Question

The distance ð‘‘ of a point ð‘ƒ to the line through points ð´ and ðµ is the length of the component of ð´ð‘ƒ that is orthogonal to ð´ðµ, as indicated in the diagram. So the distance from ð‘ƒ=(-1,0) to the line through the points ð´=(-1,4) and ðµ=(2,4) is 2. Find the volume of the parallelepiped with one vertex at (3,-5,0), and adjacent vertices at (9,-2,3), (6,-5,2), and (2,-11,-3).

Name: the distance dd of a point pp to the line through points aa and bb is the length of the component of apap that is orthogonal to abab as indicated in the diagram so the distance from p10p10 t 68191
Uploaded: 2022-09-27T08:46:31-08:00
Duration: 5 min 48 s
Channel: Adi S
Description: the distance dd of a point pp to the line through points aa and bb is the length of the component of apap that is orthogonal to abab as indicated in the diagram so the distance from p10p10 t 68191

          The distance ð‘‘ of a point ð‘ƒ to the line through points ð´ and ðµ is the length of the component of ð´ð‘ƒ that is orthogonal to ð´ðµ, as indicated in the diagram. So the distance from ð‘ƒ=(-1,0) to the line through the points ð´=(-1,4) and ðµ=(2,4) is 
2. Find the volume of the parallelepiped with one vertex at (3,-5,0), and adjacent vertices at (9,-2,3), (6,-5,2), and (2,-11,-3).

Added by Janice M.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Adi S

09/27/2022

Step 1

This vector is given by AP = (3 - 9, -5 - (-2), 0 - 3) = (-6, -3, -3). Show more…

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The distance ð‘‘ of a point ð‘ƒ to the line through points ð´ and ðµ is the length of the component of ð´ð‘ƒ that is orthogonal to ð´ðµ, as indicated in the diagram. So the distance from ð‘ƒ=(-1,0) to the line through the points ð´=(-1,4) and ðµ=(2,4) is 2. Find the volume of the parallelepiped with one vertex at (3,-5,0), and adjacent vertices at (9,-2,3), (6,-5,2), and (2,-11,-3).