Difference between revisions of "031 Review Part 2, Problem 6"
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{| class="mw-collapsible mw-collapsed" style = "text-align:left;" | {| class="mw-collapsible mw-collapsed" style = "text-align:left;" | ||
!Foundations: | !Foundations: | ||
+ | |- | ||
+ | |'''1.''' The distance between the vectors <math style="vertical-align: 0px">\vec{u}</math> and <math style="vertical-align: 0px">\vec{v}</math> is | ||
|- | |- | ||
| | | | ||
+ | ::<math>\text{dist}(\vec{u},\vec{v})=||\vec{u}-\vec{v}||.</math> | ||
+ | |- | ||
+ | |'''2.''' The orthogonal projection of <math style="vertical-align: -3px">\vec{y}</math> onto <math style="vertical-align: 0px">L</math> is | ||
+ | |- | ||
+ | | | ||
+ | ::<math>\hat{y}=\text{proj}_L \vec{y}=\frac{\vec{y}\cdot \vec{u}}{\vec{u}\cdot \vec{u}}\vec{u}.</math> | ||
|} | |} | ||
Revision as of 20:55, 11 October 2017
Let and
(a) Find a unit vector in the direction of
(b) Find the distance between and
(c) Let Compute the orthogonal projection of onto
Foundations: |
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1. The distance between the vectors and is |
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2. The orthogonal projection of onto is |
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Solution:
(a)
Step 1: |
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Step 2: |
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(b)
Step 1: |
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Step 2: |
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(c)
Step 1: |
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Step 2: |
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Final Answer: |
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(a) |
(b) |