Consider a triangle ABC whose vertices are A(0, \alpha , \alpha ), B( \alpha , 0, \alpha ) and C( \alpha , \alpha , 0), \alpha > 0. Let D be a point moving on the line x + z - 3 = 0 = y and G be the centroid of \Delta ABC. If the minimum length of GD is \sqrt 57 \over 2 , then \alpha is equal to \underline\hspace2cm .

3D Geometry PYQ — JEE Main 2022 · 30 Jun · Shift 1 · Q40 with Solution

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Correct answer: 6

Coordinates of the centroid $G$ of $\triangle ABC$

Given $A(0,\alpha,\alpha),\quad B(\alpha,0,\alpha),\quad C(\alpha,\alpha,0).$

The centroid of a triangle is the average of the coordinates of its vertices:

=\left(\frac{2\alpha}{3},\frac{2\alpha}{3},\frac{2\alpha}{3}\right).$$ 2. **Equation of the line on which $D$ moves** The point $D$ moves on the line $$x+z-3=0, \qquad y=0.$$ So a general point on this line is $$D(t,0,3-t).$$ 3. **Distance squared $GD^2$** Now, $$GD^2=\left(t-\frac{2\alpha}{3}\right)^2+\left(0-\frac{2\alpha}{3}\right)^2+\left((3-t)-\frac{2\alpha}{3}\right)^2.$$ Thus $$GD^2=\left(t-\frac{2\alpha}{3}\right)^2+\frac{4\alpha^2}{9}+\left(3-t-\frac{2\alpha}{3}\right)^2.$$ We need the minimum distance from point $G$ to the given line. 4. **Use perpendicular distance from a point to a line in 3D** The line is the intersection of the planes $y=0$ and $x+z=3$. A convenient parametrization is $$D=(0,0,3)+\lambda(1,0,-1).$$ So a point on the line is $$P=(0,0,3),$$ and its direction vector is $$\vec v=(1,0,-1).$$ Also, $$G=\left(\frac{2\alpha}{3},\frac{2\alpha}{3},\frac{2\alpha}{3}\right).$$ Hence $$\overrightarrow{PG}=G-P=\left(\frac{2\alpha}{3},\frac{2\alpha}{3},\frac{2\alpha}{3}-3\right).$$ The distance from point $G$ to the line is $$d=\frac{\|\overrightarrow{PG}\times \vec v\|}{\|\vec v\|}.$$ Compute the cross product: $$\overrightarrow{PG}\times \vec v= \begin{vmatrix} \hat i & \hat j & \hat k\\ \frac{2\alpha}{3} & \frac{2\alpha}{3} & \frac{2\alpha}{3}-3\\ 1 & 0 & -1 \end{vmatrix}.$$ This gives $$\overrightarrow{PG}\times \vec v= \left(-\frac{2\alpha}{3},\; \left(\frac{2\alpha}{3}-3\right)+\frac{2\alpha}{3},\; -\frac{2\alpha}{3}\right) = \left(-\frac{2\alpha}{3},\; \frac{4\alpha}{3}-3,\; -\frac{2\alpha}{3}\right).$$ So, $$\|\overrightarrow{PG}\times \vec v\|^2= \left(\frac{2\alpha}{3}\right)^2+\left(\frac{4\alpha}{3}-3\right)^2+\left(\frac{2\alpha}{3}\right)^2.$$ That is, $$\|\overrightarrow{PG}\times \vec v\|^2= rac{4\alpha^2}{9}+\left(\frac{4\alpha}{3}-3\right)^2+\frac{4\alpha^2}{9}.$$ Also, $$\|\vec v\|=\sqrt{1^2+0^2+(-1)^2}=\sqrt2.$$ Therefore, $$d^2=\frac{1}{2}\left[\frac{8\alpha^2}{9}+\left(\frac{4\alpha}{3}-3\right)^2\right].$$ Expand: $$\left(\frac{4\alpha}{3}-3\right)^2=\frac{16\alpha^2}{9}-8\alpha+9.$$ Hence $$d^2=\frac{1}{2}\left(\frac{8\alpha^2}{9}+\frac{16\alpha^2}{9}-8\alpha+9\right) =\frac{1}{2}\left(\frac{24\alpha^2}{9}-8\alpha+9\right).$$ So, $$d^2=\frac{1}{2}\left(\frac{8\alpha^2}{3}-8\alpha+9\right).$$ 5. **Use the given minimum length** Given minimum length of $GD$ is $$\sqrt{\frac{57}{2}}.$$ Therefore, $$d^2=\frac{57}{2}.$$ So, $$\frac{1}{2}\left(\frac{8\alpha^2}{3}-8\alpha+9\right)=\frac{57}{2}.$$ Multiply by 2: $$\frac{8\alpha^2}{3}-8\alpha+9=57.$$ Thus, $$\frac{8\alpha^2}{3}-8\alpha-48=0.$$ Multiply by 3: $$8\alpha^2-24\alpha-144=0.$$ Divide by 8: $$\alpha^2-3\alpha-18=0.$$ Factorize: $$\alpha^2-3\alpha-18=(\alpha-6)(\alpha+3)=0.$$ Since $\alpha>0$, we get $$\alpha=6.$$ 6. **Final answer** $$\boxed{6}$$

