7 bar{i}-4 bar{j}+7 bar{k}, bar{i}-6 bar{j}+1 | vedclass

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(B) Let the position vectors of points $A, B, C, D$ be $\vec{a} = 7\bar{i}-4\bar{j}+7\bar{k}$, $\vec{b} = \bar{i}-6\bar{j}+10\bar{k}$, $\vec{c} = -\ba

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$5$

Vedclass · Answer

(B) Let the position vectors of points $A, B, C, D$ be $\vec{a} = 7\bar{i}-4\bar{j}+7\bar{k}$, $\vec{b} = \bar{i}-6\bar{j}+10\bar{k}$, $\vec{c} = -\bar{i}-3\bar{j}+4\bar{k}$, and $\vec{d} = 5\bar{i}-\bar{j}+\bar{k}$.For a quadrilateral $ABCD$ to have an intersection of diagonals, the points must be coplanar. The intersection point $P$ of diagonals $AC$ and $BD$ exists if the quadrilateral is planar.The intersection point of diagonals $AC$ and $BD$ is given by the point that divides $AC$ in ratio $m:1-m$ and $BD$ in ratio $n:1-n$.$P = (1-m)\vec{a} + m\vec{c} = (1-n)\vec{b} + n\vec{d}$.Equating the coefficients of $\bar{i}, \bar{j}, \bar{k}$:$1-m(8) - m = 1-n(6) + 5n \implies 7-8m = 1-n$ (for $\bar{i}$)$-4+m = -6+3n \implies m-3n = -2$ (for $\bar{j}$)$7-3m = 10-6n \implies -3m+6n = 3 \implies -m+2n = 1$ (for $\bar{k}$).Solving the system: $m-3n = -2$ and $-m+2n = 1$. Adding gives $-n = -1 \implies n=1$. Then $m=1$.Since $m=1$ and $n=1$, the intersection point $P$ is $\vec{c}$ and $\vec{d}$, which implies the quadrilateral is degenerate or the diagonals intersect at a vertex. However, checking the midpoint of $AC$ and $BD$:Midpoint of $AC = \frac{\vec{a}+\vec{c}}{2} = \frac{6\bar{i}-7\bar{j}+11\bar{k}}{2} = 3\bar{i}-3.5\bar{j}+5.5\bar{k}$.Midpoint of $BD = \frac{\vec{b}+\vec{d}}{2} = \frac{6\bar{i}-7\bar{j}+11\bar{k}}{2} = 3\bar{i}-3.5\bar{j}+5.5\bar{k}$.Since the midpoints coincide, $ABCD$ is a parallelogram. The intersection of diagonals is the midpoint: $p=3, q=-3.5, r=5.5$.$p+q+r = 3-3.5+5.5 = 5$.

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