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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`

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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`
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