Math Problem Statement

Find the distance from the point (2, 9, 6) to the line \(\mathbf{a}(t) = \langle -1, 2, -4 \rangle t + \langle 4, -3, -2 \rangle \).

Solution

To find the distance from a point to a line in three-dimensional space, we use the formula:

Distance=∣AB⋅(v)∣∣v∣\text{Distance} = \frac{|\mathbf{AB} \cdot (\mathbf{v})|}{|\mathbf{v}|}

Where:

  • Point A=(2,9,6)A = (2, 9, 6)
  • The line is given in parametric form: a(t)=⟨−1,2,−4⟩t+⟨4,−3,−2⟩\mathbf{a}(t) = \langle -1, 2, -4 \rangle t + \langle 4, -3, -2 \rangle
  • Point B=(4,−3,−2)B = (4, -3, -2) (from the constant vector in the parametric equation)
  • Vector v=⟨−1,2,−4⟩\mathbf{v} = \langle -1, 2, -4 \rangle (from the coefficient of tt in the parametric equation)

Let's proceed step-by-step to calculate the distance.

Step 1: Find Vector AB

AB=A−B=(2−4,9−(−3),6−(−2))=(−2,12,8)\mathbf{AB} = \mathbf{A} - \mathbf{B} = (2 - 4, 9 - (-3), 6 - (-2)) = (-2, 12, 8)

Step 2: Compute the Cross Product AB×v\mathbf{AB} \times \mathbf{v}

\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 12 & 8 \\ -1 & 2 & -4 \end{vmatrix}$$ ### Step 3: Find the Magnitude of the Cross Product The magnitude of a vector $$\mathbf{u} = \langle u_1, u_2, u_3 \rangle$$ is given by: $$|\mathbf{u}| = \sqrt{u_1^2 + u_2^2 + u_3^2}$$ ### Step 4: Find the Magnitude of Vector $$\mathbf{v}$$ Now let's compute these steps.The distance from the point $$(2, 9, 6)$$ to the line defined by $$\mathbf{a}(t) = \langle -1, 2, -4 \rangle t + \langle 4, -3, -2 \rangle$$ is approximately **14.50** units. Would you like more details on any step or have any other questions? ### Related Questions: 1. How is the cross product of two vectors calculated? 2. What is the geometric interpretation of the distance formula used here? 3. How do you find the distance between a point and a plane? 4. Can we find the shortest distance between two skew lines in 3D space? 5. How is the dot product different from the cross product? ### Tip: When working with vectors, knowing when to use dot products (for projections and angles) versus cross products (for areas and perpendiculars) is crucial for solving geometry problems in 3D.

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Math Problem Analysis

Mathematical Concepts

Vector Algebra
Distance Between Point and Line
3D Geometry

Formulas

Distance = \frac{|\mathbf{AB} \cdot (\mathbf{v})|}{|\mathbf{v}|}
Cross Product for two vectors: \mathbf{u} \times \mathbf{v}
Magnitude of a vector: |\mathbf{u}| = \sqrt{u_1^2 + u_2^2 + u_3^2}

Theorems

Pythagoras' Theorem in 3D
Vector Cross Product

Suitable Grade Level

Grades 11-12 or college-level

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