If the shortest distance between the lines \fracx-\lambda3=\fracy-2-1=\fracz-11 and \fracx+2-3=\fracy+52=\fracz-44 is \frac44\sqrt30 , then the largest possible value of |\lambda| is equal to \underline\hspace2cm .

3D Geometry PYQ — JEE Main 2024 · 6 Apr · Shift 2 · Q54 with Solution

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Correct answer: 43

Let the two lines be

$L_1:\quad \frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$

and

$L_2:\quad \frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}.$

We use the formula for the shortest distance between two skew lines:

$d=\frac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times \vec{b_2})|}{|\vec{b_1}\times \vec{b_2}|}$

where:

$\vec{a_1}$ is a point on $L_1$
$\vec{a_2}$ is a point on $L_2$
$\vec{b_1},\vec{b_2}$ are direction vectors.

1. Extract points and direction vectors

From $L_1$ :

$\vec{a_1}=(\lambda,2,1), \qquad \vec{b_1}=(3,-1,1).$

From $L_2$ :

$\vec{a_2}=(-2,-5,4), \qquad \vec{b_2}=(-3,2,4).$

2. Compute cross product $\vec{b_1}\times \vec{b_2}$

\begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & -1 & 1\\ -3 & 2 & 4 \end{vmatrix}$$ $$=\hat i((-1)(4)-1\cdot 2)-\hat j(3\cdot 4-1\cdot(-3))+\hat k(3\cdot 2-(-1)(-3))$$ $$=(-6,-15,3).$$ So, $$|\vec{b_1}\times \vec{b_2}|=\sqrt{(-6)^2+(-15)^2+3^2}= \sqrt{36+225+9}=\sqrt{270}=3\sqrt{30}.$$ --- ### 3. Compute $\vec{a_2}-\vec{a_1}$ $$\vec{a_2}-\vec{a_1}=(-2-\lambda,-5-2,4-1)=(-\lambda-2,-7,3).$$ Now, $$ (\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times \vec{b_2})$$ $$= (-\lambda-2,-7,3)\cdot(-6,-15,3) $$ $$= (-\lambda-2)(-6)+(-7)(-15)+3\cdot 3 $$ $$= 6\lambda+12+105+9=6\lambda+126.$$ Hence shortest distance is $$d=\frac{|6\lambda+126|}{3\sqrt{30}}= rac{|2\lambda+42|}{\sqrt{30}}.$$ --- ### 4. Use the given distance Given, $$\frac{|2\lambda+42|}{\sqrt{30}}=\frac{44}{\sqrt{30}}.$$ So, $$|2\lambda+42|=44.$$ Thus, $$2\lambda+42=\pm 44.$$ Case 1: $$2\lambda+42=44 \implies 2\lambda=2 \implies \lambda=1.$$ Case 2: $$2\lambda+42=-44 \implies 2\lambda=-86 \implies \lambda=-43.$$ Therefore, $$|\lambda|=1 \text{ or } 43.$$ The largest possible value of $|\lambda|$ is $$\boxed{43}. $$

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View written solutionFree

Correct answer: 43

Let the two lines be

$L_1:\quad \frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$

and

$L_2:\quad \frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}.$

We use the formula for the shortest distance between two skew lines:

$d=\frac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times \vec{b_2})|}{|\vec{b_1}\times \vec{b_2}|}$

where:

$\vec{a_1}$ is a point on $L_1$
$\vec{a_2}$ is a point on $L_2$
$\vec{b_1},\vec{b_2}$ are direction vectors.

1. Extract points and direction vectors

From $L_1$ :

$\vec{a_1}=(\lambda,2,1), \qquad \vec{b_1}=(3,-1,1).$

From $L_2$ :

$\vec{a_2}=(-2,-5,4), \qquad \vec{b_2}=(-3,2,4).$

2. Compute cross product $\vec{b_1}\times \vec{b_2}$

\begin{vmatrix} \hat i & \hat j & \hat k\\ 3 & -1 & 1\\ -3 & 2 & 4 \end{vmatrix}$$ $$=\hat i((-1)(4)-1\cdot 2)-\hat j(3\cdot 4-1\cdot(-3))+\hat k(3\cdot 2-(-1)(-3))$$ $$=(-6,-15,3).$$ So, $$|\vec{b_1}\times \vec{b_2}|=\sqrt{(-6)^2+(-15)^2+3^2}= \sqrt{36+225+9}=\sqrt{270}=3\sqrt{30}.$$ --- ### 3. Compute $\vec{a_2}-\vec{a_1}$ $$\vec{a_2}-\vec{a_1}=(-2-\lambda,-5-2,4-1)=(-\lambda-2,-7,3).$$ Now, $$ (\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times \vec{b_2})$$ $$= (-\lambda-2,-7,3)\cdot(-6,-15,3) $$ $$= (-\lambda-2)(-6)+(-7)(-15)+3\cdot 3 $$ $$= 6\lambda+12+105+9=6\lambda+126.$$ Hence shortest distance is $$d=\frac{|6\lambda+126|}{3\sqrt{30}}= rac{|2\lambda+42|}{\sqrt{30}}.$$ --- ### 4. Use the given distance Given, $$\frac{|2\lambda+42|}{\sqrt{30}}=\frac{44}{\sqrt{30}}.$$ So, $$|2\lambda+42|=44.$$ Thus, $$2\lambda+42=\pm 44.$$ Case 1: $$2\lambda+42=44 \implies 2\lambda=2 \implies \lambda=1.$$ Case 2: $$2\lambda+42=-44 \implies 2\lambda=-86 \implies \lambda=-43.$$ Therefore, $$|\lambda|=1 \text{ or } 43.$$ The largest possible value of $|\lambda|$ is $$\boxed{43}. $$

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