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Question
Show that the four points A, B, C, D with position vectors \[\vec{a,} \vec{b,} \vec{c,} \vec{d}\] respectively such that \[3 \vec{a} - 2 \vec{b} + 5 \vec{c} - 6 \vec{d} = 0,\] are coplanar. Also, find the position vector of the point of intersection of the line segments AC and BD.
Solution
Let AC and BD intersects at a point P We have,
\[3 \vec{a} - 2 b^\rightharpoonup + 5 c^\rightharpoonup - 6 d^\rightharpoonup = \vec{0} . \]
\[ \Rightarrow 3 \vec{a} + 5 \vec{c} = 2 \vec{b} + 6 \vec{d} \]
Since sum of coefficients on both sides of the above equation is 8 .
so we divide the equation on both sides by 8 .
\[ \Rightarrow \frac{3 \vec{a} + 5 \vec{c}}{8} = \frac{2 \vec{b} + 6 \vec{d}}{8}\]
\[ \Rightarrow \frac{3 \vec{a} + 5 \vec{c}}{3 + 5} = \frac{2 \vec{b} + 6 \vec{d}}{2 + 6}\]
Therefore, P divides AC in the ratio of 3: 5 and P divides BD in the ratio of 2:6.
Therefore, position vector of the point of intersection of AC and BD will be \[\frac{3 \vec{a} + 5 \vec{c}}{8} = \frac{2 \vec{b} + 6 \vec{d}}{8}\]
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