Let the shortest distance between the lines L: \fracx-5-2=\fracy-\lambda0=\fracz+\lambda1, \lambda \geq 0 and L_1: x+1=y-1=4-z be 2 \sqrt6 . If (\alpha, \beta, \gamma) lies on L , then which of the following is NOT possible?

3D Geometry PYQ — JEE Main 2023 · 31 Jan · Shift 1 · Q34 with Solution

View written solutionFree

Correct answer: A

Write the lines in parametric form

Given $L: \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1},\quad \lambda\ge 0$

Let the common parameter be $t$ . Then $x=5-2t,\quad y=\lambda,\quad z=-\lambda+t$ So a point on $L$ is $A=(5,\lambda,-\lambda)$ and its direction vector is $\vec d_1=(-2,0,1).$

For $L_1: x+1=y-1=4-z$ let the common parameter be $s$ . Then $x=s-1,\quad y=s+1,\quad z=4-s.$ So a point on $L_1$ is $B=(-1,1,4)$ and its direction vector is $\vec d_2=(1,1,-1).$

Use formula for shortest distance between two skew lines

Shortest distance between $\vec r=\vec a+t\vec d_1,\qquad \vec r=\vec b+s\vec d_2$ is $D=\frac{|(\vec b-\vec a)\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.$

Here $\vec a=(5,\lambda,-\lambda),\qquad \vec b=(-1,1,4).$ So $\vec b-\vec a=(-6,1-\lambda,4+\lambda).$

Now,

\begin{vmatrix} \hat i & \hat j & \hat k\\ -2 & 0 & 1\\ 1 & 1 & -1 \end{vmatrix} =(-1,-1,-2).$$ Hence $$|\vec d_1\times \vec d_2|=\sqrt{1+1+4}=\sqrt6.$$ Also, $$(\vec b-\vec a)\cdot(\vec d_1\times \vec d_2) =(-6,1-\lambda,4+\lambda)\cdot(-1,-1,-2).$$ Compute: $$=6-(1-\lambda)-2(4+\lambda) =6-1+\lambda-8-2\lambda =-3-\lambda.$$ Therefore $$D=\frac{|{-3-\lambda}|}{\sqrt6}= rac{\lambda+3}{\sqrt6}$$ since $\lambda\ge 0$. Given shortest distance is $2\sqrt6$, so $$\frac{\lambda+3}{\sqrt6}=2\sqrt6$$ $$\lambda+3=12$$ $$\lambda=9.$$ --- 3. **Equation of line $L$ for $\lambda=9$** Substitute $\lambda=9$: $$x=5-2t,\quad y=9,\quad z=-9+t.$$ If $(\alpha,\beta,\gamma)$ lies on $L$, then for some $t$, $$\alpha=5-2t,\qquad \beta=9,\qquad \gamma=-9+t.$$ Eliminate $t$ using $$t=\gamma+9.$$ Then $$\alpha=5-2(\gamma+9)=-13-2\gamma.$$ So every point on $L$ satisfies $$\alpha+2\gamma=-13.$$ Thus any relation inconsistent with this is **not possible**. --- 4. **Check options** ### Option A $$\alpha+2\gamma=24$$ But on the line, $$\alpha+2\gamma=-13,$$ so this is impossible. ### Option B $$2\alpha+\gamma=7$$ Using $\alpha=5-2t,\ \gamma=-9+t$, $$2\alpha+\gamma=2(5-2t)+(-9+t)=1-3t.$$ Set equal to $7$: $$1-3t=7\Rightarrow t=-2.$$ This gives a valid point on the line. So possible. ### Option C $$\alpha-2\gamma=19$$ $$\alpha-2\gamma=(5-2t)-2(-9+t)=23-4t.$$ Set equal to $19$: $$23-4t=19\Rightarrow t=1.$$ Valid. So possible. ### Option D $$2\alpha-\gamma=9$$ $$2\alpha-\gamma=2(5-2t)-(-9+t)=19-5t.$$ Set equal to $9$: $$19-5t=9\Rightarrow t=2.$$ Valid. So possible. --- 5. **Conclusion** The only relation which is **not possible** is $$\boxed{\text{A}}.$$

