91Ó°ÊÓ

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

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

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

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

Now

$\stackrel{→}{PQ}=⟨-2-2,1-5,-5-7âŸ\copyright$

$=⟨-4,-4,-12âŸ\copyright$

$\stackrel{→}{PR}=⟨-3-2,0-5,4-7âŸ\copyright$

$=⟨-5,-5,-3âŸ\copyright$

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

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

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

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

$\text{Therefore,}$

${proj}_{\stackrel{→}{PR}}\stackrel{→}{PQ}=\frac{\stackrel{→}{PQ}·\stackrel{→}{PR}}{‖\stackrel{→}{PR}{‖}^2}\stackrel{→}{PR}$

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

$=\frac{20+20+36}{25+25+9}⟨-5,-5,-3âŸ\copyright$

$=\frac{76}{59}⟨-5,-5,-3âŸ\copyright$

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

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

$\stackrel{→}{PQ}-{proj}_{\stackrel{→}{PR}}\stackrel{→}{PQ}=⟨-4,-4,-12âŸ\copyright -\frac{76}{59}⟨-5,-5,-3âŸ\copyright$

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

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

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

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

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