Answer:
a. [tex]2a+3b=(11,-4,-1)\\[/tex]
b. [tex]|b|=\sqrt{14}[/tex].
c. [tex]a.b=-1[/tex]
d. [tex]comp_{a}b=\frac{-1}{\sqrt{6}}[/tex]
e. [tex]proj_{a}b= \frac{-1}{6}[1,1-2]\\[/tex]
f. [tex]alpha=96^{0}[/tex]
Step-by-step explanation:
The vector representation of A,B,C is written as
a=[1,1,-2]
b=[3,-2,1]
c=[0,1,-5].
a.To determine the scalar product of 2a+3b we have
[tex]2a+3b=2(1,1,-2)+3(3,-2,1)\\2a+3b=(2,2,-4)+(9,-6,3)\\[/tex]
we add component by component w arrive at
[tex]2a+3b=(11,-4,-1)\\[/tex]
b. we determine the magnitude of vector b i.e |b|. since b=[3,-2,1]
[tex]|b|=\sqrt{3^{2}+(-2)^{2} +1^{2}} \\|b|=\sqrt{9+4+1}\\ |b|=\sqrt{14}[/tex].
c.a.b is the dot product of vector a and vector b.
[tex]a.b=[1,1,-2].[3,-2,1]\\a.b=(1*3)+(1*-2)+(-2*1)\\a.b=3-2-2\\a.b=-1[/tex]
d. to find the component of b on a we use the expression
[tex]comp_{a}b=\frac{b.a}{|a|}[/tex]
where [tex]|a|=\sqrt{1^{2}+1^{2}+(-2)^{2}+ } \\|a|=\sqrt{6} \\[/tex]
Hence if we substitute values we have
[tex]comp_{a}b=\frac{[3,-2,1].[1,1,-2]}{\sqrt{6}}[/tex]
but earlier a.b=b.a=-1
[tex]comp_{a}b=\frac{-1}{\sqrt{6}}[/tex]
e. to find the projection of b on a we have the expression
[tex]proj_{a}b= \frac{a.b}{|a|^{2}}a \\[/tex]
If we substitute values we arrive at
[tex]proj_{a}b= \frac{[3,-2,1].[1,1,-2]}{|a|^{2}}a \\[/tex]
also recall that earlier we calculated a.b=b.a=-1 and
[tex]|a|=\sqrt{6} \\[/tex]
Hence [tex]proj_{a}b= \frac{-1}{\sqrt{6}^{2}}[1,1-2] \\[/tex]
Hence [tex]proj_{a}b= \frac{-1}{6}[1,1-2] \\[/tex]
f. to determine the angle between vector "a" and "b" we use
[tex]a.b=|a||b|cos\alpha[/tex]
where ∝ is the required angle . If we substitute values into the expression we arrive at
[tex]-1=\sqrt{14*6}cos\alpha\\\alpha =cos^{-1}(-1/\sqrt{84})\\\alpha=96^{0}[/tex]
