Let ABCD be a parallelogram and 2hat{i}+hat{j | vedclass

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(B) In a parallelogram $ABCD$,the diagonals $AC$ and $BD$ bisect each other at their midpoint $M$. Given position vectors: $\vec{A} = 2\hat{i}+\hat{j}

Vedclass · Accepted Answer

$\frac{1}{3}(5\hat{i}+2\hat{j}+\hat{k})$

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(B) In a parallelogram $ABCD$,the diagonals $AC$ and $BD$ bisect each other at their midpoint $M$. Given position vectors: $\vec{A} = 2\hat{i}+\hat{j}+0\hat{k}$,$\vec{B} = 4\hat{i}+5\hat{j}+4\hat{k}$,$\vec{D} = -\hat{i}-4\hat{j}-3\hat{k}$. The midpoint $M$ of diagonal $BD$ is given by: $M = \frac{\vec{B}+\vec{D}}{2} = \frac{(4-1)\hat{i} + (5-4)\hat{j} + (4-3)\hat{k}}{2} = \frac{3}{2}\hat{i} + \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k}$. Since $M$ is also the midpoint of $AC$,we have $\frac{\vec{A}+\vec{C}}{2} = M$. $\vec{A}+\vec{C} = 2M = 3\hat{i} + \hat{j} + \hat{k}$. $\vec{C} = (3\hat{i} + \hat{j} + \hat{k}) - (2\hat{i} + \hat{j} + 0\hat{k}) = \hat{i} + 0\hat{j} + \hat{k}$. The points of trisection $T_1$ and $T_2$ of diagonal $AC$ divide it in the ratio $1:2$ and $2:1$ respectively. For $T_1$ (ratio $1:2$): $\vec{T_1} = \frac{1(\vec{C}) + 2(\vec{A})}{1+2} = \frac{1(\hat{i}+\hat{k}) + 2(2\hat{i}+\hat{j})}{3} = \frac{5\hat{i}+2\hat{j}+\hat{k}}{3} = \frac{1}{3}(5\hat{i}+2\hat{j}+\hat{k})$. For $T_2$ (ratio $2:1$): $\vec{T_2} = \frac{2(\vec{C}) + 1(\vec{A})}{2+1} = \frac{2(\hat{i}+\hat{k}) + 1(2\hat{i}+\hat{j})}{3} = \frac{4\hat{i}+\hat{j}+2\hat{k}}{3}$. Comparing with the options,the correct position vector is $\frac{1}{3}(5\hat{i}+2\hat{j}+\hat{k})$.

Let $ABCD$ be a parallelogram and $2\hat{i}+\hat{j}$,$4\hat{i}+5\hat{j}+4\hat{k}$ and $-\hat{i}-4\hat{j}-3\hat{k}$ be the position vectors of the vertices $A$,$B$,and $D$ respectively. Then the position vector of one of the points of trisection of the diagonal $AC$ is

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