Minimum value of $PA+PB$ is

Question

If $P(x,y,z)$ lie on line $\displaystyle \frac{x+2}{2}=\frac{y+7}{2}=\frac{z-2}{1}$ and $A(5,3,4)$ and $B(1,-1,2)$ . Then minimum value of $PA+PB$ is

what i try

let $\displaystyle \frac{x+2}{2}=\frac{y+7}{2}=\frac{z-2}{1}=\lambda$

Then $P(2\lambda-2,2\lambda-7,\lambda+2)$

$PA+PB=\sqrt{(2\lambda-7)^2+(2\lambda-10)^2+(\lambda-2)^2}+\sqrt{(2\lambda-3)^2+(2\lambda-8)^2+\lambda^2)}$

$PA+PB=\sqrt{9\lambda^2-72\lambda+153}+\sqrt{ 9\lambda^2-36\lambda+45}$

how do i minimize it b3cause derivative method is very tedious help me please

Answer 1

Note that $$ |PA|+|PB|=3\left(\sqrt{(\lambda-4)^2+1}+\sqrt{(\lambda-2)^2+1}\right). $$ The expression in the parenthesis is equal to the sum of distance between $X=(\lambda,0)$ and $Y=(4,1)$ and distance between $X=(\lambda,0)$ and $Z=(2,-1)$. Then, by the triangle inequality, $|XY|+|XZ| $ is minimized when $X$ is on the line segment between $Y$ and $Z$. This gives $$|XY|+|XZ|\ge |YZ|=\sqrt{8}=2\sqrt 2$$ hence giving $$ |PA|+|PB|\ge 6\sqrt{2}. $$

Answer 2

By Minkowski we obtain: $$\sqrt{9\lambda^2-72\lambda+153}+\sqrt{ 9\lambda^2-36\lambda+45}=$$ $$=3\left(\sqrt{(4-\lambda)^2+1}+\sqrt{(\lambda-2)^2+1}\right)\geq$$ $$\geq3\sqrt{(4-2)^2+(1+1)^2}=6\sqrt2.$$ The equality occurs for $\lambda=3,$ which says that we got a minimal value.