# Minimum value of $PA+PB$ is

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

## 2 Answers

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 $$|PA|+|PB|\ge 6\sqrt{2}.$$

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.