JEE Main 2022 — Mathematics Vectors & 3D Geometry

We know that, shortest distance between two line is given by $\frac{|({a}_{2}-{a}_{1})\cdot ({b}_{1}\times {b}_{2})|}{|{b}_{1}\times {b}_{2}|}$∣b1​×b2​∣∣(a2​−a1​)⋅(b1​×b2​)∣​

Here, ${a}_{1}=(1,2,3)$a1​=(1,2,3) ${a}_{2}=(2,4,5)$a2​=(2,4,5)

${\vec{b}}_{2}=2\hat{i}+3\hat{j}+\lambda \hat{k}$b2​=2i^+3j^​+λk^ and ${\vec{b}}_{2}=\hat{i}+4\hat{j}+5\hat{k}$b2​=i^+4j^​+5k^

So, S.D. $=\frac{|((2-1)\hat{i}+(4-2)\hat{j}+(5-3)\hat{k})\cdot ({\vec{b}}_{1}\times {\vec{b}}_{2})|}{|{b}_{1}\times {b}_{2}|}$=∣b1​×b2​∣∣((2−1)i^+(4−2)j^​+(5−3)k^)⋅(b1​×b2​)∣​

Calculating ${\vec{b}}_{1}\times {\vec{b}}_{2}=|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & \lambda \\ 1 & 4 & 5\end{matrix}|$b1​×b2​=∣i^21​j^​34​k^λ5​∣

$$=\hat{i}(15-4\lambda )+\hat{j}(\lambda -10)+\hat{k}(5)$$=i^(15−4λ)+j^​(λ−10)+k^(5)

$$=(15-4\lambda )\hat{i}+(\lambda -10)\hat{j}+5\hat{k}$$=(15−4λ)i^+(λ−10)j^​+5k^

So, $|{\vec{b}}_{1}\times {\vec{b}}_{2}|=\sqrt{{(15-4\lambda )}^{2}+{(\lambda -10)}^{2}+25}$∣b1​×b2​∣=(15−4λ)2+(λ−10)2+25​

Now again,

S.D. $=\frac{|(\hat{i}+2\hat{j}+2\hat{k})\cdot [(15-4\lambda )\hat{i}+(\lambda -10)\hat{j}+5k]|}{\sqrt{{(15-4\lambda )}^{2}+{(\lambda -10)}^{2}+25}}$=(15−4λ)2+(λ−10)2+25​∣(i^+2j^​+2k^)⋅[(15−4λ)i^+(λ−10)j^​+5k]∣​

$$\frac{|15-4\lambda +2\lambda -20+10|}{\sqrt{{(15-4\lambda )}^{2}+{(\lambda -10)}^{2}+25}}=\frac{1}{\sqrt{3}}$$(15−4λ)2+(λ−10)2+25​∣15−4λ+2λ−20+10∣​=3​1​

squaring both side we get,

$$3{(5-2\lambda )}^{2}=225+16{\lambda }^{2}-120\lambda +{\lambda }^{2}+100-20\lambda +25$$3(5−2λ)2=225+16λ2−120λ+λ2+100−20λ+25

$$12{\lambda }^{2}+75-60\lambda =17{\lambda }^{2}-140\lambda +350$$12λ2+75−60λ=17λ2−140λ+350

$$5{\lambda }^{2}-80\lambda +275=0$$5λ2−80λ+275=0

$${\lambda }^{2}-16\lambda +55=0$$λ2−16λ+55=0

$$(\lambda -5)(\lambda -11)=0$$(λ−5)(λ−11)=0

$$\Rightarrow \lambda =5,11$$⇒λ=5,11

So sum of $5+11=16$5+11=16