Respuesta :
Answer:
The scalars are s=-1 and t=5 such that sa+tb=<12,−18>
Step-by-step explanation:
Let a=<3,3> and b=<3,-3>, s and t are scalars such that sa+tb=<12,-18>
Let d be a scalar quantity and v=<a, b> be a vector quantity.
Now to multiply the vector v=<a, b> by the scalar quantity d, that is multiply each of its components by d as below:
d<a, b>=<ad,bd>
Given that sa+tb=s<3,3>+t<3,-3>
=<3s,3s>+<3t,-3t>
sa+tb=<3s+3t,3s-3t>
Now equating it to <12,-18> we get
3s+3t=12 and 3s-3t=-18
3(s+t)=12
[tex]s+t=\frac{12}{3}[/tex]
s+t=4
s=4-t
Now substitute the value s=4-t in 3s-3t=-18 we get
3(4-t)-3t=-18
12-3t-3t=-18
-6t=-18-12
-6t=-30
[tex]t=\frac{30}{6}[/tex]
Therefore t=5
And s=4-5=-1
Therefore s=-1 and t=5 such that sa+tb=<12,−18>
The sa+tb = <12, -18> can be written as [tex]-1<3, 3> +\ 5<3, -3> = <12, -18>[/tex].
Given to us,
a = <3, 3>
b = <3, -3>
resultant = <12, -18>
[tex]sa + tb = <12, -18>\\s<3, 3> + t<3, -3> = <12, -18>\\<3s, 3s> + <3t, -3t> = <12, -18>\\[/tex]
calculating corresponding sides,
Solving for x side,
[tex]3s + 3t = 12\\3(s + t) = 12\\(s+t) = \dfrac{12}{3}\\s+t = 4\\s = 4-t[/tex]
solving for y side,
[tex]3s - 3t = -18\\3(s-t) = -18\\s-t = \dfrac{-18}{3}\\s-t = -6[/tex]
substituting the value of s,
[tex]s-t = 6\\(4-t)-t = -6\\4-t-t = -6\\-2t = -6-4\\-2t = -10\\t=\dfrac{10}{2} = 5[/tex]
substituting the value of t,
[tex]s= 4-t\\s= 4-5\\s=-1[/tex]
Hence, the sa+tb = <12, -18> can be written as [tex]-1<3, 3> +\ 5<3, -3> = <12, -18>[/tex].
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