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Question Number 72832 by necxxx last updated on 03/Nov/19

In a parallelogram OABC, OA^⇁ =a^(−⇁) , OC^→ =c^→ , D is a point such that AD^→ :DB^→ =1:2 Express the following in terms of a and c (i)CB^→ (ii)BC^→ (iii)AB^→ (iv) AD^→ (v)OD^→ (vi)DC^→

$${In}\:{a}\:{parallelogram}\:{OABC},\:{O}\overset{\rightharpoondown} {{A}}=\overset{−\rightharpoondown} {{a}}, \\ $$$${O}\overset{\rightarrow} {{C}}=\overset{\rightarrow} {{c}},\:{D}\:{is}\:{a}\:{point}\:{such}\:{that}\:{A}\overset{\rightarrow} {{D}}:{D}\overset{\rightarrow} {{B}}=\mathrm{1}:\mathrm{2} \\ $$$${Express}\:{the}\:{following}\:{in}\:{terms}\:{of}\:{a}\:{and}\:{c} \\ $$$$\left({i}\right){C}\overset{\rightarrow} {{B}}\:\left({ii}\right){B}\overset{\rightarrow} {{C}}\:\left({iii}\right){A}\overset{\rightarrow} {{B}}\:\left({iv}\right)\:{A}\overset{\rightarrow} {{D}}\:\left({v}\right){O}\overset{\rightarrow} {{D}} \\ $$$$\left({vi}\right){D}\overset{\rightarrow} {{C}} \\ $$

Commented by necxxx last updated on 03/Nov/19

please help. Thanks in advance.

$${please}\:{help}.\:{Thanks}\:{in}\:{advance}. \\ $$

Answered by mind is power last updated on 03/Nov/19

CB^→ =OA^→ =a^→ BC^→ =−CB^→ =−a^→ AB^→ =OC^→ =c^→ AD^→ =(1/2).DB^→ AD^→ +DB^→ =AB^→ 3AD^→ =AB^→ ⇒AD^→ =((AB^→ )/3)=(c^→ /3) OD^→ =OA^→ +AD^→ =a^→ +(c^→ /3)

$$\mathrm{C}\overset{\rightarrow} {\mathrm{B}}=\mathrm{O}\overset{\rightarrow} {\mathrm{A}}=\overset{\rightarrow} {\mathrm{a}} \\ $$$$\mathrm{B}\overset{\rightarrow} {\mathrm{C}}=−\mathrm{C}\overset{\rightarrow} {\mathrm{B}}=−\overset{\rightarrow} {\mathrm{a}} \\ $$$$\mathrm{A}\overset{\rightarrow} {\mathrm{B}}=\mathrm{O}\overset{\rightarrow} {\mathrm{C}}=\overset{\rightarrow} {\mathrm{c}} \\ $$$$\mathrm{A}\overset{\rightarrow} {\mathrm{D}}=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{D}\overset{\rightarrow} {\mathrm{B}} \\ $$$$\mathrm{A}\overset{\rightarrow} {\mathrm{D}}+\mathrm{D}\overset{\rightarrow} {\mathrm{B}}=\mathrm{A}\overset{\rightarrow} {\mathrm{B}} \\ $$$$\mathrm{3A}\overset{\rightarrow} {\mathrm{D}}=\mathrm{A}\overset{\rightarrow} {\mathrm{B}}\Rightarrow\mathrm{A}\overset{\rightarrow} {\mathrm{D}}=\frac{\mathrm{A}\overset{\rightarrow} {\mathrm{B}}}{\mathrm{3}}=\frac{\overset{\rightarrow} {\mathrm{c}}}{\mathrm{3}} \\ $$$$\mathrm{O}\overset{\rightarrow} {\mathrm{D}}=\mathrm{O}\overset{\rightarrow} {\mathrm{A}}+\mathrm{A}\overset{\rightarrow} {\mathrm{D}}=\overset{\rightarrow} {\mathrm{a}}+\frac{\overset{\rightarrow} {\mathrm{c}}}{\mathrm{3}} \\ $$

Commented by necxxx last updated on 03/Nov/19

I now get it. Thank you so much sir

$${I}\:{now}\:{get}\:{it}.\:{Thank}\:{you}\:{so}\:{much}\:{sir} \\ $$

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