Question 1049379
1.(4,2,1)=a(0,2,1)+b(2,1,1)+c(1,3,-1)
(4,2,1)=(0a+2b+c,2a+b+3c,a+b-c)
Using Cramer's rule,
{{{A=(matrix(3,3,
0,2,1,
2,1,3,
1,1,-1))}}}
{{{abs(A)=11}}}
.
.
.
{{{A[a]=(matrix(3,3,
4,2,1,
2,1,3,
1,1,-1))}}}
{{{abs(A[a])=-5}}}
.
.
.
{{{A[b]=(matrix(3,3,
0,4,1,
2,2,3,
1,1,-1))}}}
{{{abs(A[b])=20}}}
.
.
.
{{{A[c]=(matrix(3,3,
0,2,4,
2,1,2,
1,1,1))}}}
{{{abs(A[c])=4}}}
.
.
.
{{{a=abs(A[a])/abs(A)=-5/11}}}
{{{b=abs(A[b])/abs(A)=20/11}}}
{{{c=abs(A[c])/abs(A)=4/11}}}
So then,
(4,2,1)=(-5/11)*(0,2,1)+(20/11)*(2,1,1)+(4/11)*(1,3,-1)
.
.
.
2.i) {{{abs(V[1])=sqrt(0^2+(-4)^2+7^2)}}}
ii) V[1]+V[2]=(0+3,-4-2,7-1)=(3,-6,6)
iii){{{abs(V[1]+V[2])=sqrt(3^2+(-6)^2+6^2)}}}
iv) Use the dot product,
{{{V[1]*V[2]=0*3+(-4)(-2)+7(-1)=8-7=1}}}
{{{cos(theta)=(V[1]*V[2])/(abs(V[1])abs(V[2]))}}}
{{{abs(V[2])=sqrt(3^2+(-2)^2+(-1)^2)}}}
Plug in the values to get the answer.
v) From previous, {{{V[1]*V[2]=1}}}
vi)
{{{V[1]xV[2]=(matrix(3,3,
i,j,k,
0,-4,7,
3,-2,-1))}}}
{{{V[1]xV[2]=(matrix(1,3,
18,21,12))}}}