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Question Number 18284 by tawa tawa last updated on 18/Jul/17

$$\mathrm{If}\mathrm{k}\mathrm{is}\mathrm{any}\mathrm{possible}\mathrm{number}.\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{size}\mathrm{of}\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{vectors}$$$$\mathrm{a}(\mathrm{K},\mathrm{k})\mathrm{and}\mathrm{b}(-3,4)$$

Answered by Arnab Maiti last updated on 18/Jul/17

$$\mathrm{let}\overrightarrow{\mathrm{A}}=\mathrm{K}\hat{\mathrm{i}}+\mathrm{k}\hat{\mathrm{j}}$$$$\overrightarrow{\mathrm{B}}=-3\hat{\mathrm{i}}+4\hat{\mathrm{j}}$$$$\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}=-3\mathrm{K}+4\mathrm{k}$$$$\overrightarrow{\mathrm{B}}.\overrightarrow{\mathrm{B}}={\mathrm{B}}^2=9+16=25$$$$\mathrm{B}=5$$$$\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{A}}={\mathrm{A}}^2={\mathrm{K}}^2+{\mathrm{k}}^2$$$$\mathrm{A}=\sqrt{{\mathrm{K}}^2+{\mathrm{k}}^2}$$$$\overrightarrow{\mathrm{A}}.\overrightarrow{\mathrm{B}}=\mathrm{ABcos}\theta$$$$\cos \theta =\frac{-3\mathrm{K}+4\mathrm{k}}{5\sqrt{{\mathrm{K}}^2+{\mathrm{k}}^2}}$$$$\theta ={\cos }^{-1}\left(\frac{-3\mathrm{K}+4\mathrm{k}}{5\sqrt{{\mathrm{K}}^2+{\mathrm{k}}^2}}\right)$$

Commented by tawa tawa last updated on 18/Jul/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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