Menu Close

Question-210843

Tinku Tara 0 Comments
Pin




Question Number 210843 by peter frank last updated on 20/Aug/24
Answered by Spillover last updated on 20/Aug/24
let u,v be a two unit vectors with direction cosine (l_(1 ) ,m_1 ,n_1 ) and (l_(2 ) ,m_2 ,n_2 ) respectively cos θ=((u.v)/(∣u∣∣v∣)) u=(l_(1 ) ,m_1 ,n_1 ) v=(l_(2 ) ,m_2 ,n_2 ) u.v=(l_1 l_(2 ) +m_1 m_2 +n_1 n_2 ) ∣u∣=∣v∣=1 (unit vector) cos θ=((u.v)/(∣u∣∣v∣)) cos θ=l_1 l_(2 ) +m_1 m_2 +n_1 n_2 ∣u×v∣=∣u∣∣v∣sin θ ∣u×v∣=sin θ u×v= determinant ((i,j,k),(l_1 ,m_1 ,n_1 ),(l_2 ,n_1 ,n_2 )) u×v=(m_1 n_2 −m_2 n_1 )i+(n_1 l_2 −n_2 l_1 )j+(l_1 m_2 −l_2 m_1 )k ∣u×v∣=sin θ=(√((m_1 n_2 −m_2 n_1 )^2 +(n_1 l_2 −n_2 l_1 )^2 +(l_1 m_2 −l_2 m_1 )^2 )) sin θ=(√((m_1 n_2 −m_2 n_1 )^2 +(n_1 l_2 −n_2 l_1 )^2 +(l_1 m_2 −l_2 m_1 )^2 )) sin^2 θ=(m_1 n_2 −m_2 n_1 )^2 +(n_1 l_2 −n_2 l_1 )^2 +(l_1 m_2 −l_2 m_1 )^2
$${let}\:{u},{v}\:{be}\:{a}\:{two}\:{unit}\:{vectors}\:{with}\:{direction} \\ $$$${cosine}\:\left({l}_{\mathrm{1}\:} ,{m}_{\mathrm{1}} ,{n}_{\mathrm{1}} \right)\:{and}\:\left({l}_{\mathrm{2}\:} ,{m}_{\mathrm{2}} ,{n}_{\mathrm{2}} \right)\:{respectively} \\ $$$$\mathrm{cos}\:\theta=\frac{{u}.{v}}{\mid{u}\mid\mid{v}\mid} \\ $$$${u}=\left({l}_{\mathrm{1}\:} ,{m}_{\mathrm{1}} ,{n}_{\mathrm{1}} \right)\:\:\:\:\:{v}=\left({l}_{\mathrm{2}\:} ,{m}_{\mathrm{2}} ,{n}_{\mathrm{2}} \right) \\ $$$${u}.{v}=\left({l}_{\mathrm{1}} {l}_{\mathrm{2}\:} +{m}_{\mathrm{1}} {m}_{\mathrm{2}} +{n}_{\mathrm{1}} {n}_{\mathrm{2}} \right) \\ $$$$\mid{u}\mid=\mid{v}\mid=\mathrm{1}\:\:\left({unit}\:{vector}\right) \\ $$$$\mathrm{cos}\:\theta=\frac{{u}.{v}}{\mid{u}\mid\mid{v}\mid} \\ $$$$\mathrm{cos}\:\theta={l}_{\mathrm{1}} {l}_{\mathrm{2}\:} +{m}_{\mathrm{1}} {m}_{\mathrm{2}} +{n}_{\mathrm{1}} {n}_{\mathrm{2}} \\ $$$$\mid{u}×{v}\mid=\mid{u}\mid\mid{v}\mid\mathrm{sin}\:\theta\:\:\:\:\:\:\:\mid{u}×{v}\mid=\mathrm{sin}\:\theta \\ $$$${u}×{v}=\begin{vmatrix}{{i}}&{{j}}&{{k}}\\{{l}_{\mathrm{1}} }&{{m}_{\mathrm{1}} }&{{n}_{\mathrm{1}} }\\{{l}_{\mathrm{2}} }&{{n}_{\mathrm{1}} }&{{n}_{\mathrm{2}} }\end{vmatrix} \\ $$$${u}×{v}=\left({m}_{\mathrm{1}} {n}_{\mathrm{2}} −{m}_{\mathrm{2}} {n}_{\mathrm{1}} \right){i}+\left({n}_{\mathrm{1}} {l}_{\mathrm{2}} −{n}_{\mathrm{2}} {l}_{\mathrm{1}} \right){j}+\left({l}_{\mathrm{1}} {m}_{\mathrm{2}} −{l}_{\mathrm{2}} {m}_{\mathrm{1}} \right){k} \\ $$$$\mid{u}×{v}\mid=\mathrm{sin}\:\theta=\sqrt{\left({m}_{\mathrm{1}} {n}_{\mathrm{2}} −{m}_{\mathrm{2}} {n}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({n}_{\mathrm{1}} {l}_{\mathrm{2}} −{n}_{\mathrm{2}} {l}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({l}_{\mathrm{1}} {m}_{\mathrm{2}} −{l}_{\mathrm{2}} {m}_{\mathrm{1}} \right)^{\mathrm{2}} }\: \\ $$$$\mathrm{sin}\:\theta=\sqrt{\left({m}_{\mathrm{1}} {n}_{\mathrm{2}} −{m}_{\mathrm{2}} {n}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({n}_{\mathrm{1}} {l}_{\mathrm{2}} −{n}_{\mathrm{2}} {l}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({l}_{\mathrm{1}} {m}_{\mathrm{2}} −{l}_{\mathrm{2}} {m}_{\mathrm{1}} \right)^{\mathrm{2}} }\: \\ $$$$\mathrm{sin}\:^{\mathrm{2}} \theta=\left({m}_{\mathrm{1}} {n}_{\mathrm{2}} −{m}_{\mathrm{2}} {n}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({n}_{\mathrm{1}} {l}_{\mathrm{2}} −{n}_{\mathrm{2}} {l}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({l}_{\mathrm{1}} {m}_{\mathrm{2}} −{l}_{\mathrm{2}} {m}_{\mathrm{1}} \right)^{\mathrm{2}} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *