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NCERT Exemplar · Q4

Q.If a⃗\vec{a} and b⃗\vec{b} are the position vectors of A and B, respectively, find the position vector of a point C in BA produced such that BC=1.5 BABC = 1.5\,BA.

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With CC on BABA produced and BC=1.5 BABC=1.5\,BA, the position vector is c⃗=3a⃗−b⃗2\vec{c}=\dfrac{3\vec{a}-\vec{b}}{2}.

Reading the problem

AA and BB have position vectors a⃗\vec{a} and b⃗\vec{b}. "BABA produced" means we travel from BB towards AA and keep going past AA. So CC lies on that ray, in the same direction as BA⃗\vec{BA}, at a distance BC=1.5 BABC=1.5\,BA.

Set up the displacement

The step from BB to AA is

BA⃗=a⃗−b⃗.\vec{BA}=\vec{a}-\vec{b}.

Because CC is along this same direction with BC=1.5 BABC=1.5\,BA,

BC⃗=1.5 (a⃗−b⃗).\vec{BC}=1.5\,(\vec{a}-\vec{b}).

Find the position vector of C

c⃗=b⃗+BC⃗=b⃗+1.5(a⃗−b⃗)=1.5a⃗+(1−1.5)b⃗=1.5a⃗−0.5b⃗\vec{c}=\vec{b}+\vec{BC}=\vec{b}+1.5(\vec{a}-\vec{b})=1.5\vec{a}+(1-1.5)\vec{b}=1.5\vec{a}-0.5\vec{b}

c⃗=3a⃗−b⃗2\vec{c}=\frac{3\vec{a}-\vec{b}}{2}

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