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I was just messing around on this calculator when I pressed shift and found that the pound sign had a button on the calculator. Since then, I have been wondering about what it does. Could someone please answer this?

helperid1839321  Apr 6, 2017
 #1
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Compute the following cross product:
(1, 2, 3)x(4, 5, 6), where # = x

Construct a matrix where the first row contains unit vectors i^^, j^^, and k^^; and the second and third rows are made of vectors (1, 2, 3) and (4, 5, 6):
(i^^ | j^^ | k^^
1 | 2 | 3
4 | 5 | 6)

Take the determinant of this matrix:
 left bracketing bar i^^ | j^^ | k^^
1 | 2 | 3
4 | 5 | 6 right bracketing bar 

Expand with respect to row 1:
 left bracketing bar i^^ | j^^ | k^^
1 | 2 | 3
4 | 5 | 6 right bracketing bar 

The determinant of the matrix (i^^ | j^^ | k^^
1 | 2 | 3
4 | 5 | 6) is given by i^^ left bracketing bar 2 | 3
5 | 6 right bracketing bar + (-j^^) left bracketing bar 1 | 3
4 | 6 right bracketing bar + k^^ left bracketing bar 1 | 2
4 | 5 right bracketing bar :
i^^ left bracketing bar 2 | 3
5 | 6 right bracketing bar + (-j^^) left bracketing bar 1 | 3
4 | 6 right bracketing bar + k^^ left bracketing bar 1 | 2
4 | 5 right bracketing bar 

i^^ left bracketing bar 2 | 3
5 | 6 right bracketing bar = i^^ (2 6 - 3 5) = i^^ (-3) = -3 i^^:
-3 i^^ + (-j^^) left bracketing bar 1 | 3
4 | 6 right bracketing bar + k^^ left bracketing bar 1 | 2
4 | 5 right bracketing bar 

-j^^ left bracketing bar 1 | 3
4 | 6 right bracketing bar = -j^^ (1 6 - 3 4) = -j^^ (-6) = 6 j^^:
-3 i^^ + 6 j^^ + k^^ left bracketing bar 1 | 2
4 | 5 right bracketing bar 

k^^ left bracketing bar 1 | 2
4 | 5 right bracketing bar = k^^ (1 5 - 2 4) = k^^ (-3) = -3 k^^:
-3 i^^ + 6 j^^ - 3 k^^

-3 i^^ + 6 j^^ - 3 k^^ = (-3, 6, -3):
Answer: | (-3, 6, -3)

Guest Apr 6, 2017

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