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Question Number 111006 by pete last updated on 01/Sep/20

$$\mathrm{The}\mathrm{vectors}\mathbf{p},\mathbf{q}\mathrm{and}\mathbf{r}\mathrm{are}\mathrm{mutially}\mathrm{perpendicularwith}$$$$\mid \mathbf{q}\mid =3\mathrm{and}\mid \mathbf{r}\mid =\sqrt{5.4}.\mathrm{If}\mathrm{X}=7\mathbf{p}+5\mathbf{q}+7\mathbf{r}\mathrm{and}$$$$\mathrm{Y}=2\mathbf{p}+3\mathbf{q}-5\mathbf{r}\mathrm{are}\mathrm{perpendicular},\mathrm{find}\mid \mathbf{p}\mid .$$

Commented by kaivan.ahmadi last updated on 01/Sep/20

$$p.q=p.r=q.r=0$$$$X.Y=0\Rightarrow (7p+5q+7r).(2p+3q-5r)=$$$$14p.p+15q.q-35r.r=14\mid p{\mid }^2+15\mid q{\mid }^2-35\mid r{\mid }^2=$$$$14\mid p{\mid }^2+15\times 9-35\times 5.4=14\mid p{\mid }^2+135-189=$$$$14\mid p{\mid }^2-=0\Rightarrow \mid p{\mid }^2=\frac{54}{14}=\frac{27}{7}\Rightarrow \mid p\mid =\sqrt{\frac{27}{7}}$$

Commented by pete last updated on 01/Sep/20

$$\mathrm{I}\mathrm{appreciate}\mathrm{your}\mathrm{effort}\mathrm{sir}$$

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