If the shortest distance between the lines \beginarrayll L_1: \vecr=(2+\lambda) \hati+(1-3 \lambda) \hatj+(3+4 \lambda) \hatk, & \lambda \in \mathbbR L_2: \vecr=2(1+\mu) \hati+3(1+\mu) \hatj+(5+\mu) \hatk, & \mu \in \mathbbR \endarray is \fracm\sqrtn , where \operatornamegcd(m, n)=1 , then the value of m+n equals

3D Geometry PYQ — JEE Main 2024 · 8 Apr · Shift 1 · Q49 with Solution

View written solutionFree

Correct answer: B

Write the lines in vector form

For $L_1: \vec r=(2+\lambda)\hat i+(1-3\lambda)\hat j+(3+4\lambda)\hat k$ we identify

a point on $L_1$ : $A=(2,1,3)$
direction vector of $L_1$ : $\vec d_1=(1,-3,4)$

For $L_2: \vec r=2(1+\mu)\hat i+3(1+\mu)\hat j+(5+\mu)\hat k$ we simplify: $\vec r=(2+2\mu)\hat i+(3+3\mu)\hat j+(5+\mu)\hat k$ So,

a point on $L_2$ : $B=(2,3,5)$
direction vector of $L_2$ : $\vec d_2=(2,3,1)$

Use the formula for shortest distance between two skew lines

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

Here, $\vec b-\vec a=B-A=(2,3,5)-(2,1,3)=(0,2,2)$

Compute the cross product

\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -3 & 4 \\ 2 & 3 & 1 \end{vmatrix}$$ $$=\hat i((-3)(1)-4(3)) - \hat j((1)(1)-4(2)) + \hat k((1)(3)-(-3)(2))$$ $$=\hat i(-3-12)-\hat j(1-8)+\hat k(3+6)$$ $$=-15\hat i+7\hat j+9\hat k$$ So, $$\vec d_1\times \vec d_2=(-15,7,9)$$ Its magnitude is $$|\vec d_1\times \vec d_2|=\sqrt{(-15)^2+7^2+9^2}$$ $$=\sqrt{225+49+81}=\sqrt{355}$$ --- 4. **Compute the scalar triple product** $$ (\vec b-\vec a)\cdot (\vec d_1\times \vec d_2)=(0,2,2)\cdot(-15,7,9) $$ $$=0(-15)+2(7)+2(9)=14+18=32$$ Therefore, $$D=\frac{32}{\sqrt{355}}$$ So, $$m=32, \qquad n=355$$ Since $\gcd(32,355)=1$, this is already in lowest terms. Hence, $$m+n=32+355=387$$ --- 5. **Check with options** - A: $384$ - B: $387$ ✅ - C: $390$ - D: $377$ So the correct option is **B**.

More from 3D Geometry

View written solutionFree

Correct answer: B

Write the lines in vector form

For $L_1: \vec r=(2+\lambda)\hat i+(1-3\lambda)\hat j+(3+4\lambda)\hat k$ we identify

a point on $L_1$ : $A=(2,1,3)$
direction vector of $L_1$ : $\vec d_1=(1,-3,4)$

For $L_2: \vec r=2(1+\mu)\hat i+3(1+\mu)\hat j+(5+\mu)\hat k$ we simplify: $\vec r=(2+2\mu)\hat i+(3+3\mu)\hat j+(5+\mu)\hat k$ So,

a point on $L_2$ : $B=(2,3,5)$
direction vector of $L_2$ : $\vec d_2=(2,3,1)$

Use the formula for shortest distance between two skew lines

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

Here, $\vec b-\vec a=B-A=(2,3,5)-(2,1,3)=(0,2,2)$

Compute the cross product

\begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -3 & 4 \\ 2 & 3 & 1 \end{vmatrix}$$ $$=\hat i((-3)(1)-4(3)) - \hat j((1)(1)-4(2)) + \hat k((1)(3)-(-3)(2))$$ $$=\hat i(-3-12)-\hat j(1-8)+\hat k(3+6)$$ $$=-15\hat i+7\hat j+9\hat k$$ So, $$\vec d_1\times \vec d_2=(-15,7,9)$$ Its magnitude is $$|\vec d_1\times \vec d_2|=\sqrt{(-15)^2+7^2+9^2}$$ $$=\sqrt{225+49+81}=\sqrt{355}$$ --- 4. **Compute the scalar triple product** $$ (\vec b-\vec a)\cdot (\vec d_1\times \vec d_2)=(0,2,2)\cdot(-15,7,9) $$ $$=0(-15)+2(7)+2(9)=14+18=32$$ Therefore, $$D=\frac{32}{\sqrt{355}}$$ So, $$m=32, \qquad n=355$$ Since $\gcd(32,355)=1$, this is already in lowest terms. Hence, $$m+n=32+355=387$$ --- 5. **Check with options** - A: $384$ - B: $387$ ✅ - C: $390$ - D: $377$ So the correct option is **B**.

3D Geometry question

3D Geometry question

More from 3D Geometry

3D Geometry question

3D Geometry question

More from 3D Geometry