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Question Number 118277 by hmh2k20 last updated on 16/Oct/20

(23) Let a^→ = ((( 7)),((24)) ) and b^→ = ((( 3)),((−4)) ) a^→ .b^→ = (7)(3) + (24)(−4) = 21−96 = −75 a = (√(7^2 +24^2 )) = (√(49+576)) = (√(625)) = 25 b = (√(3^2 +(−4)^2 )) = (√(9+16)) = (√(25)) = 5 a.b = 25 . 5 = 125 Let θ be the angle between a^→ and b^→ . cosθ = ((a^→ .b^→ )/(a.b)) = ((−75)/(125)) = −0.6 = cos126.87° ∴ θ = 126.87°

$$\left(\mathrm{23}\right)\:\mathrm{Let}\:\overset{\rightarrow} {\mathrm{a}}\:=\:\begin{pmatrix}{\:\:\mathrm{7}}\\{\mathrm{24}}\end{pmatrix}\:\:\:\mathrm{and}\:\:\overset{\rightarrow} {\mathrm{b}}\:=\:\begin{pmatrix}{\:\:\:\mathrm{3}}\\{−\mathrm{4}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}\:=\:\left(\mathrm{7}\right)\left(\mathrm{3}\right)\:+\:\left(\mathrm{24}\right)\left(−\mathrm{4}\right)\:=\:\mathrm{21}−\mathrm{96}\:=\:−\mathrm{75} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}\:=\:\:\sqrt{\mathrm{7}^{\mathrm{2}} +\mathrm{24}^{\mathrm{2}} }\:=\:\sqrt{\mathrm{49}+\mathrm{576}}\:=\:\sqrt{\mathrm{625}}\:=\:\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:=\:\:\sqrt{\mathrm{3}^{\mathrm{2}} +\left(−\mathrm{4}\right)^{\mathrm{2}} }\:=\:\sqrt{\mathrm{9}+\mathrm{16}}\:=\:\sqrt{\mathrm{25}}\:=\:\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{a}.\mathrm{b}\:=\:\mathrm{25}\:.\:\mathrm{5}\:=\:\:\mathrm{125}\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{Let}\:\theta\:\mathrm{be}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\overset{\rightarrow} {\mathrm{a}}\:\mathrm{and}\:\overset{\rightarrow} {\mathrm{b}}. \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{cos}\theta\:=\:\frac{\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}}{\mathrm{a}.\mathrm{b}}\:=\:\frac{−\mathrm{75}}{\mathrm{125}}\:=\:−\mathrm{0}.\mathrm{6}\:=\:\mathrm{cos126}.\mathrm{87}° \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore\:\:\theta\:=\:\mathrm{126}.\mathrm{87}°\:\:\:\: \\ $$

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