Consider the line $L$ defined by the following parametric equations
$x= 3+2t$ $y= 4+t$ $z=5-6t$
Find the point $Q$ on $L$ that is closest to $(4,1,7)$.
Note: I do not really remember the formulas, I need help!
Consider the line $L$ defined by the following parametric equations
$x= 3+2t$ $y= 4+t$ $z=5-6t$
Find the point $Q$ on $L$ that is closest to $(4,1,7)$.
Note: I do not really remember the formulas, I need help!
The plane containing this closest point will have normal vector $n = 2i + j - 6k. $ Since the point $(4,1,7)$ is in the plane and the plane's equation is $ 2x + y - z = 2\cdot4 + 1 - 7 = 1.$ The closest point will lie where the point and line intersect. Find this, then the distance to the point $(4,1,7)$.
The distance between $(x,y,z)$ and $(4,1,7)$ is $\sqrt{(x-4)^2+(y-1)^2+(z-7)^2}$. That follows from the Pythagorean theorem. This is the same as $\sqrt{(2+2t-4)^2+(4+t-1)^2+(5-6t-7)^2}.$ That simplifies to $ \sqrt{(2t-2)^2+(3+t)^2+(-2-6t)^2}. $ The value of $t$ that minimizes the distance is the same as the value of $t$ that minimizes the square of the distance, i.e. $(2t-2)^2+(3+t)^2+(-2-6t)^2$. If you multiply that out, you get $(\bullet t^2) + (\bullet t) + (\bullet)$. Find the three numbers and then complete the square.
Thanks to Michael Hardy, in (poor) spanish:
La distancia entre $(x,y,z)$ y $(4,1,7)$ es $\sqrt{(x-4)^2+(y-1)^2+(z-7)^2}$. Este sigue de la teorema de Pythagorean. Esto es lo mismo que $\sqrt{(2+2t-4)^2+(4+t-1)^2+(5-6t-7)^2}.$ Simplifica a $ \sqrt{(2t-2)^2+(3+t)^2+(-2-6t)^2}. $ El valor de $t$ que minimiza la distancia es lo mismo que el valor de $t$ que minimiza el el cuadrado de la distancia: $(2t-2)^2+(3+t)^2+(-2-6t)^2$. Si multiplica eso, consique $(\bullet t^2) + (\bullet t) + (\bullet)$. Encuentra estos numeros y completa el cuadrado.
Tambien, disculpa mi traduccion mala.
Direction cosines of given line are $(2,1,-6)$. If $P=(4,1,7)$ and $Q=(3+2t,4+t,5-6t)$ be the closest point on the line to P, then line joining these two points must be normal to direction cosines of given line. Hence, dot product of vectors $PQ=(2t-1,3+t,-6t-2)$ and $(2,1,-6)$ would be $0 \implies$ $4t-2+3+t+36t+12=0 \implies 41t=-13 \implies t=-13/41$ for $Q$.Thus, the point is $(97/41,151/41,283/41)$.