Question

find a,b,c for which the matrix (top: a, 1/sqr2, -1/sqr2; middle: b,1/sqr6, 1/sqr6; bottom: c, 1/sqr3, 1/sqr3 ) is orthogonal. Are the values of a, b,c unique? Explain. Help, please

Answer 1

\[\left[\begin{matrix}a & \frac{ 1 }{ \sqrt{2} }& \frac{ -1 }{ \sqrt{2} } \\ b & \frac{ 1 }{ \sqrt{6} }& \frac{ 1 }{ \sqrt{6} }\\c & \frac{ 1 }{ \sqrt{3} }& \frac{ 1 }{ \sqrt{3} }\end{matrix}\right]\]

Answer 2

the columns should be orthogonal unit vectors..
so, the last two columns:
\(-{1\over2}+{1\over6}+{1\over3}=0\)
and \({1\over2}+{1\over6}+{1\over3}=1\)
so, the last two columns are orthogonal unit vectors.

Answer 3

the first column has to be
(1) a unit vector so,
\(\qquad a^2+b^2+c^2=1\)
(2) orthogonal to column 2 so,
\(\qquad {a\over\sqrt{2}}+{b\over\sqrt{6}}+{c\over\sqrt{3}}=0\)
(3) orthogonal to column 3 so,
\(\qquad {-a\over\sqrt{2}}+{b\over\sqrt{6}}+{c\over\sqrt{3}}=0\)

Answer 4

you mean orthonormal?

Answer 5

eq (2) - eq(3) gives:
\(\large2{a\over\sqrt{2}}=0\implies a=0\)

Answer 6

then we have
\[b^2=1-c^2\\
{\rm and}\qquad{b\over\sqrt{6}}=-{c\over\sqrt{3}}\implies b^2=2c^2\\
\therefore 3c^2=1\implies \boxed{\large\color{red} {c=\pm{1\over\sqrt{3}}}}\]

Answer 7

and
\[\boxed{\large\color{red} {b=\pm{\sqrt{2\over3}}}}\]

Answer 8

so, there is no unique asnwer but "4" possible solutions

Answer 9

\[(a,b,c) = (0,\sqrt{2/3},1/\sqrt{3})\\
(a,b,c) = (0,\sqrt{2/3},-1/\sqrt{3})\\
(a,b,c) = (0,-\sqrt{2/3},1/\sqrt{3})\\
(a,b,c) = (0,-\sqrt{2/3},-1/\sqrt{3})
\]

Answer 10

now, which of these gives the orthogonal componetns?

Answer 11

@Hoa you follow?

Answer 12

we notice that when "b" is positive, "c" HAS to be negative and vice-versa
so, only two answers!!!
\[(a,b,c) = (0,\sqrt{2/3},-1/\sqrt{3})\qquad{\rm and}\\
(a,b,c) = (0,-\sqrt{2/3},1/\sqrt{3})
\]

Answer 13

why, is it not sum of them =1

Answer 14

not just sum,
sum of squares!

Answer 15

sum of square, so no matter how the sign is , they satisfy

Answer 16

exactly. but that is not enough!

Answer 17

check these results:
(1) http://wolfr.am/12a2cVi
(2) http://wolfr.am/XDGI4v

Answer 18

ok, let it there, I'll be back to check, is it done?

Answer 19

these are only two solutions. at the wolfram site, try to set them both positve and both negative and see if the result is still oorthogonal!