the parallelogram formed by the original vectors. This is particularly useful in the field of physics for calculations involving torque and in geometry for finding distances between points and lines or planes, as demonstrated in the textbook exercise.

Magnitude of Vectors

The magnitude of a vector represents its length or size, and it's one of the most fundamental aspects of vectors. It is a scalar quantity, which means it has only magnitude and no direction. You can picture a vector's magnitude as the distance from the starting point of the vector to its tip when drawn in space.

Mathematically, the magnitude of a vector \( \underline{\mathbf{u}} = (u_1, u_2, u_3) \) in three-dimensional space is calculated with the formula:
\[ |\underline{\mathbf{u}}| = \sqrt{u_1^2 + u_2^2 + u_3^2} \]
Where \( u_1 \) is the x-component, \( u_2 \) is the y-component, and \( u_3 \) is the z-component of the vector. In the context of the textbook problem, knowing how to calculate the magnitude of a vector is crucial for finding the length of line segments or the distance between points in space. For example, computing the magnitude of a cross product gives the area of a parallelogram spanned by the vectors, which relates to the distance calculation between a point and a line.

Geometry in Vector Algebra

Vector algebra is an essential tool in geometry, providing a way to precisely describe points, lines, planes, and distances. Position vectors serve as the foundation for this, representing the location of points in space relative to an origin. Using vectors, we can perform various geometric operations such as finding the distance between points, determining the equation of a line or plane, and calculating areas and volumes of geometric figures.

When we deal with the geometry of vectors, we often use operations like addition, subtraction, and the cross product to find distances or angles between geometrical shapes. For instance, the cross product helps us find a vector perpendicular to a plane, which can be used to calculate the distance from a point not in the plane, like in the given exercise where we find the length of the segment from point A to the line segment BC. Understanding the geometric interpretations of vector algebra enhances our ability to visualize and solve complex spatial problems, making it a powerful tool in fields such as physics, engineering, and computer graphics.