Explained: "How to find the distance from the point..."

taiko3r4

Answered question

2023-01-09

How to find the distance from the point A(3,-5,5) to the line x=2+3t , y=1-2t, z=-1+t?

milanduy19

Beginner2023-01-10Added 8 answers

We have equation

x + 2 y + z - 3 = 0

use distance formula

= \frac{(1 \cdot 3 - 5 \cdot 2 + 5 \cdot 1) - 3}{\sqrt{1^{2} + 2^{2} + 1^{2}}}

= - \frac{5}{\sqrt{6}}

| - \frac{5}{\sqrt{6}} |

= \frac{5}{\sqrt{6}}

wezergoat420l2k

Beginner2023-01-11Added 1 answers

Defining

p_{0} = {2, 1, - 1}

\vec{v} = {3, - 2, 1}

p_{A} = {3, - 5, 5}

We must measure the separation between the lines.

r \to p_{0} + t \vec{v}

and the point

p_{A}

Using Pitagoras we have

a = ‖ p_{a} - p_{0} ‖

b = | ⟨ p_{A} - p_{0}, \frac{\vec{v}}{‖ \vec{v} ‖} ⟩ |

d = \sqrt{a^{2} - b^{2}}

which is the sought distance

a = \sqrt{{(3 - 2)}^{2} + {(- 5 - 1)}^{2} + {(5 + 1)}^{2}}

\frac{\vec{v}}{‖ \vec{v} ‖} = \frac{{3, - 2, 1}}{\sqrt{3^{2} + 2^{2} + 1}}

b = | \frac{(3 - 2) \cdot 3 + (5 + 1) \cdot 2 + (5 + 1) \cdot 1}{\sqrt{3^{2} + 2^{2} + 1}} |

Then

d = \sqrt{\frac{83}{2}}

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