Show that points P_{1} \left(5,2,-4\right), P_{2} \left(1,1,2\right), and P_{3} \left(-3,0,8\right) all lie on a straight line. Determine the shortest distance between the line and point P_{4} \left(3,-1,0\right).
Show that points P_{1} \left(5,2,-4\right), P_{2} \left(1,1,2\right), and P_{3} \left(-3,0,8\right) all lie on a straight line. Determine the shortest distance between the line and point P_{4} \left(3,-1,0\right).
The distance vector r_{P_{1},P_{2}} is given by
r_{P_{1},P_{2}}=r_{P_{2}}-r_{P_{1}}=\left(1,1,2\right)-\left(5,2,-4\right)=\left(-4,-1,6\right)
Similarly,
r_{P_{1},P_{3}}=r_{P_{3}}-r_{P_{1}}=\left(-3,0,8\right)-\left(5,2,-4\right)=\left(-8,-2,12\right)
r_{P_{1},P_{4}}=r_{P_{4}}-r_{P_{1}}=\left(3,-1,0\right)-\left(5,2,-4\right)=\left(-2,-3,4\right)
r_{P_{1}P_{2}}\times r_{P_{1}P_{3}}=\left | \begin{matrix} a_{x} & a_{y} & a_{z} \\ -4 & -1 & 6 \\ -8 & -2 & 12 \end{matrix} \right | =\left(0,0,0\right)
showing that the angle between r_{P_{1} P_{2}} and r_{P_{1} P_{3}} is zero (sinθ = 0). This implies that P_{1},P_{2}, and P_{3} lie on a straight line.
Alternatively, the vector equation of the straight line is easily determined from Figure1.12(a). For any point P on the line joining P_{1} and P_{2}
r_{P_{1} P}=\lambda r_{P_{1} P_{2}}
where λ is a constant. Hence the position vector r_{P} of the point P must satisfy
r_{P}-r_{P_{1}}=\lambda \left(r_{P_{2}}-r_{P_{1}}\right)
that is,
r_{P}=r_{P_{1}}+\lambda \left(r_{P_{2}}-r_{P_{1}}\right) =\left(5,2,-4\right)-\lambda \left(4,1,-6\right)
r_{P}=\left(5-4\lambda,2-\lambda,-4+6\lambda\right)
This is the vector equation of the straight line joining P_{1} and P_{2}. If P_{3} is on this line, the position vector of P_{3} must satisfy the equation; r_{3} does satisfy the equation when λ = 2.
The shortest distance between the line and point P_{4}\left(3,-1,0\right), is the perpendicular distance from the point to the line. From Figure1.12(b), it is clear that
d=r_{P_{1}P_{4}}\sin \theta=\left|r_{P_{1}P_{4}}\times a_{P_{1}P_{2}}\right|=\frac{\left|\left(-2,-3,4\right)\times\left(-4,-1,6\right)\right|}{\left|\left(-4,-1,6\right) \right| }
=\frac{\sqrt{312} }{\sqrt{53} } =2.426
Any point on the line may be used as a reference point. Thus, instead of using P_{1} as a reference point, we could use P_{3}. If
\angle{P_{4}P_{3}P_{2}}=\theta^{'}
then
d=\left|r_{P_{3}P_{4}}\right|\sin \theta^{'}=\left|r_{P_{3}P_{4}}\times a_{P_{3}P_{2}}\right|