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Question Number 28640 by sk9236421@gmail.com last updated on 28/Jan/18

$$\mathrm{Each}\mathrm{of}\mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{vectors}\mathbf{a},$$$$\mathbf{b}\mathrm{and}\mathbf{c}\mathrm{is}\mathrm{equal}\mathrm{to}60°.\mathrm{If}\mid \mathbf{a}\mid =4,\mid \mathbf{b}\mid =2$$$$\mathrm{and}\mid \mathbf{c}\mid =6,\mathrm{then}\mathrm{the}\mathrm{modulus}\mathrm{of}$$$$\mathbf{a}+\mathbf{b}+\mathbf{c}\mathrm{is}$$

Commented by ajfour last updated on 28/Jan/18

Commented by ajfour last updated on 28/Jan/18

Answered by ajfour last updated on 28/Jan/18

$$let\overline{b}=6\hat{i}and$$$$\overline{a}=2\hat{i}+2\sqrt{3}\hat{j}$$$$\overline{c}=x\hat{i}+y\hat{j}+z\hat{k};\mid \overline{c}\mid =2$$$$\overline{b}.\overline{c}=\mid \overline{b}\mid \mid \overline{c}\mid \cos 60°$$$$\Rightarrow 6x=6\times 2\times \frac{1}{2}$$$$\Rightarrow x=1$$$$\overline{a}.\overline{c}=\mid \overline{a}\mid \mid \overline{c}\mid \cos 60°$$$$\Rightarrow 2x+2\sqrt{3}y=4\times 2\times \frac{1}{2}$$$$\Rightarrow 2+2\sqrt{3}y=4$$$$\Rightarrow y=\frac{1}{\sqrt{3}}$$$$as\mid \overline{c}\mid =2\Rightarrow x^2+y^2+z^2=4$$$$so1+\frac{1}{3}+z^2=4$$$$\Rightarrow z=\pm \frac{2\sqrt{2}}{\sqrt{3}}$$$$\mid \overline{a}+\overline{b}+\overline{c}\mid =\mid 6\hat{i}+2\hat{i}+2\sqrt{3}\hat{j}+\hat{i}+\frac{1}{\sqrt{3}}\hat{j}\pm \frac{2\sqrt{2}}{\sqrt{3}}\hat{k}\mid$$$$=\sqrt{81+\frac{49}{3}+\frac{8}{3}}=\sqrt{81+19}$$$$\mid \overline{a}+\overline{b}+\overline{c}\mid =10.$$

Answered by ajfour last updated on 28/Jan/18

$$\mid \overline{a}+\overline{b}+\overline{c}{\mid }^2=a^2+b^2+c^2+2(\overline{a}.\overline{b}+\overline{b}.\overline{c}+\overline{c}.\overline{a})$$$$=16+4+36+(12+8+24)=100$$$$\mid \overline{a}+\overline{b}+\overline{c}\mid =10.$$

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