Vectors — Maths STD 12 Science — Question

$\begin{aligned} & \therefore-25 \ln -15 n^2-2 n|-30|^2-18 \ln =0 \\ & \therefore-\left.30\right|^2-45 \ln -15 n^2=0 \\ & \left.\therefore 2\right|^2+3 \ln +n^2=0 \\ & \left.\therefore 2\right|^2+2 \ln +\ln +n^2=0\end{aligned}$

∴ 2l(l + n) + n(l + n) = 0
∴ (l + n)(2l + n) = 0
∴ l + n = 0 or 2l + n = 0
l = -n or n = -2l
Now, m = -(5l + 3n), therefore, if l = -n,
m = -(-5n + 3n) = 2n

$\begin{aligned} & \therefore-1=\frac{m}{2}=n \\ & \therefore \frac{l}{-1}=\frac{m}{2}=\frac{n}{1}\end{aligned}$

∴ the direction ratios of the first line are

$\begin{aligned} & a_1=-1, b_1=2, c_1=1 \\ & \text { If } n=-2 l_1 m=-(51-6 l)-1\end{aligned}$

$\begin{aligned} & \therefore 1=m=\frac{n}{-2} \\ & \therefore \frac{l}{1}=\frac{m}{1}=\frac{n}{-2}\end{aligned}$

$\begin{aligned} & \therefore 1=m=\frac{n}{-2} \\ & \therefore \frac{l}{1}=\frac{m}{1}=\frac{n}{-2}\end{aligned}$

$\therefore$ the direction ratios of the second line are

$a_2=-1, b_2=1, c_2=-2$

Let θ be the angle between the lines.

Then $\cos \theta=\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1{ }^2+c_1^2} \cdot \sqrt{a_2{ }^2+b_2{ }^2+c_2^2}}\right|$

$=\left|\frac{(-1)(1)+2(1)+1(-2)}{\sqrt{(-1)^2+2^2+1^2} \cdot \sqrt{1^2+1^2+(-2)^2}}\right|$

$=\left|\frac{-1+2-2}{\sqrt{6} \cdot \sqrt{6}}\right|=\left|\frac{-1}{6}\right|=\frac{1}{6}$

$\therefore \theta=\cos ^{-1}\left(\frac{1}{6}\right)$