Question

Let P(0,3,4) and Q(5,12,0) be 2 given points. A is point on x axis for which PA+QA is minimum. B is a point on Y axis for which PB+QB is minimum. C is a point on z axis for which PC+QC is minimum. then find eq of plane through ABC and volume of tetrahedron OABC

Answer 1

Plane: $\displaystyle \frac{17x}{25}+\frac{y}{7}+\frac{4z}{13}=1$

Volume of Tetrahedron OABC: $\displaystyle \frac{2275}{408}$

Answer 2

We are given

$$P(0,3,4),\quad Q(5,12,0)$$

and we need to choose

• A on the x‐axis, i.e. $A=(a,0,0)$ so that $PA+QA$ is minimum,
• B on the y‐axis, i.e. $B=(0,b,0)$ so that $PB+QB$ is minimum,
• C on the z‐axis, i.e. $C=(0,0,c)$ so that $PC+QC$ is minimum.

Then we find (i) the equation of the plane through A, B, C and (ii) the volume of tetrahedron OABC.

We now show a brief solution.

Step 1. Find A on the x–axis

Let

$$A=(a,0,0).$$

Then

$$PA=\sqrt{(a-0)^2+(0-3)^2+(0-4)^2}=\sqrt{a^2+9+16}=\sqrt{a^2+25},$$ $$QA=\sqrt{(a-5)^2+(0-12)^2+(0-0)^2}=\sqrt{(a-5)^2+144}.$$

So the distance function is

$$f(a)=\sqrt{a^2+25}+\sqrt{(a-5)^2+144}.$$

Differentiate with respect to $a$ and set $f'(a)=0$. (A standard trick is to set up the derivative:)

$$\frac{d}{da}\sqrt{a^2+25}=\frac{a}{\sqrt{a^2+25}},\quad \frac{d}{da}\sqrt{(a-5)^2+144}=\frac{a-5}{\sqrt{(a-5)^2+144}}.$$

Thus the condition is

$$\frac{a}{\sqrt{a^2+25}}+\frac{a-5}{\sqrt{(a-5)^2+144}}=0.$$

Solve by isolating one term and then squaring:

$$\frac{a}{\sqrt{a^2+25}}=\frac{5-a}{\sqrt{(a-5)^2+144}},$$

Squaring gives:

$$\frac{a^2}{a^2+25}=\frac{(5-a)^2}{(a-5)^2+144}.$$

Noting that $(5-a)^2=(a-5)^2,$ cross–multiply:

$$a^2[(a-5)^2+144]= (a-5)^2(a^2+25).$$

Expanding and canceling the common term we get:

$$144a^2=25(a-5)^2.$$

Taking positive square roots (we expect the optimum to occur with $a>0$) gives

$$12a=5|a-5|.$$

(For $a$ between 0 and 5, $a-5$ is negative so $|a-5|=5-a$.) Thus

$$12a=5(5-a)\quad\Longrightarrow\quad 12a=25-5a\quad\Longrightarrow\quad 17a=25,$$

so

$$a=\frac{25}{17}.$$

Thus,

$$A=\left(\frac{25}{17},\,0,\,0\right).$$

Step 2. Find B on the y–axis

Let

$$B=(0,b,0).$$

Then

$$PB=\sqrt{(0-0)^2+(b-3)^2+(0-4)^2}=\sqrt{(b-3)^2+16},$$ $$QB=\sqrt{(5-0)^2+(12-b)^2+(0-0)^2}=\sqrt{25+(12-b)^2}.$$

Define

$$g(b)=\sqrt{(b-3)^2+16}+\sqrt{(12-b)^2+25}.$$

Differentiating and setting $g'(b)=0$ leads to

$$\frac{b-3}{\sqrt{(b-3)^2+16}}-\frac{12-b}{\sqrt{(12-b)^2+25}}=0.$$

So

$$\frac{b-3}{\sqrt{(b-3)^2+16}}=\frac{12-b}{\sqrt{(12-b)^2+25}}.$$

Squaring both sides yields:

$$\frac{(b-3)^2}{(b-3)^2+16}=\frac{(12-b)^2}{(12-b)^2+25}.$$

Cross–multiplying,

$$(b-3)^2[(12-b)^2+25] = (12-b)^2[(b-3)^2+16].$$

Cancel the common term and simplify to obtain:

$$25(b-3)^2 = 16(12-b)^2.$$

Taking square roots (and for $3<b<12$ we have $b-3>0$ and $12-b>0$):

$$5(b-3)=4(12-b).$$

Solve:

$$5b-15 = 48-4b\quad\Longrightarrow\quad9b=63\quad\Longrightarrow\quad b=7.$$

Thus,

$$B=(0,7,0).$$

Step 3. Find C on the z‐axis

Let

$$C=(0,0,c).$$

Then

$$PC=\sqrt{(0-0)^2+(0-3)^2+(c-4)^2}=\sqrt{9+(c-4)^2},$$ $$QC=\sqrt{(5-0)^2+(12-0)^2+(0-c)^2}=\sqrt{25+144+c^2}=\sqrt{c^2+169}.$$

