91Ó°ÊÓ

let P = (2, 5, 7), Q = (−2, 1, −5), and R = (−3, 0, 4)

$\text{Consider the points}P=(2,5,7)Q=(-2,1,-5)\text{and}R=(-3,0,4)\text{.}$

$\text{Using result, if there are two points}P=\left(x_0,y_0,z_0\right)\text{and}Q=\left(x_1,y_1,z_1\right)\text{, then}$

$\stackrel{¯}{PQ}=\left(x_1-x_0,y_1-y_0,z_1-z_0\right).$

Now,

$\stackrel{→}{QP}=(2+2,5-1,7+5)$

$=⟨4,4,12âŸ\copyright$

$\stackrel{→}{QR}=⟨-3+2,0-1,4-(-5)âŸ\copyright$

$=⟨-1,-1,9âŸ\copyright$

$\text{Consider the vectors}\stackrel{¯}{QP}\text{and}\stackrel{¯}{QR}\text{.}$

$\text{First find}{proj}_{\stackrel{¯}{QR}}\stackrel{→}{QP}\text{.}$

$\text{Using result, let}\mathbf{u}\text{be any non- zero vector, then the vector projection of}\mathbf{v}\text{onto}\mathbf{u}\text{is given by}$

${proj}_{\mathbf{u}}\mathbf{v}=\frac{\mathbf{u}·\mathbf{v}}{‖\mathbf{u}{‖}^2}\mathbf{u}$

${proj}_{\stackrel{→}{QR}}\stackrel{→}{QP}=\frac{\stackrel{→}{QP}·\stackrel{→}{QR}}{‖\stackrel{→}{QR}{‖}^2}\stackrel{→}{QR}$

$=\frac{⟨4,4,12âŸ\copyright ·⟨-1,-1,9âŸ\copyright }{\sqrt{(-1{)}^2+(-1{)}^2+(9{)}^2}}⟨-1,-1,9âŸ\copyright$

$=\frac{-4-4+108}{{\sqrt{1+1+81}}^2}⟨-1,-1,9âŸ\copyright$

$=\frac{100}{83}⟨-1,-1,9âŸ\copyright$

$\text{Now, the vector component of}\stackrel{→}{QP}\text{orthogonal to}\stackrel{→}{QR}\text{is given by}\stackrel{→}{QP}-pro{j}_{\stackrel{¯}{QR}}\stackrel{→}{QP}\text{.}$

$\text{Now, calculate}\stackrel{→}{QP}-{\text{proj}}_{\stackrel{¯}{QR}}\stackrel{→}{QP}$

$\stackrel{→}{QP}-{\text{proj}}_{\stackrel{→}{QK}}\stackrel{→}{QP}=⟨4,4,12âŸ\copyright -\frac{100}{83}⟨-1,-1,9âŸ\copyright$

$\text{Now, the distance from}P\text{to the line determined by the points Qand}R\text{is the magnitude of the}$

$||\frac{432}{83},\frac{432}{83},\frac{1896}{83}||=\sqrt{{\left(\frac{432}{83}\right)}^2+{\left(\frac{432}{83}\right)}^2+{\left(\frac{1896}{83}\right)}^2}$

$=\frac{1}{83}\sqrt{186624+186624+9}$

$=\frac{1}{83}\sqrt{382464}$

$=\frac{48}{83}\sqrt{166}$

$\text{Hence, the distance between the point}P\text{and the line determined by the point}Q\text{and}R\text{is}$

$=\frac{48}{83}\sqrt{166}$