Q .38. - Calculus

Question

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

$\text{ Find the altitude of triangle }∆PQR\text{ from vertex }R\text{ to side }$$\overline{PQ}.$

Step-by-Step Solution

Answer

$\text{ Hence, the altitude of the triangle }PQR\text{ from the vertex }R\text{ to the side }\overline{PQ}\text{ is }\frac{48}{59}\sqrt{118}\text{. }$

Step 1:Given information

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

Step 2:Simplification

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

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

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

Now

$\overrightarrow{PQ}=⟨-2-2,1-5,-5-7⟩$

$=⟨-4,-4,-12⟩$

$\overrightarrow{PR}=⟨-3-2,0-5,4-7⟩$

$=⟨-5,-5,-3⟩$

$\text{ Consider the vectors }\overrightarrow{PQ}\text{ and }\overrightarrow{PR}\text{. }$

$\text{ First find }{proj}_{PR}\overrightarrow{PQ}\text{. }$

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

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

$\text{ Therefore, }$

${proj}_{\overrightarrow{PR}}\overrightarrow{PQ}=\frac{\overrightarrow{PQ}·\overrightarrow{PR}}{\Vert \overrightarrow{PR}{\Vert }^2}\overrightarrow{PR}$

$=\frac{⟨-4,-4,-12⟩·⟨-5,-5,-3⟩}{{\left(\sqrt{(-5{)}^2+(-5{)}^2+(-3{)}^2}\right)}^2}⟨-5,-5,-3⟩$

$=\frac{20+20+36}{25+25+9}⟨-5,-5,-3⟩$

$=\frac{76}{59}⟨-5,-5,-3⟩$

$\text{ Now, the vector component of }\overrightarrow{PQ}\text{ orthogonal to }\overrightarrow{PR}\text{ is given by }\overrightarrow{PQ}-{proj}_{\overrightarrow{PR}}\overrightarrow{PQ}\text{. }$

$\text{ Now, calculate the value of }\overrightarrow{PQ}-{proj}_{\overrightarrow{PR}}\overrightarrow{PQ}\text{. }$

$\overrightarrow{PQ}-{proj}_{\overrightarrow{PR}}\overrightarrow{PQ}=⟨-4,-4,-12⟩-\frac{76}{59}⟨-5,-5,-3⟩$

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

$=\frac{1}{59}\sqrt{20736+20736+230400}$

$=\frac{1}{59}\sqrt{271872}$

$=\frac{48}{59}\sqrt{118}$

$\text{ Hence, the altitude of the triangle }PQR\text{ from the vertex }R\text{ to the side }\overline{PQ}\text{ is }\frac{48}{59}\sqrt{118}\text{. }$