Let overline{i}-2 overline{j}+overline{k}, ov | vedclass

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Question

(B) Let the position vectors be $\vec{a} = \overline{i}-2 \overline{j}+\overline{k}$,$\vec{b} = \overline{i}+\overline{j}-2 \overline{k}$,$\vec{c} = 2

Vedclass · Accepted Answer

$-2:1$

Vedclass · Answer

(B) Let the position vectors be $\vec{a} = \overline{i}-2 \overline{j}+\overline{k}$,$\vec{b} = \overline{i}+\overline{j}-2 \overline{k}$,$\vec{c} = 2 \overline{i}-\overline{j}-\overline{k}$,and $\vec{d} = \overline{i}+\overline{j}+\overline{k}$.Point $P$ divides $AB$ in ratio $2:1$ internally: $\vec{p} = \frac{2\vec{b} + 1\vec{a}}{2+1} = \frac{2(\overline{i}+\overline{j}-2 \overline{k}) + (\overline{i}-2 \overline{j}+\overline{k})}{3} = \frac{3\overline{i} - 3\overline{k}}{3} = \overline{i} - \overline{k}$.Point $Q$ divides $CD$ in ratio $1:2$ externally: $\vec{q} = \frac{1\vec{d} - 2\vec{c}}{1-2} = \frac{(\overline{i}+\overline{j}+\overline{k}) - 2(2 \overline{i}-\overline{j}-\overline{k})}{-1} = \frac{-3\overline{i} + 3\overline{j} + 3\overline{k}}{-1} = 3\overline{i} - 3\overline{j} - 3\overline{k}$.Let the point $R$ with position vector $\vec{r} = 5\overline{i}-6\overline{j}-5\overline{k}$ divide $PQ$ in ratio $k:1$.Then $\vec{r} = \frac{k\vec{q} + 1\vec{p}}{k+1} \implies (k+1)(5\overline{i}-6\overline{j}-5\overline{k}) = k(3\overline{i}-3\overline{j}-3\overline{k}) + (\overline{i}-\overline{k})$.Comparing components:$x$-component: $5k+5 = 3k+1 \implies 2k = -4 \implies k = -2$.Thus,the ratio is $-2:1$.

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