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Sagot :
Given:
A: (3, -6, 2)
B: (6, -7, -1)
C: (0, -1, 5)
Find:
a. component form of the vectors u (from A to B) and v (from A to C)
b. The angle between vectors u and v.
Solution:
a. In order to get the component of the vector u (from A to B), we simply have to subtract each corresponding component of A from B. Thus, we have:
[tex](6-3,-7--6,-1-2)[/tex][tex]VectorU=(3,-1,-3)[/tex]For Vector V which is from A to C, we simply have to subtract each corresponding component of A from C.
[tex](0-3,-1--6,5-2)[/tex][tex]VectorV=(-3,5,3)[/tex]The component form of Vector U is (3, -1, -3) while Vector V is (-3, 5, 3).
b. To determine the angle between them, here are the steps:
Calculate the dot product of Vector U and V by getting the sum of the product of each corresponding component of U and V.
[tex](3\times-3)+(-1\times5)+(-3\times3)[/tex][tex]-9+(-5)+(-9)\Rightarrow-23[/tex]The dot product is -23.
Next, calculate the magnitude of each vector.
To get the magnitude of a vector, square each component of the vector and add them. After that, get the square root of the sum.
[tex]|u|=\sqrt{x^2+y^2+z^2}\Rightarrow|u|=\sqrt{3^2+(-1)^2+(-3)^2}[/tex][tex]|u|=\sqrt{9+1+9}\Rightarrow|u|=\sqrt{19}[/tex]The magnitude of vector u is √19.
[tex]|v|=\sqrt{(-3)^2+5^2+3^2}\Rightarrow|v|=\sqrt{9+25+9}\Rightarrow|v|=\sqrt{43}[/tex]The magnitude of vector v is √43.
Multiply the magnitude of vectors u and v.
[tex]\sqrt{19}\times\sqrt{43}=\sqrt{817}[/tex]Divide the dot product by the product of the two magnitudes.
[tex]-23\div\sqrt{817}=-0.8046681428[/tex]Then, multiply the inverse of cosine by the result above.
[tex]\begin{gathered} \theta=cos^{-1}(-0.8046681428) \\ \end{gathered}[/tex][tex]\theta=143.578\approx143.58\degree[/tex]The angle between Vector U and V is approximately 143.58°.
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