More from 3D Geometry

View written solutionFree

Correct answer: A

Write the lines in parametric form

Given $L: \frac{x-5}{-2}=\frac{y-\lambda}{0}=\frac{z+\lambda}{1},\quad \lambda\ge 0$

Let the common parameter be $t$ . Then $x=5-2t,\quad y=\lambda,\quad z=-\lambda+t$ So a point on $L$ is $A=(5,\lambda,-\lambda)$ and its direction vector is $\vec d_1=(-2,0,1).$

For $L_1: x+1=y-1=4-z$ let the common parameter be $s$ . Then $x=s-1,\quad y=s+1,\quad z=4-s.$ So a point on $L_1$ is $B=(-1,1,4)$ and its direction vector is $\vec d_2=(1,1,-1).$

Use formula for shortest distance between two skew lines

Shortest distance between $\vec r=\vec a+t\vec d_1,\qquad \vec r=\vec b+s\vec d_2$ is $D=\frac{|(\vec b-\vec a)\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.$

Here $\vec a=(5,\lambda,-\lambda),\qquad \vec b=(-1,1,4).$ So $\vec b-\vec a=(-6,1-\lambda,4+\lambda).$

Now,

\begin{vmatrix} \hat i & \hat j & \hat k\\ -2 & 0 & 1\\ 1 & 1 & -1 \end{vmatrix} =(-1,-1,-2).$$ Hence $$|\vec d_1\times \vec d_2|=\sqrt{1+1+4}=\sqrt6.$$ Also, $$(\vec b-\vec a)\cdot(\vec d_1\times \vec d_2) =(-6,1-\lambda,4+\lambda)\cdot(-1,-1,-2).$$ Compute: $$=6-(1-\lambda)-2(4+\lambda) =6-1+\lambda-8-2\lambda =-3-\lambda.$$ Therefore $$D=\frac{|{-3-\lambda}|}{\sqrt6}= rac{\lambda+3}{\sqrt6}$$ since $\lambda\ge 0$. Given shortest distance is $2\sqrt6$, so $$\frac{\lambda+3}{\sqrt6}=2\sqrt6$$ $$\lambda+3=12$$ $$\lambda=9.$$ --- 3. **Equation of line $L$ for $\lambda=9$** Substitute $\lambda=9$: $$x=5-2t,\quad y=9,\quad z=-9+t.$$ If $(\alpha,\beta,\gamma)$ lies on $L$, then for some $t$, $$\alpha=5-2t,\qquad \beta=9,\qquad \gamma=-9+t.$$ Eliminate $t$ using $$t=\gamma+9.$$ Then $$\alpha=5-2(\gamma+9)=-13-2\gamma.$$ So every point on $L$ satisfies $$\alpha+2\gamma=-13.$$ Thus any relation inconsistent with this is **not possible**. --- 4. **Check options** ### Option A $$\alpha+2\gamma=24$$ But on the line, $$\alpha+2\gamma=-13,$$ so this is impossible. ### Option B $$2\alpha+\gamma=7$$ Using $\alpha=5-2t,\ \gamma=-9+t$, $$2\alpha+\gamma=2(5-2t)+(-9+t)=1-3t.$$ Set equal to $7$: $$1-3t=7\Rightarrow t=-2.$$ This gives a valid point on the line. So possible. ### Option C $$\alpha-2\gamma=19$$ $$\alpha-2\gamma=(5-2t)-2(-9+t)=23-4t.$$ Set equal to $19$: $$23-4t=19\Rightarrow t=1.$$ Valid. So possible. ### Option D $$2\alpha-\gamma=9$$ $$2\alpha-\gamma=2(5-2t)-(-9+t)=19-5t.$$ Set equal to $9$: $$19-5t=9\Rightarrow t=2.$$ Valid. So possible. --- 5. **Conclusion** The only relation which is **not possible** is $$\boxed{\text{A}}.$$

3D Geometry question

3D Geometry question

More from 3D Geometry

3D Geometry question

3D Geometry question

More from 3D Geometry