So

$$h(c)=\sqrt{(c-4)^2+9}+\sqrt{c^2+169}.$$

Differentiate with respect to $c$ to get:

$$\frac{d}{dc}\sqrt{(c-4)^2+9}=\frac{c-4}{\sqrt{(c-4)^2+9}},\quad \frac{d}{dc}\sqrt{c^2+169}=\frac{c}{\sqrt{c^2+169}}.$$

Setting the derivative to zero,

$$\frac{c-4}{\sqrt{(c-4)^2+9}}+\frac{c}{\sqrt{c^2+169}}=0.$$

That is,

$$\frac{c-4}{\sqrt{(c-4)^2+9}} = -\frac{c}{\sqrt{c^2+169}}.$$

Square both sides:

$$\frac{(c-4)^2}{(c-4)^2+9}=\frac{c^2}{c^2+169}.$$

Cross–multiply:

$$(c-4)^2(c^2+169)= c^2[(c-4)^2+9].$$

A cancellation leads to

$$169(c-4)^2= 9c^2.$$

Thus,

$$(c-4)^2=\frac{9}{169}c^2.$$

Taking square roots (and noting that for the optimum we expect $c<4$ so that $4-c>0$):

$$4-c=\frac{3}{13}c.$$

So,

$$4=\left(1+\frac{3}{13}\right)c=\frac{16}{13}c\quad\Longrightarrow\quad c=\frac{13}{4}.$$

Thus,

$$C=\Bigl(0,\,0,\,\frac{13}{4}\Bigr).$$

Step 4. Equation of the Plane Through A, B, C

We now have

$$A=\left(\frac{25}{17},0,0\right),\quad B=(0,7,0),\quad C=\Bigl(0,0,\frac{13}{4}\Bigr).$$

A very neat form is to use the intercept form of a plane. Indeed, a plane with intercepts $l, m, n$ on the $x,y,z$–axes has equation

$$\frac{x}{l}+\frac{y}{m}+\frac{z}{n}=1.$$

Here, the intercepts are exactly the coordinates of A, B, C so that

$$l=\frac{25}{17},\quad m=7,\quad n=\frac{13}{4}.$$

Thus the plane is

$$\frac{17x}{25}+\frac{y}{7}+\frac{4z}{13}=1.$$

Step 5. Volume of Tetrahedron OABC

The tetrahedron has vertices O (origin), A, B, C. Its volume is

$$V=\frac{1}{6}\left|\det\begin{pmatrix} \frac{25}{17} & 0 & 0\\[1mm] 0 & 7 & 0\\[1mm] 0 & 0 & \frac{13}{4} \end{pmatrix}\right|.$$

The determinant is the product of the diagonal entries:

$$\det = \frac{25}{17}\times 7\times\frac{13}{4}=\frac{25\cdot7\cdot13}{17\cdot4}=\frac{2275}{68}.$$

Thus,

$$V=\frac{1}{6}\cdot\frac{2275}{68}=\frac{2275}{408}.$$

Minimal –to–the–point explanation

1. Write the expression for $PA+QA$ when $A=(a,0,0)$ and minimize to get $a=\frac{25}{17}$.
2. Similarly, by writing $PB+QB$ for $B=(0,b,0)$ you get $b=7$.
3. Writing $PC+QC$ for $C=(0,0,c)$ leads to $c=\frac{13}{4}$.
4. The plane through A, B, C is written in intercept form as $$\frac{x}{25/17}+\frac{y}{7}+\frac{z}{13/4}=1,\quad \text{or}\quad \frac{17x}{25}+\frac{y}{7}+\frac{4z}{13}=1.$$
5. The volume of tetrahedron OABC is $V=\frac{1}{6}\left(\frac{25}{17}\cdot7\cdot\frac{13}{4}\right)=\frac{2275}{408}$.