JEE Main 2022 (Online) 30th June Morning Shift Mathematics - 3D Geometry Question 200 English Explanation

Given, G is the centroid of $$\Delta$$ABC

$$\therefore$$ $$G = \left( {{{0 + \alpha + \alpha } \over 3},\,{{\alpha + 0 + \alpha } \over 3},\,{{\alpha + \alpha + 0} \over 3}} \right)$$

$$ = \left( {{{2\alpha } \over 3},\,{{2\alpha } \over 3},\,{{2\alpha } \over 3}} \right)$$

Also given, D is a point moving on the line $$x + z - 3 = 0 = y$$

Let $$D = (h,\,0,\,k)$$

$$x + z - 3 = 0 \Rightarrow h + k - 3 = 0 \Rightarrow h = 3 - k$$

$$\therefore$$ $$D = (3 - k,\,0,\,k)$$

Now, length of $$GD = d$$

$$ = \sqrt {{{\left( {{{2\alpha } \over 3} - 3 + k} \right)}^2} + {{\left( {{{2\alpha } \over 3}} \right)}^2} + {{\left( {{{2\alpha } \over 3} - k} \right)}^2}} $$

$$ \Rightarrow {d^2} = {\left( {{{2\alpha } \over 3} - 3 + k} \right)^2} + {\left( {{{2\alpha } \over 3}} \right)^2} + {\left( {{{2\alpha } \over 3} - k} \right)^2}$$

Differentiating both side with respect to k, we get

$$2dd' = 2\left( {{{2\alpha } \over 3} - 3 + k} \right) + 0 + 2\left( {{{2\alpha } \over 3} - k} \right) \times ( - 1)$$

For maximum or minimum value of k, $$d' = 0$$

$$\therefore$$ $$0 = 2\left( {{{2\alpha } \over 3} - 3 + k} \right) - 2\left( {{{2\alpha } \over 3} - k} \right)$$

$$ \Rightarrow 2\left( {{{2\alpha } \over 3} - 3 + k} \right) = 2\left( {{{2\alpha } \over 3} - k} \right)$$

$$ \Rightarrow - 3 + k = - k$$

$$ \Rightarrow k = {3 \over 2}$$

$$\therefore$$ $$G{D^2} = {\left( {{{2\alpha } \over 3} - 3 + {3 \over 2}} \right)^2} + {\left( {{{2\alpha } \over 3}} \right)^2} + {\left( {{{2\alpha } \over 3} - {3 \over 2}} \right)^2} = {{57} \over 2}$$

$$ \Rightarrow {{{{(4\alpha - 9)}^2}} \over {36}} + {{16{\alpha ^2}} \over {36}} + {{{{(4\alpha - 9)}^2}} \over {36}} = {{57} \over 2}$$

$$ \Rightarrow {(4\alpha - 9)^2} + 16{\alpha ^2} + {(4\alpha - 9)^2} = 57 \times 18$$

$$ \Rightarrow 16{\alpha ^2} - 72\alpha + 81 + 16{\alpha ^2} + 16{\alpha ^2} - 72\alpha + 81 = 57 \times 18$$

$$ \Rightarrow 48{\alpha ^2} - 144\alpha + 162 = 1026$$

$$ \Rightarrow 24{\alpha ^2} - 72\alpha + 81 - 513 = 0$$

$$ \Rightarrow 24{\alpha ^2} - 72\alpha - 432 = 0$$

$$ \Rightarrow {\alpha ^2} - 3\alpha - 18 = 0$$

$$ \Rightarrow {\alpha ^2} - 6\alpha + 3\alpha - 18 = 0$$

$$ \Rightarrow \alpha (\alpha - 6) + 3(\alpha - 6) = 0$$

$$ \Rightarrow (\alpha - 6)(\alpha + 3) = 0$$

$$ \Rightarrow \alpha = 6,\, - 3$$

Given, $$\alpha > 0$$

$$\therefore$$ Possible value of $$\alpha = 6$$.

3d Geometry

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