How do you evaluate vector magnitudes which include multiplication?
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For the vectors u= (1,3,0) and v=(3,0,2), how do I find the magnitude of u+4v? For letters u and v, I try adding all 3 components of the vectors, squaring each component, then square rooting it. For v, I multiple everything by 4. Then, I had the results together. Yet, this is incorrect. What is the proper way to evaluate this? A picture is attached so that the original problem may be referenced. Also, please feel free to edit my post as needed for clarity. I am still trying to figure out the layout of this site. enter image description here
calculus vectors
asked Jan 17 at 6:15
KvotheKvothe
84
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add a comment |
$begingroup$
For the vectors u= (1,3,0) and v=(3,0,2), how do I find the magnitude of u+4v? For letters u and v, I try adding all 3 components of the vectors, squaring each component, then square rooting it. For v, I multiple everything by 4. Then, I had the results together. Yet, this is incorrect. What is the proper way to evaluate this? A picture is attached so that the original problem may be referenced. Also, please feel free to edit my post as needed for clarity. I am still trying to figure out the layout of this site. enter image description here
calculus vectors
asked Jan 17 at 6:15
KvotheKvothe
84
$endgroup$
add a comment |
$begingroup$
For the vectors u= (1,3,0) and v=(3,0,2), how do I find the magnitude of u+4v? For letters u and v, I try adding all 3 components of the vectors, squaring each component, then square rooting it. For v, I multiple everything by 4. Then, I had the results together. Yet, this is incorrect. What is the proper way to evaluate this? A picture is attached so that the original problem may be referenced. Also, please feel free to edit my post as needed for clarity. I am still trying to figure out the layout of this site. enter image description here
calculus vectors
asked Jan 17 at 6:15
KvotheKvothe
84
$endgroup$
For the vectors u= (1,3,0) and v=(3,0,2), how do I find the magnitude of u+4v? For letters u and v, I try adding all 3 components of the vectors, squaring each component, then square rooting it. For v, I multiple everything by 4. Then, I had the results together. Yet, this is incorrect. What is the proper way to evaluate this? A picture is attached so that the original problem may be referenced. Also, please feel free to edit my post as needed for clarity. I am still trying to figure out the layout of this site. enter image description here
calculus vectors
calculus vectors
asked Jan 17 at 6:15
KvotheKvothe
84
asked Jan 17 at 6:15
KvotheKvothe
84
asked Jan 17 at 6:15
KvotheKvothe
84
asked Jan 17 at 6:15
KvotheKvothe
84
asked Jan 17 at 6:15
KvotheKvothe
84
84
add a comment |
add a comment |
2 Answers
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$begingroup$
The norm (or magnitude) of $(a,b,c)$ is $sqrt {a^{2}+b^{2}+c^{2}}$. First square the components, then add them and take the square root at the end. Ex: $u+4v=(13,3,8)$ and its magnitude) is $sqrt {169+9+64}=sqrt {242}$.
answered Jan 17 at 6:18
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
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$begingroup$
$$|u+4v|=|(1,3,0)+4(3,0,2)|=|(13,3,8)|=...$$
I got $11sqrt2.$
answered Jan 17 at 6:20
Michael RozenbergMichael Rozenberg
110k1896201
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2 Answers
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active
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2 Answers
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$begingroup$
The norm (or magnitude) of $(a,b,c)$ is $sqrt {a^{2}+b^{2}+c^{2}}$. First square the components, then add them and take the square root at the end. Ex: $u+4v=(13,3,8)$ and its magnitude) is $sqrt {169+9+64}=sqrt {242}$.
answered Jan 17 at 6:18
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
$endgroup$
add a comment |
$begingroup$
The norm (or magnitude) of $(a,b,c)$ is $sqrt {a^{2}+b^{2}+c^{2}}$. First square the components, then add them and take the square root at the end. Ex: $u+4v=(13,3,8)$ and its magnitude) is $sqrt {169+9+64}=sqrt {242}$.
answered Jan 17 at 6:18
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
$endgroup$
add a comment |
$begingroup$
The norm (or magnitude) of $(a,b,c)$ is $sqrt {a^{2}+b^{2}+c^{2}}$. First square the components, then add them and take the square root at the end. Ex: $u+4v=(13,3,8)$ and its magnitude) is $sqrt {169+9+64}=sqrt {242}$.
answered Jan 17 at 6:18
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
$endgroup$
The norm (or magnitude) of $(a,b,c)$ is $sqrt {a^{2}+b^{2}+c^{2}}$. First square the components, then add them and take the square root at the end. Ex: $u+4v=(13,3,8)$ and its magnitude) is $sqrt {169+9+64}=sqrt {242}$.
answered Jan 17 at 6:18
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
answered Jan 17 at 6:18
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
answered Jan 17 at 6:18
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
answered Jan 17 at 6:18
Kavi Rama MurthyKavi Rama Murthy
74.4k53270
74.4k53270
add a comment |
add a comment |
$begingroup$
$$|u+4v|=|(1,3,0)+4(3,0,2)|=|(13,3,8)|=...$$
I got $11sqrt2.$
answered Jan 17 at 6:20
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
$$|u+4v|=|(1,3,0)+4(3,0,2)|=|(13,3,8)|=...$$
I got $11sqrt2.$
answered Jan 17 at 6:20
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
add a comment |
$begingroup$
$$|u+4v|=|(1,3,0)+4(3,0,2)|=|(13,3,8)|=...$$
I got $11sqrt2.$
answered Jan 17 at 6:20
Michael RozenbergMichael Rozenberg
110k1896201
$endgroup$
$$|u+4v|=|(1,3,0)+4(3,0,2)|=|(13,3,8)|=...$$
I got $11sqrt2.$
answered Jan 17 at 6:20
Michael RozenbergMichael Rozenberg
110k1896201
answered Jan 17 at 6:20
Michael RozenbergMichael Rozenberg
110k1896201
answered Jan 17 at 6:20
Michael RozenbergMichael Rozenberg
110k1896201
answered Jan 17 at 6:20
Michael RozenbergMichael Rozenberg
110k1896201
110k1896201
add a comment |
add a comment |
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