Question

Find the distance between the point P and the line determined by the points Q and R

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

Short answer

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

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

Step-by-step solution

Step 1:Given information

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

Step 2:Simplification

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

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

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

Now,

$\overrightarrow{QP}=(2+2,5-1,7+5)$

$=⟨4,4,12⟩$

$\overrightarrow{QR}=⟨-3+2,0-1,4-(-5)⟩$

$=⟨-1,-1,9⟩$

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

$\text{First find}{proj}_{\overline{QR}}\overrightarrow{QP}\text{.}$

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

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

${proj}_{\overrightarrow{QR}}\overrightarrow{QP}=\frac{\overrightarrow{QP}·\overrightarrow{QR}}{\Vert \overrightarrow{QR}{\Vert }^2}\overrightarrow{QR}$

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

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

$=\frac{100}{83}⟨-1,-1,9⟩$

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

$\text{Now, calculate}\overrightarrow{QP}-{\text{proj}}_{\overline{QR}}\overrightarrow{QP}$

$\overrightarrow{QP}-{\text{proj}}_{\overrightarrow{QK}}\overrightarrow{QP}=⟨4,4,12⟩-\frac{100}{83}⟨-1,-1,9⟩$

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

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

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

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

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

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

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