Find 3-dimensional vectors `vec v_1,vec v_2,vec v_3` satisfying `vec v_1* vec v_1=4,vec v_1* vec v_2=-2,vec v_1* vec v_3=6,`
`vec v_2* vec v_2=2 , vec v_2 *vec v_3=-5,vec v_3* vec v_3=29`

Explanation

Given data are insufficient to uniquely determine the three vectors as there are only six equations involving nine variable.
Therefore, we can obtain infinite number of sets of three vectors, `vecv_(1),vecv_(2) and vecv_(3)` ,satisfying these conditions,
from the given data, we get
`vecv_(1).vecv_(2)=4Rightarrow |vecv_(1)|=2`
`vecv_(2).vecv_(2)=2 Rightarrow|vecv_(2)|=sqrt2`
`vecv_(3).vecv_(3) =29Rightarrow |vecv_(3)|=sqrt29`
Also `vecv_(1).vecv_(2)=-2`
` Rightarrow |v_(1)||v_(2)| cos theta=-2`
(where `theta` is the angle between `vecv_(1) and vecv_(2)`)
`or cos theta = (-1)/sqrt2`
`or theta-135^(@)`
Since any two vectors are always coplanar, let us suppose that `vecv_(1) and vecv_(2)` are in the x-y plane. Let `vecv_(1) ` be along the possitive direaction of the x-axis. then
`vecv_(1) = - hati +- hatj`
As `vecv_(2)` makes an angle `135^(@)` with `vecv_(1)` and lies in the x-y plane, also keeping in mind `|vecv_(2)| = sqrt2` we obtain
`vecv_(2) =-hati +- hatj`
Again let `vecv_(3) = alphahati + betahatj + gammahatk`
`vecv_(3) .vecv_(1)=6 Rightarrow2 alpha=6or alpha=3`
`and vecv_(3).vecv_(2)=-5 Rightarrow alpha+- beta=-5 or beta=+-2`
Also `|vecv_(3)|=sqrt29Rightarrowalpha^(2)+beta^(2)+gamma^(2)=29`
`Rightarrow gamma=+-4`
Hence `vecv_(3)=3hati+-2hatj+-34hatk`