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}.$

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

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

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

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

Now

$\overrightarrow{PQ}=⟨-2-2,1-5,-5-7⟩$

$=⟨-4,-4,-12⟩$

$\overrightarrow{PR}=⟨-3-2,0-5,4-7⟩$

$=⟨-5,-5,-3⟩$

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

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

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

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

$\text{Therefore,}$

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

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

$=\frac{20+20+36}{25+25+9}⟨-5,-5,-3⟩$

$=\frac{76}{59}⟨-5,-5,-3⟩$

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

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

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

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

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

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

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

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