Question

Calculate the given quantity if

a=[1,1,-2]
b=[3,-2,1]
c=[0,1,-5]
(a) 2a + 3b
(b) |b|
(c) a.b
(d) compab
(e) projab
(f) The angle between a and b (correct to the nearest degree)

Answer 1

a.
`2a+3b=(11,-4,-1)\\`

b.
`|b|=√(14)`.

c.
`a.b=-1`

d.
`comp_(a)b=(-1)/(√(6))`

e.
`proj_(a)b= (-1)/(6)[1,1-2]\\`

f.
`alpha=96^(0)`

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

`2a+3b=2(1,1,-2)+3(3,-2,1)\\2a+3b=(2,2,-4)+(9,-6,3)\\`

we add component by component w arrive at

`2a+3b=(11,-4,-1)\\`

b. we determine the magnitude of vector b i.e |b|. since b=[3,-2,1]

`|b|=\sqrt{3^(2)+(-2)^(2) +1^(2)} \\|b|=√(9+4+1)\\ |b|=√(14)`.

c.a.b is the dot product of vector a and vector b.

`a.b=[1,1,-2].[3,-2,1]\\a.b=(1*3)+(1*-2)+(-2*1)\\a.b=3-2-2\\a.b=-1`

d. to find the component of b on a we use the expression

`comp_(a)b=(b.a)/(|a|)`

where
`|a|=\sqrt{1^(2)+1^(2)+(-2)^(2)+ } \\|a|=√(6) \\`

Hence if we substitute values we have

`comp_(a)b=([3,-2,1].[1,1,-2])/(√(6))`

but earlier a.b=b.a=-1

`comp_(a)b=(-1)/(√(6))`

e. to find the projection of b on a we have the expression

`proj_(a)b= (a.b)/(|a|^(2))a \\`

If we substitute values we arrive at

`proj_(a)b= ([3,-2,1].[1,1,-2])/(|a|^(2))a \\`

also recall that earlier we calculated a.b=b.a=-1 and

`|a|=√(6) \\`

Hence
`proj_(a)b= (-1)/(√(6)^(2))[1,1-2] \\`

Hence
`proj_(a)b= (-1)/(6)[1,1-2] \\`

f. to determine the angle between vector "a" and "b" we use

`a.b=|a||b|cos\alpha`

where ∝ is the required angle . If we substitute values into the expression we arrive at

`-1=√(14*6)cos\alpha\\\alpha =cos^(-1)(-1/√(84))\\\alpha=96^(0)`