Question

Let $A\equiv (3,-1,-1)$ and $B\equiv (3,1,0)$ be the points in the plane $x+y+z=1$ and $2x-y-z=5$ respectively. C is a variable point lying in both the planes such that perimeter of $△ABC$ is minimum then coordinates of point C is.

• A. $(2,\frac{-1}{\sqrt{3}(1+\sqrt{2})},\frac{-1}{\sqrt{2}})$
• B. $(2,\frac{-1}{\sqrt{2}(1+\sqrt{2})},\frac{-1}{\sqrt{2}})$
• C. $(1,\frac{-1}{\sqrt{2}(1+\sqrt{3})},\frac{-1}{\sqrt{3}})$
• D. $(1,\frac{-1}{\sqrt{3}(1+\sqrt{2})},\frac{-1}{\sqrt{3}})$

Answer 1

Answer: (A)

$(2,\frac{-1}{\sqrt{3}(1+\sqrt{2})},\frac{-1}{\sqrt{2}})$

Given the points $A(3,-1,-1)$ and $B(3,1,0)$ lying on the plane $x+y+z=1$ and $2x-y-z=5$ respectively.

Now rotating the plane $x+y+z=1$ so that it lies on the other plane and finding the coordinates of the point $A(3,-1,-1)$ say $A^{{}^{\prime }}$ on the rotated plane $(2x-y-z=5)$.

Now the equation of $A^{{}^{\prime }}B$ with the intersection of the given two planes is

$\frac{x-2}{0}=\frac{y+1}{1}=\frac{z-0}{-1}=t\left(say\right)$

Now $x=2,y=t-1,z=-t$

Hence the coordinate of point $C(2,t-1,-t)$

Using the distance formula, we calculate $\overline{AC}$ and $\overline{BC}$ , and adding their sum, we get;

$\sqrt{2+2t+2t^2}+\sqrt{3+2t^2}=y\left(say\right)$

For perimeter to be minimum, we differentiate the above equation w.r.t. 't', we get

$\frac{dy}{dt}=\frac{1+2t}{\sqrt{2+2t+2t^2}}+\frac{2t}{\sqrt{3+2t^2}}=0$

On solving the above equation, we get

$t=\frac{1}{\sqrt{2}}$

Hence the cordinate of C is $(2,\frac{1}{\sqrt{2}}-1,-\frac{1}{\sqrt{2}})$

$⟹C(2,\frac{-2-\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$

$⟹C(2,\frac{-1}{\sqrt{2}(1+\sqrt{2})},-\frac{1}{\sqrt{2}})$

Option (b) is correct.