Math Problem Statement

Find the distance from the point (8, 0, 0) to the line x = 8 + t, y = 2t, z = 21√5 + 2t.

Solution

We need to calculate the distance from the point (8,0,0)(8, 0, 0) to the line given by the parametric equations:

x=8+t,y=2t,z=215+2t.x = 8 + t, \quad y = 2t, \quad z = 21\sqrt{5} + 2t.

Step-by-step solution:

  1. Direction vector of the line:

    The direction vector of the line can be extracted from the parametric equations. It is given by the coefficients of tt:

    d=1,2,2.\vec{d} = \langle 1, 2, 2 \rangle.

  2. Point on the line:

    When t=0t = 0, the point on the line is:

    P1=(8,0,215).P_1 = (8, 0, 21\sqrt{5}).

  3. Vector from the point (8,0,0)(8, 0, 0) to the point on the line P1P_1:

    Let P2=(8,0,0)P_2 = (8, 0, 0) be the given point. The vector from P2P_2 to P1P_1 is:

    v=P1P2=88,00,2150=0,0,215.\vec{v} = P_1 - P_2 = \langle 8 - 8, 0 - 0, 21\sqrt{5} - 0 \rangle = \langle 0, 0, 21\sqrt{5} \rangle.

  4. Cross product of v\vec{v} and d\vec{d}:

    We need the cross product v×d\vec{v} \times \vec{d} to find the perpendicular distance. The cross product of 0,0,215\langle 0, 0, 21\sqrt{5} \rangle and 1,2,2\langle 1, 2, 2 \rangle is:

    v×d=i^j^k^00215122=i^(022152)j^(022151)+k^(0201)\vec{v} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 21\sqrt{5} \\ 1 & 2 & 2 \end{vmatrix} = \hat{i}(0 \cdot 2 - 21\sqrt{5} \cdot 2) - \hat{j}(0 \cdot 2 - 21\sqrt{5} \cdot 1) + \hat{k}(0 \cdot 2 - 0 \cdot 1)

    Simplifying, we get:

    v×d=425,215,0.\vec{v} \times \vec{d} = \langle -42\sqrt{5}, -21\sqrt{5}, 0 \rangle.

  5. Magnitude of the cross product:

    The magnitude of v×d\vec{v} \times \vec{d} is:

    v×d=(425)2+(215)2+02=17645+4415=8820+2205=11025=105.|\vec{v} \times \vec{d}| = \sqrt{(-42\sqrt{5})^2 + (-21\sqrt{5})^2 + 0^2} = \sqrt{1764 \cdot 5 + 441 \cdot 5} = \sqrt{8820 + 2205} = \sqrt{11025} = 105.

  6. Magnitude of the direction vector:

    The magnitude of d=1,2,2\vec{d} = \langle 1, 2, 2 \rangle is:

    d=12+22+22=1+4+4=9=3.|\vec{d}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3.

  7. Distance from the point to the line:

    The distance DD from the point to the line is given by:

    D=v×dd=1053=35.D = \frac{|\vec{v} \times \vec{d}|}{|\vec{d}|} = \frac{105}{3} = 35.

Thus, the distance from the point (8,0,0)(8, 0, 0) to the line is 35.

Let me know if you need any further clarifications!

Related Questions:

  1. How do you find the distance from a point to a plane?
  2. What is the geometrical interpretation of the cross product?
  3. How can you find the shortest distance between two skew lines?
  4. What role does the direction vector play in parametric equations?
  5. Can you derive the distance formula for two points in 3D space?

Tip:

Always check that you're working with the correct direction vector for a line when finding distances or projections!

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Math Problem Analysis

Mathematical Concepts

Vector Geometry
Cross Product
Distance from Point to Line

Formulas

Distance from point to line formula: D = |v × d| / |d|
Cross product of vectors
Magnitude of a vector formula

Theorems

Cross Product Theorem
Distance from Point to Line in 3D

Suitable Grade Level

Grade 11-12 (Pre-Calculus/Calculus)