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Question Number 105356 by bemath last updated on 28/Jul/20

Given a cube ABCD.EFGH.  if α is the angle BEH & BEG ,  find the value of cos α ?

$$\mathcal{G}{iven}\:{a}\:{cube}\:{ABCD}.{EFGH}. \\ $$$${if}\:\alpha\:{is}\:{the}\:{angle}\:{BEH}\:\&\:{BEG}\:, \\ $$$${find}\:{the}\:{value}\:{of}\:\mathrm{cos}\:\alpha\:? \\ $$

Answered by john santu last updated on 28/Jul/20

Commented by john santu last updated on 28/Jul/20

with normal vector n_1 ^→  & n_2 ^→   n_1 ^→ = determinant (((−2       0       0)),((    0        2    −2)))=(0,−4,−4)   n_2 ^→ =  determinant (((   0      2     −2)),((−2    2         0)))=(4,4,4)  cos α = ((∣n_1 ^→ ○n_2 ^→ ∣)/(∣n_1 ^→ ∣.∣n_2 ^→ ∣)) = ((32)/(4(√2) .4(√3))) = ((√6)/3)  (JS ♠⧫)

$${with}\:{normal}\:{vector}\:\overset{\rightarrow} {{n}}_{\mathrm{1}} \:\&\:\overset{\rightarrow} {{n}}_{\mathrm{2}} \\ $$$$\overset{\rightarrow} {{n}}_{\mathrm{1}} =\begin{vmatrix}{−\mathrm{2}\:\:\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\mathrm{0}}\\{\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:\:\mathrm{2}\:\:\:\:−\mathrm{2}}\end{vmatrix}=\left(\mathrm{0},−\mathrm{4},−\mathrm{4}\right)\: \\ $$$$\overset{\rightarrow} {{n}}_{\mathrm{2}} =\:\begin{vmatrix}{\:\:\:\mathrm{0}\:\:\:\:\:\:\mathrm{2}\:\:\:\:\:−\mathrm{2}}\\{−\mathrm{2}\:\:\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\mathrm{0}}\end{vmatrix}=\left(\mathrm{4},\mathrm{4},\mathrm{4}\right) \\ $$$$\mathrm{cos}\:\alpha\:=\:\frac{\mid\overset{\rightarrow} {{n}}_{\mathrm{1}} \circ\overset{\rightarrow} {{n}}_{\mathrm{2}} \mid}{\mid\overset{\rightarrow} {{n}}_{\mathrm{1}} \mid.\mid\overset{\rightarrow} {{n}}_{\mathrm{2}} \mid}\:=\:\frac{\mathrm{32}}{\mathrm{4}\sqrt{\mathrm{2}}\:.\mathrm{4}\sqrt{\mathrm{3}}}\:=\:\frac{\sqrt{\mathrm{6}}}{\mathrm{3}} \\ $$$$\left({JS}\:\spadesuit\blacklozenge\right)\: \\ $$

Commented by bemath last updated on 28/Jul/20

great..

$${great}.. \\ $$

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