JEE Main 2021 — Mathematics Vectors & 3D Geometry

The given lines are $3(x-1)=6(y-2)=2(z-1)$3(x−1)=6(y−2)=2(z−1) and $4(x-2)=2(y-\lambda )=(z-3)$4(x−2)=2(y−λ)=(z−3)

And, they can be written as ${L}_{1}:\frac{(x-1)}{2}=\frac{(y-2)}{1}=\frac{(z-1)}{3}$L1​:2(x−1)​=1(y−2)​=3(z−1)​ and

$${L}_{2}:\frac{(x-2)}{1}=\frac{y-\lambda }{2}=\frac{z-3}{4}.$$L2​:1(x−2)​=2y−λ​=4z−3​.

The lines are in the form $\frac{x-{x}_{1}}{l}=\frac{y-{y}_{1}}{m}=\frac{z-{z}_{1}}{n},$lx−x1​​=my−y1​​=nz−z1​​, where $({x}_{1},{y}_{1},{z}_{1})$(x1​,y1​,z1​) is a point on the line and $l,m,n$l,m,n are the direction ratios of the vector parallel to the line.

Thus, the position vectors of the points on the line are respectively \vec{{a}_{1}}=\hat{i}+2\hat{j}+\hat{k}&\vec{{a}_{2}}=2\hat{i}+\lambda \hat{j}+3\hat{k} and direction of the lines are along the vectors ${\vec{r}}_{1}=2\hat{i}+\hat{j}+3\hat{k}$r1​=2i^+j^​+3k^ and $\vec{{r}_{2}}=\hat{i}+2\hat{j}+4\hat{k}.$r2​​=i^+2j^​+4k^.

Thus, we have $\vec{a}=\vec{{a}_{2}}-\vec{{a}_{1}}=\hat{i}+(\lambda -2)\hat{j}+2\hat{k}.$a=a2​​−a1​​=i^+(λ−2)j^​+2k^.

Shortest distance $=$= Projection of $\vec{a}$a on $\vec{{r}_{1}}\times \vec{{r}_{2}}$r1​​×r2​​

$$=\frac{|\vec{a}\cdot (\vec{{r}_{1}}\times \vec{{r}_{2}})|}{|\vec{{r}_{1}}\times \vec{{r}_{2}}|}$$=∣r1​​×r2​​∣∣a⋅(r1​​×r2​​)∣​

Now $\vec{{r}_{1}}\times \vec{{r}_{2}}=|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & 2 & 4\end{matrix}|$r1​​×r2​​=∣i^21​j^​12​k^34​∣

$$\Rightarrow \vec{{r}_{1}}\times \vec{{r}_{2}}=\hat{i}(4-6)-\hat{j}(8-3)+\hat{k}(4-1)$$⇒r1​​×r2​​=i^(4−6)−j^​(8−3)+k^(4−1)

$$\Rightarrow \vec{{r}_{1}}\times \vec{{r}_{2}}=-2\hat{i}-5\hat{j}+3\hat{k}$$⇒r1​​×r2​​=−2i^−5j^​+3k^

$$\Rightarrow |\vec{{r}_{1}}\times \vec{{r}_{2}}|=\sqrt{{(-2)}^{2}+{(-5)}^{2}+{3}^{2}}=\sqrt{4+25+9}=\sqrt{38}$$⇒∣r1​​×r2​​∣=(−2)2+(−5)2+32​=4+25+9​=38​

And, $|\vec{a}\cdot (\vec{{r}_{1}}\times {\vec{r}}_{2})|=|\begin{matrix}1 & \lambda -2 & 2 \\ 2 & 1 & 3 \\ 1 & 2 & 4\end{matrix}|$∣a⋅(r1​​×r2​)∣=∣121​λ−212​234​∣

$$\Rightarrow |\vec{a}\cdot (\vec{{r}_{1}}\times {\vec{r}}_{2})|=|1(4-6)-(\lambda -2)(8-3)+2(4-1)|$$⇒∣a⋅(r1​​×r2​)∣=∣1(4−6)−(λ−2)(8−3)+2(4−1)∣

$$\Rightarrow |\vec{a}\cdot (\vec{{r}_{1}}\times {\vec{r}}_{2})|=|-2-5\lambda +10+6|$$⇒∣a⋅(r1​​×r2​)∣=∣−2−5λ+10+6∣

$$\Rightarrow |\vec{a}\cdot (\vec{{r}_{1}}\times {\vec{r}}_{2})|=|14-5\lambda |$$⇒∣a⋅(r1​​×r2​)∣=∣14−5λ∣

Given, the shortest distance is $\frac{1}{\sqrt{38}},$38​1​, hence, we have

$$\frac{1}{\sqrt{38}}=\frac{|14-5\lambda |}{\sqrt{38}}$$38​1​=38​∣14−5λ∣​

$$\Rightarrow |14-5\lambda |=1$$⇒∣14−5λ∣=1

$\Rightarrow 14-5\lambda =1$⇒14−5λ=1 or $14-5\lambda =-1$14−5λ=−1

$\Rightarrow \lambda =\frac{13}{5}$⇒λ=513​ or $3$3

$\therefore$∴ Integral value is $\lambda =3.$λ=3.