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View written solutionFree

Correct answer: 6

Coordinates of the centroid $G$ of $\triangle ABC$

Given $A(0,\alpha,\alpha),\quad B(\alpha,0,\alpha),\quad C(\alpha,\alpha,0).$

The centroid of a triangle is the average of the coordinates of its vertices:

=\left(\frac{2\alpha}{3},\frac{2\alpha}{3},\frac{2\alpha}{3}\right).$$ 2. **Equation of the line on which $D$ moves** The point $D$ moves on the line $$x+z-3=0, \qquad y=0.$$ So a general point on this line is $$D(t,0,3-t).$$ 3. **Distance squared $GD^2$** Now, $$GD^2=\left(t-\frac{2\alpha}{3}\right)^2+\left(0-\frac{2\alpha}{3}\right)^2+\left((3-t)-\frac{2\alpha}{3}\right)^2.$$ Thus $$GD^2=\left(t-\frac{2\alpha}{3}\right)^2+\frac{4\alpha^2}{9}+\left(3-t-\frac{2\alpha}{3}\right)^2.$$ We need the minimum distance from point $G$ to the given line. 4. **Use perpendicular distance from a point to a line in 3D** The line is the intersection of the planes $y=0$ and $x+z=3$. A convenient parametrization is $$D=(0,0,3)+\lambda(1,0,-1).$$ So a point on the line is $$P=(0,0,3),$$ and its direction vector is $$\vec v=(1,0,-1).$$ Also, $$G=\left(\frac{2\alpha}{3},\frac{2\alpha}{3},\frac{2\alpha}{3}\right).$$ Hence $$\overrightarrow{PG}=G-P=\left(\frac{2\alpha}{3},\frac{2\alpha}{3},\frac{2\alpha}{3}-3\right).$$ The distance from point $G$ to the line is $$d=\frac{\|\overrightarrow{PG}\times \vec v\|}{\|\vec v\|}.$$ Compute the cross product: $$\overrightarrow{PG}\times \vec v= \begin{vmatrix} \hat i & \hat j & \hat k\\ \frac{2\alpha}{3} & \frac{2\alpha}{3} & \frac{2\alpha}{3}-3\\ 1 & 0 & -1 \end{vmatrix}.$$ This gives $$\overrightarrow{PG}\times \vec v= \left(-\frac{2\alpha}{3},\; \left(\frac{2\alpha}{3}-3\right)+\frac{2\alpha}{3},\; -\frac{2\alpha}{3}\right) = \left(-\frac{2\alpha}{3},\; \frac{4\alpha}{3}-3,\; -\frac{2\alpha}{3}\right).$$ So, $$\|\overrightarrow{PG}\times \vec v\|^2= \left(\frac{2\alpha}{3}\right)^2+\left(\frac{4\alpha}{3}-3\right)^2+\left(\frac{2\alpha}{3}\right)^2.$$ That is, $$\|\overrightarrow{PG}\times \vec v\|^2= rac{4\alpha^2}{9}+\left(\frac{4\alpha}{3}-3\right)^2+\frac{4\alpha^2}{9}.$$ Also, $$\|\vec v\|=\sqrt{1^2+0^2+(-1)^2}=\sqrt2.$$ Therefore, $$d^2=\frac{1}{2}\left[\frac{8\alpha^2}{9}+\left(\frac{4\alpha}{3}-3\right)^2\right].$$ Expand: $$\left(\frac{4\alpha}{3}-3\right)^2=\frac{16\alpha^2}{9}-8\alpha+9.$$ Hence $$d^2=\frac{1}{2}\left(\frac{8\alpha^2}{9}+\frac{16\alpha^2}{9}-8\alpha+9\right) =\frac{1}{2}\left(\frac{24\alpha^2}{9}-8\alpha+9\right).$$ So, $$d^2=\frac{1}{2}\left(\frac{8\alpha^2}{3}-8\alpha+9\right).$$ 5. **Use the given minimum length** Given minimum length of $GD$ is $$\sqrt{\frac{57}{2}}.$$ Therefore, $$d^2=\frac{57}{2}.$$ So, $$\frac{1}{2}\left(\frac{8\alpha^2}{3}-8\alpha+9\right)=\frac{57}{2}.$$ Multiply by 2: $$\frac{8\alpha^2}{3}-8\alpha+9=57.$$ Thus, $$\frac{8\alpha^2}{3}-8\alpha-48=0.$$ Multiply by 3: $$8\alpha^2-24\alpha-144=0.$$ Divide by 8: $$\alpha^2-3\alpha-18=0.$$ Factorize: $$\alpha^2-3\alpha-18=(\alpha-6)(\alpha+3)=0.$$ Since $\alpha>0$, we get $$\alpha=6.$$ 6. **Final answer** $$\boxed{6}$$

3D Geometry question

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