# rona55

#1
+16
0

To find the perpendicular distance from the given line to the point (2,1,7), we need to find a point on the line that is closest to the given point. This point will be on the line connecting the given point and the line, and will be the point where a perpendicular line from the given point intersects the line.

Let's first find the vector equation of the line:
r(t) = (-1,4,3) + t(4,-10,10)

Now we need to find the point on this line closest to the given point (2,1,7). Let P be this point. The vector connecting P to (2,1,7) will be perpendicular to the line. So, we can find P by projecting the vector between (2,1,7) and (-1,4,3) onto the direction vector of the line.

The direction vector of the line is d = (4,-10,10). So, we can find the projection of the vector between (2,1,7) and (-1,4,3) onto d as follows:

proj_d(v) = ((v · d) / ||d||^2) d,

where · denotes the dot product, and ||d|| is the magnitude of d.

v = (2-(-1), 1-4, 7-3) = (3,-3,4)

||d|| = sqrt(4^2 + (-10)^2 + 10^2) = sqrt(216) = 6*sqrt(6)

So, proj_d(v) = ((34 - 3(-10) + 410) / (6sqrt(6))^2) (4,-10,10) = (37/216) (4,-10,10) = (37/54) (2,-5,5)

The point P on the line closest to (2,1,7) is given by:

P = (-1,4,3) + proj_d(v) = (-1,4,3) + (37/54) (2,-5,5) = (-1 + 37/27, 4 - 185/54, 3 + 185/54) = (4/27, 83/54, 229/54)

Now we can find the distance between P and (2,1,7):

dist = ||P - (2,1,7)|| = ||(4/27 - 2, 83/54 - 1, 229/54 - 7)|| = ||(22/27, 29/54, -97/54)|| = sqrt((22/27)^2 + (29/54)^2 + (-97/54)^2) = sqrt(2325/1458) = (5/6) sqrt(93)

Therefore, the perpendicular distance from the line to the point (2,1,7) is (5/6) sqrt(93).

Apr 24, 2023
#1
+16
0

Let the side length of the inscribed cube be $s$. Since $\angle AOB = \angle AOC = \angle BOC = 90^\circ$, the triangle $ABC$ is a right triangle. Without loss of generality, assume $AB > AC$ and $BC > AC$. Let $D$ be the foot of the altitude from $O$ to $\triangle ABC$. Then $AD = BD = CD$ because $O$ is the circumcenter of $\triangle ABC$, and $OD = s\sqrt{3}/2$ because $O$ is the centroid of $\triangle ABD$.

[asy]
unitsize(1.5 cm);

pair A, B, C, D, E, F, O, P, Q, R;

A = (0,1);
B = (1,1);
C = (1,0);
O = (1/2,1/2);
D = (1/2,0);
E = (1/2,1);
F = (0,1/2);
P = (O + F)/2;
Q = (O + D)/2;
R = (O + E)/2;

draw(A--B--C--cycle);
draw(A--O);
draw(B--O);
draw(C--O);
draw(D--O);
draw(O--E,dashed);
draw(O--F,dashed);
draw(O--P,dashed);
draw(O--Q,dashed);
draw(O--R,dashed);
draw(E--F);
draw(P--Q);
draw(P--R);
draw(D--E,dashed);
draw(D--F,dashed);

label("$A$", A, N);
label("$B$", B, N);
label("$C$", C, E);
label("$D$", D, S);
label("$E$", E, N);
label("$F$", F, W);
label("$O$", O, N);
label("$P$", P, W);
label("$Q$", Q, S);
label("$R$", R, NE);
label("$s$", (O + R)/2, NE);
[/asy]

Let $E$ be the point on $\overline{AB}$ such that $BE = s$. Then $DE = BD - BE = AD - AE$, so
\begin{align*}
s^2 &= DE^2 \
\end{align*}Solving for $s$, we find $s = AB/3$. It follows that $s = AC/3$ and $s = BC/3$ as well, so $s$ is the side length of the tetrahedron's other inscribed cube, and the desired side length is $s = AB/3 = AC/3 = BC/3 = \boxed{\frac{1}{3}}$.