Question

a distancia do ponto P(5,1,3) a reta que passa por A(3,1,3) e B(4,-1,1) é:

Answer 1

$\mathrm{A(x_0,y_0,z_0)=A(3,1,3)\ \ \|\ \ B(x_1,y_1,z_1)=B(4,-1,1)}\\\\ r:\begin{cases} x=x_0+(x_1-x_0)t\\ y=y_0+(y_1-y_0)t\\ z=z_0+(z_1-z_0)t\end{cases}\ \to\ r:\begin{cases} x=3+t\\ y=1-2t\\ z=3-2t\\ \end{cases}\\\\\\ \mathrm{\vec{d_r}=(1,-2,-2)\ \ \|\ \ A(3,1,3)\ \ \|\ \ P(5,1,3)}\\\\ \mathrm{\vec{AP}=P-A=(5,1,3)-(3,1,3)=(2,0,0)}\\\\ \mathrm{\vec{AP}\times\vec{d_r}=\left|\begin{array}{ccc}\vec{i}&\vec{j}&\vec{k}\\2&0&0\\1&-2&-2\end{array}\right|=(0,4,-4)}$

$\mathrm{d(P,r)=\dfrac{\|\vec{AP}\times\vec{d_r}\|}{\|\vec{d_r}\|}=\dfrac{\sqrt{0^2+4^2+(-4)^2}}{\sqrt{1^2+(-2)^2+(-2)^2}}}\\\\\\ \mathrm{d(P,r)=\dfrac{\sqrt{16+16}}{\sqrt{1+4+4}}=\dfrac{\sqrt{32}}{\sqrt{9}}=\dfrac{4\sqrt{2}}{3}}\\\\\\ \boxed{\mathbf{d(P,r)=\dfrac{4\sqrt{2}}{3}\ u.c.}}}$