JEE Main 2023 — Mathematics Vectors & 3D Geometry

Given,

A line $L$L pass through the origin and be perpendicular to the lines ${L}_{1}:\vec{r}=(\hat{i}-11\hat{j}-7\hat{k})+\lambda (\hat{i}+2\hat{j}+3\hat{k}),\lambda \in \mathbb{R}$L1​:r=(i^−11j^​−7k^)+λ(i^+2j^​+3k^),λ∈R and

$${L}_{2}:\vec{r}=(-\hat{i}+\hat{k})+\mu (2\hat{i}+2\hat{j}+\hat{k}),\mu \in \mathbb{R}$$L2​:r=(−i^+k^)+μ(2i^+2j^​+k^),μ∈R

So, direction ratio of line will be,

$$|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 2 & 1\end{matrix}|$$∣i^12​j^​22​k^31​∣

$$=\hat{i}(-4)-\hat{j}(-5)+\hat{k}(-2)$$=i^(−4)−j^​(−5)+k^(−2)

$$=-4\hat{i}+5\hat{j}-2\hat{k}$$=−4i^+5j^​−2k^

Hence, the equation of line $L$L which passes through origin will be,

$$L:\vec{r}=\sigma (-4\hat{i}+5\hat{j}-2\hat{k})$$L:r=σ(−4i^+5j^​−2k^)

Now finding the intersection point $P$P we get,

$$1+\lambda =-4\sigma ....(1)$$1+λ=−4σ....(1)

$$-11+2\lambda =5\sigma ....(2)$$−11+2λ=5σ....(2)

$$-7+3\lambda =-2\sigma ....(3)$$−7+3λ=−2σ....(3)

So, from equation (1)&(3) we get, $1+\lambda =-14+6\lambda \Rightarrow \lambda =3,\sigma =-1$1+λ=−14+6λ⇒λ=3,σ=−1

Hence, point $P\equiv (4,-5,2)$P≡(4,−5,2)

Now finding the point $Q$Q which is foot of perpendicular from point $P(4,-5,2)$P(4,−5,2) on line ${L}_{2}:(-\hat{i}+\hat{k})+\mu (2\hat{i}+2\hat{j}+\hat{k})$L2​:(−i^+k^)+μ(2i^+2j^​+k^) we get,

Now any point on line ${L}_{2}$L2​ will be $Q((2\mu -1)\hat{i}+2\mu \hat{j}+(\mu +1)\hat{k})$Q((2μ−1)i^+2μj^​+(μ+1)k^)

Now direction ratio of $PQ$PQ will be $(-5+2\mu )\hat{i}+(2\mu +5)\hat{j}+(\mu -1)\hat{k}=0$(−5+2μ)i^+(2μ+5)j^​+(μ−1)k^=0

Now using the perpendicular condition we get,

$$2(-5+2\mu )+2(2\mu +5)+1(\mu -1)=0$$2(−5+2μ)+2(2μ+5)+1(μ−1)=0

$$\Rightarrow \mu =\frac{1}{9}$$⇒μ=91​

Hence, $\alpha +\beta +\gamma =2\mu -1+2\mu +\mu +1=5\mu =\frac{5}{9}$α+β+γ=2μ−1+2μ+μ+1=5μ=95​

$$\Rightarrow 9(\alpha +\beta +\gamma )=5$$⇒9(α+β+γ)=5