Q .37. - Calculus

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) 

Step-by-Step Solution

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 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,

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

$=⟨4,4,12⟩$

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

$=⟨-1,-1,9⟩$

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

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

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

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

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

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

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

$=\frac{100}{83}⟨-1,-1,9⟩$

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

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

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

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

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

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

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

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

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

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