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Question Number 111006 by pete last updated on 01/Sep/20
The vectors p,q and r are mutially perpendicularwith  ∣q∣=3 and ∣r∣=(√(5.4 )) .If X= 7p+5q+7r and  Y=2p+3q−5r are perpendicular, find∣p∣.
$$\mathrm{The}\:\mathrm{vectors}\:\boldsymbol{\mathrm{p}},\boldsymbol{\mathrm{q}}\:\mathrm{and}\:\boldsymbol{\mathrm{r}}\:\mathrm{are}\:\mathrm{mutially}\:\mathrm{perpendicularwith} \\ $$$$\mid\boldsymbol{\mathrm{q}}\mid=\mathrm{3}\:\mathrm{and}\:\mid\boldsymbol{\mathrm{r}}\mid=\sqrt{\mathrm{5}.\mathrm{4}\:}\:.\mathrm{If}\:\mathrm{X}=\:\mathrm{7}\boldsymbol{\mathrm{p}}+\mathrm{5}\boldsymbol{\mathrm{q}}+\mathrm{7}\boldsymbol{\mathrm{r}}\:\mathrm{and} \\ $$$$\mathrm{Y}=\mathrm{2}\boldsymbol{\mathrm{p}}+\mathrm{3}\boldsymbol{\mathrm{q}}−\mathrm{5}\boldsymbol{\mathrm{r}}\:\mathrm{are}\:\mathrm{perpendicular},\:\mathrm{find}\mid\boldsymbol{\mathrm{p}}\mid. \\ $$
Commented by kaivan.ahmadi last updated on 01/Sep/20
p.q=p.r=q.r=0  X.Y=0⇒(7p+5q+7r).(2p+3q−5r)=  14p.p+15q.q−35r.r=14∣p∣^2 +15∣q∣^2 −35∣r∣^2 =  14∣p∣^2 +15×9−35×5.4=14∣p∣^2 +135−189=  14∣p∣^2 −=0⇒∣p∣^2 =((54)/(14))=((27)/7)⇒∣p∣=(√((27)/7))
$${p}.{q}={p}.{r}={q}.{r}=\mathrm{0} \\ $$$${X}.{Y}=\mathrm{0}\Rightarrow\left(\mathrm{7}{p}+\mathrm{5}{q}+\mathrm{7}{r}\right).\left(\mathrm{2}{p}+\mathrm{3}{q}−\mathrm{5}{r}\right)= \\ $$$$\mathrm{14}{p}.{p}+\mathrm{15}{q}.{q}−\mathrm{35}{r}.{r}=\mathrm{14}\mid{p}\mid^{\mathrm{2}} +\mathrm{15}\mid{q}\mid^{\mathrm{2}} −\mathrm{35}\mid{r}\mid^{\mathrm{2}} = \\ $$$$\mathrm{14}\mid{p}\mid^{\mathrm{2}} +\mathrm{15}×\mathrm{9}−\mathrm{35}×\mathrm{5}.\mathrm{4}=\mathrm{14}\mid{p}\mid^{\mathrm{2}} +\mathrm{135}−\mathrm{189}= \\ $$$$\mathrm{14}\mid{p}\mid^{\mathrm{2}} −=\mathrm{0}\Rightarrow\mid{p}\mid^{\mathrm{2}} =\frac{\mathrm{54}}{\mathrm{14}}=\frac{\mathrm{27}}{\mathrm{7}}\Rightarrow\mid{p}\mid=\sqrt{\frac{\mathrm{27}}{\mathrm{7}}} \\ $$
Commented by pete last updated on 01/Sep/20
I appreciate your effort sir
$$\mathrm{I}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort}\:\mathrm{sir} \\ $$

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