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Question Number 168086 by mokys last updated on 02/Apr/22

Answered by som(math1967) last updated on 03/Apr/22

4. Angle between 2x+y−2z=5  and 3x−6y−2z=7   θ=cos^(−1) ((2×3+1×(−6)+(−2)×(−2))/( (√(2^2 +1^2 +(−2)^2 ))(√(3^2 +(−6)^2 +(−2)^2 ))))  =cos^(−1) (4/( (√9)(√(49))))=cos^(−1) (4/(21))

$$\mathrm{4}.\:{Angle}\:{between}\:\mathrm{2}{x}+{y}−\mathrm{2}{z}=\mathrm{5} \\ $$$${and}\:\mathrm{3}{x}−\mathrm{6}{y}−\mathrm{2}{z}=\mathrm{7} \\ $$$$\:\theta=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{2}×\mathrm{3}+\mathrm{1}×\left(−\mathrm{6}\right)+\left(−\mathrm{2}\right)×\left(−\mathrm{2}\right)}{\:\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} +\left(−\mathrm{2}\right)^{\mathrm{2}} }\sqrt{\mathrm{3}^{\mathrm{2}} +\left(−\mathrm{6}\right)^{\mathrm{2}} +\left(−\mathrm{2}\right)^{\mathrm{2}} }} \\ $$$$=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{4}}{\:\sqrt{\mathrm{9}}\sqrt{\mathrm{49}}}=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{4}}{\mathrm{21}} \\ $$

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