Question and Answers Forum

All Questions      Topic List

Vector Questions

Previous in All Question      Next in All Question      

Previous in Vector      Next in Vector      

Question Number 50755 by ajfour last updated on 19/Dec/18

Commented by ajfour last updated on 19/Dec/18

Find the angle between the two triangular (coloured) planes.

$${Find}\:{the}\:{angle}\:{between}\:{the}\:{two} \\ $$$${triangular}\:\left({coloured}\right)\:{planes}. \\ $$

Answered by mr W last updated on 20/Dec/18

let θ=angle between both coloured triangles Δ_(abc) =((√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)))/4) Δ_(pqb) =((√((p+q+b)(−p+q+b)(p−q+b)(p+q−b)))/4) Δ_(pqb) =((bh_b )/2) ⇒h_b =((2Δ_(pqb) )/b)=altitude to base line b ⇒h_b =((√((p+q+b)(−p+q+b)(p−q+b)(p+q−b)))/(2b)) V=((Δ_(abc) h)/3) V=((√(a^2 q^2 (−a^2 −q^2 +b^2 +l^2 +c^2 +p^2 )+b^2 l^2 (a^2 +q^2 −b^2 −l^2 +c^2 +p^2 )+c^2 p^2 (a^2 +q^2 +b^2 +l^2 −c^2 −p^2 )−a^2 b^2 c^2 −a^2 l^2 p^2 −b^2 q^2 p^2 −c^2 q^2 l^2 ))/(12)) ⇒h=((3V)/Δ_(abc) )=altitude to base triangle abc h=h_b sin θ ⇒θ=sin^(−1) (h/h_b )=sin^(−1) ((3bV)/(2Δ_(abc) Δ_(pqb) )) ⇒θ=sin^(−1) ((2b(√(a^2 q^2 (−a^2 −q^2 +b^2 +l^2 +c^2 +p^2 )+b^2 l^2 (a^2 +q^2 −b^2 −l^2 +c^2 +p^2 )+c^2 p^2 (a^2 +q^2 +b^2 +l^2 −c^2 −p^2 )−a^2 b^2 c^2 −a^2 l^2 p^2 −b^2 q^2 p^2 −c^2 q^2 l^2 )))/(√((a+b+c)(−a+b+c)(a−b+c)(a+b−c)(p+q+b)(−p+q+b)(p−q+b)(p+q−b))))

$${let}\:\theta={angle}\:{between}\:{both}\:{coloured}\:{triangles} \\ $$$$\Delta_{{abc}} =\frac{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)}}{\mathrm{4}} \\ $$$$\Delta_{{pqb}} =\frac{\sqrt{\left({p}+{q}+{b}\right)\left(−{p}+{q}+{b}\right)\left({p}−{q}+{b}\right)\left({p}+{q}−{b}\right)}}{\mathrm{4}} \\ $$$$\Delta_{{pqb}} =\frac{{bh}_{{b}} }{\mathrm{2}} \\ $$$$\Rightarrow{h}_{{b}} =\frac{\mathrm{2}\Delta_{{pqb}} }{{b}}={altitude}\:{to}\:{base}\:{line}\:{b} \\ $$$$\Rightarrow{h}_{{b}} =\frac{\sqrt{\left({p}+{q}+{b}\right)\left(−{p}+{q}+{b}\right)\left({p}−{q}+{b}\right)\left({p}+{q}−{b}\right)}}{\mathrm{2}{b}} \\ $$$${V}=\frac{\Delta_{{abc}} {h}}{\mathrm{3}} \\ $$$${V}=\frac{\sqrt{{a}^{\mathrm{2}} {q}^{\mathrm{2}} \left(−{a}^{\mathrm{2}} −{q}^{\mathrm{2}} +{b}^{\mathrm{2}} +{l}^{\mathrm{2}} +{c}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)+{b}^{\mathrm{2}} {l}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{q}^{\mathrm{2}} −{b}^{\mathrm{2}} −{l}^{\mathrm{2}} +{c}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)+{c}^{\mathrm{2}} {p}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{q}^{\mathrm{2}} +{b}^{\mathrm{2}} +{l}^{\mathrm{2}} −{c}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} −{a}^{\mathrm{2}} {l}^{\mathrm{2}} {p}^{\mathrm{2}} −{b}^{\mathrm{2}} {q}^{\mathrm{2}} {p}^{\mathrm{2}} −{c}^{\mathrm{2}} {q}^{\mathrm{2}} {l}^{\mathrm{2}} }}{\mathrm{12}} \\ $$$$\Rightarrow{h}=\frac{\mathrm{3}{V}}{\Delta_{{abc}} }={altitude}\:{to}\:{base}\:{triangle}\:{abc} \\ $$$${h}={h}_{{b}} \:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\theta=\mathrm{sin}^{−\mathrm{1}} \frac{{h}}{{h}_{{b}} }=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}{bV}}{\mathrm{2}\Delta_{{abc}} \Delta_{{pqb}} } \\ $$$$\Rightarrow\theta=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{2}{b}\sqrt{{a}^{\mathrm{2}} {q}^{\mathrm{2}} \left(−{a}^{\mathrm{2}} −{q}^{\mathrm{2}} +{b}^{\mathrm{2}} +{l}^{\mathrm{2}} +{c}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)+{b}^{\mathrm{2}} {l}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{q}^{\mathrm{2}} −{b}^{\mathrm{2}} −{l}^{\mathrm{2}} +{c}^{\mathrm{2}} +{p}^{\mathrm{2}} \right)+{c}^{\mathrm{2}} {p}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{q}^{\mathrm{2}} +{b}^{\mathrm{2}} +{l}^{\mathrm{2}} −{c}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} {b}^{\mathrm{2}} {c}^{\mathrm{2}} −{a}^{\mathrm{2}} {l}^{\mathrm{2}} {p}^{\mathrm{2}} −{b}^{\mathrm{2}} {q}^{\mathrm{2}} {p}^{\mathrm{2}} −{c}^{\mathrm{2}} {q}^{\mathrm{2}} {l}^{\mathrm{2}} }}{\sqrt{\left({a}+{b}+{c}\right)\left(−{a}+{b}+{c}\right)\left({a}−{b}+{c}\right)\left({a}+{b}−{c}\right)\left({p}+{q}+{b}\right)\left(−{p}+{q}+{b}\right)\left({p}−{q}+{b}\right)\left({p}+{q}−{b}\right)}} \\ $$

Commented by mr W last updated on 20/Dec/18

Commented by ajfour last updated on 20/Dec/18

Great Sir! Thanks.

$${Great}\:{Sir}!\:{Thanks}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com