JEE Main 2024 — Mathematics Vectors & 3D Geometry

Let,

$${L}_{1}:\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}=\lambda$$L1​:4x−5​=1y−4​=3z−5​=λ

$$\Rightarrow M(4\lambda +5,\lambda +4,3\lambda +5)$$⇒M(4λ+5,λ+4,3λ+5)

And ${L}_{2}:\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}=\mu$L2​:12x+8​=5y+2​=9z+11​=μ

$$\Rightarrow N(12\mu -8,5\mu -2,9\mu -11)$$⇒N(12μ−8,5μ−2,9μ−11)

Now, finding $\vec{MN}=(4\lambda -12\mu +13,\lambda -5\mu +6,3\lambda -9\mu +16)..\ldots (1)$MN=(4λ−12μ+13,λ−5μ+6,3λ−9μ+16)..…(1)

Now finding cross product of direction ratios of line we get,

$${\vec{b}}_{1}\times {\vec{b}}_{2}=|\begin{matrix}\hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & 3 \\ 12 & 5 & 9\end{matrix}|=-6\hat{i}+8\hat{k}\ldots ...(2)$$b1​×b2​=∣i^412​j^​15​k^39​∣=−6i^+8k^…...(2)

Now, solving the equation (1)&(2) as they are parallel vectors, we get,

$$\therefore \frac{4\lambda -12\mu +13}{-6}=\frac{\lambda -5\mu +6}{0}=\frac{3\lambda -9\mu +16}{8}$$∴−64λ−12μ+13​=0λ−5μ+6​=83λ−9μ+16​

Taking $\frac{4\lambda -12\mu +13}{-6}=\frac{\lambda -5\mu +6}{0}$−64λ−12μ+13​=0λ−5μ+6​

$$\Rightarrow \lambda -5\mu +6=0\ldots (3)$$⇒λ−5μ+6=0…(3)

Taking $\frac{4\lambda -12\mu +13}{-6}=\frac{3\lambda -9\mu +16}{8}$−64λ−12μ+13​=83λ−9μ+16​

$$\lambda -3\mu +4=0\ldots (4)$$λ−3μ+4=0…(4)

Solve $(3)$(3) and $(4)$(4) we get

$$\lambda =-1,\mu =1$$λ=−1,μ=1

So, $M(1,3,2)$M(1,3,2) and $N(4,3,-2)$N(4,3,−2)

Hence, $\vec{OM}\cdot \vec{ON}=4+9-4=9$OM⋅ON=4+9−4=9