Answer :
To find the magnitude of vector [tex]\(\vec{A}\)[/tex], we will use the formula for the magnitude of a vector in a plane (2-dimensional space).
Given:
[tex]\[ \vec{A} = 3\hat{i} + 2\hat{j} \][/tex]
The magnitude of a vector [tex]\(\vec{A} = a\hat{i} + b\hat{j}\)[/tex] is calculated using the formula:
[tex]\[ |\vec{A}| = \sqrt{a^2 + b^2} \][/tex]
For [tex]\(\vec{A}\)[/tex], [tex]\(a = 3\)[/tex] and [tex]\(b = 2\)[/tex]:
[tex]\[ |\vec{A}| = \sqrt{(3)^2 + (2)^2} \][/tex]
Let's calculate each term inside the square root:
[tex]\[ (3)^2 = 9 \][/tex]
[tex]\[ (2)^2 = 4 \][/tex]
Now, sum these:
[tex]\[ 9 + 4 = 13 \][/tex]
Finally, take the square root of the sum:
[tex]\[ |\vec{A}| = \sqrt{13} \approx 3.605551275463989 \][/tex]
Therefore, the magnitude of [tex]\(\vec{A} = 3\hat{i} + 2\hat{j}\)[/tex] is approximately [tex]\(3.605551275463989\)[/tex].
Given:
[tex]\[ \vec{A} = 3\hat{i} + 2\hat{j} \][/tex]
The magnitude of a vector [tex]\(\vec{A} = a\hat{i} + b\hat{j}\)[/tex] is calculated using the formula:
[tex]\[ |\vec{A}| = \sqrt{a^2 + b^2} \][/tex]
For [tex]\(\vec{A}\)[/tex], [tex]\(a = 3\)[/tex] and [tex]\(b = 2\)[/tex]:
[tex]\[ |\vec{A}| = \sqrt{(3)^2 + (2)^2} \][/tex]
Let's calculate each term inside the square root:
[tex]\[ (3)^2 = 9 \][/tex]
[tex]\[ (2)^2 = 4 \][/tex]
Now, sum these:
[tex]\[ 9 + 4 = 13 \][/tex]
Finally, take the square root of the sum:
[tex]\[ |\vec{A}| = \sqrt{13} \approx 3.605551275463989 \][/tex]
Therefore, the magnitude of [tex]\(\vec{A} = 3\hat{i} + 2\hat{j}\)[/tex] is approximately [tex]\(3.605551275463989\)[/tex].
