The position vectors of the points P and Q ar | vedclass

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(D) Let the point $A$ with position vector $\vec{a} = \frac{-9}{2} \bar{i} - 6 \bar{j} + \frac{1}{2} \bar{k}$ divide the line segment $PQ$ in the rati

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$-1 : 3$

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(D) Let the point $A$ with position vector $\vec{a} = \frac{-9}{2} \bar{i} - 6 \bar{j} + \frac{1}{2} \bar{k}$ divide the line segment $PQ$ in the ratio $\lambda : 1$.Using the section formula,the position vector of $A$ is given by:$\vec{a} = \frac{\lambda \vec{q} + 1 \vec{p}}{\lambda + 1}$Substituting the given vectors $\vec{p} = -2 \bar{i} - 3 \bar{j} + \bar{k}$ and $\vec{q} = 3 \bar{i} + 3 \bar{j} + 2 \bar{k}$:$\frac{-9}{2} \bar{i} - 6 \bar{j} + \frac{1}{2} \bar{k} = \frac{\lambda(3 \bar{i} + 3 \bar{j} + 2 \bar{k}) + 1(-2 \bar{i} - 3 \bar{j} + \bar{k})}{\lambda + 1}$Comparing the $x$-coordinates:$\frac{-9}{2} = \frac{3 \lambda - 2}{\lambda + 1}$$-9(\lambda + 1) = 2(3 \lambda - 2)$$-9 \lambda - 9 = 6 \lambda - 4$$-15 \lambda = 5$$\lambda = -\frac{5}{15} = -\frac{1}{3}$Since $\lambda$ is negative,the point $A$ divides the line segment $PQ$ externally in the ratio $1 : 3$.

The position vectors of the points $P$ and $Q$ are respectively $-2 \bar{i}-3 \bar{j}+\bar{k}$ and $3 \bar{i}+3 \bar{j}+2 \bar{k}$. The ratio in which the point having position vector $\frac{-9}{2} \bar{i}-6 \bar{j}+\frac{1}{2} \bar{k}$ divides the line segment joining $P$ and $Q$ is

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