Answer:
Given points are A (3, 4, 1) and B (5, 1, 6). The cartesian equation of line AB passing through the points A and B is \[\frac{x-3}{5-3}=\frac{y-4}{1-4}=\frac{z-1}{6-1}\] \[\left[ \text{by}\,\,\text{using}\,\,\frac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\frac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}} \right]\] \[\Rightarrow \] \[\frac{x-3}{2}=\frac{y-4}{-\,3}=\frac{z-1}{5}=\lambda \] (say) ?(i) Any point P on the line is \[P(2\lambda +3,\,\,-3\lambda +4,\,\,5\lambda +1)\] ?(ii) According to the question, the line crosses the XY-plane, so it will intersect the plane at a point whose z-coordinate will be zero. i.e. \[5\lambda +1=0\] \[\Rightarrow \] \[\lambda =-1/5\] On putting the value of X, in Eq. (ii), we get the coordinates of \[P=\left( -\frac{2}{5}+3,\,\,\frac{3}{5}+4,\,\,-\frac{5}{5}+1 \right)\] \[=\left( \frac{13}{5},\,\,\frac{23}{5},\,\,0 \right),\]
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