Question

Find the values of \({\rm{x}}\)such that the vectors \(\langle {\rm{3,2,x}}\rangle \)and \(\langle {\rm{2x,4,x}}\rangle \)are orthogonal.

Short answer

The value of \({\rm{x}}\)is\({\rm{( - 4, - 2)}}\).

Step-by-step solution

Find the value of \({\rm{x}}\).

\(\langle {\rm{3,2,x}}\rangle \cdot \langle {\rm{2x,4,x}}\rangle = 0\)The dot product two orthogonal vectors is\(0\).

Known \(\left\langle {{{\rm{a}}_{\rm{1}}}{\rm{,}}{{\rm{a}}_{\rm{2}}}{\rm{,}}{{\rm{a}}_{\rm{3}}}} \right\rangle \cdot \left\langle {{{\rm{b}}_{\rm{1}}}{\rm{,}}{{\rm{b}}_{\rm{2}}}{\rm{,}}{{\rm{b}}_{\rm{3}}}} \right\rangle {\rm{ = }}{{\rm{a}}_{\rm{1}}}{{\rm{b}}_{\rm{1}}}{\rm{ + }}{{\rm{a}}_{\rm{2}}}{{\rm{b}}_{\rm{2}}}{\rm{ + }}{{\rm{a}}_{\rm{3}}}{{\rm{b}}_{\rm{3}}}\)

\({\rm{6x + 8 + }}{{\rm{x}}^{\rm{2}}}{\rm{ = 0}}\)

Re-arrange,

\({{\rm{x}}^{\rm{2}}}{\rm{ + 6x + 8 = 0}}\)

\({\rm{(x + 4)(x + 2) = 0}}\)

Factor the trinomial to solve\({\rm{x}}\).

\({\rm{x = - 4 or x = - 2}}\)

Result.

Therefore, the value of \({\rm{x}}\)is\({\rm{( - 4, - 2)}}\).