Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Given equations are $6mn-2nl+5lm=0$ and $3l+m+5n=0$.

From the second equation, we get $m=-3l-5n$.

Substituting $m$ into the first equation:

$6(-3l-5n)n-2nl+5l(-3l-5n)=0$

$-18ln-30n^2-2nl-15l^2-25ln=0$

$-15l^2-45ln-30n^2=0$

Dividing by $-15$, we get:

$l^2+3ln+2n^2=0$

$(l+n)(l+2n)=0$

This gives $l=-n$ or $l=-2n$.

Case 1: If $l=-n$, then $m=-3(-n)-5n=-2n$.
The direction ratios of the first line are proportional to $(-n,-2n,n)$, which simplifies to $(-1,-2,1)$. Let $\vec{a}=-\hat{i}-2\hat{j}+\hat{k}$.

Case 2: If $l=-2n$, then $m=-3(-2n)-5n=n$.
The direction ratios of the second line are proportional to $(-2n,n,n)$, which simplifies to $(-2,1,1)$. Let $\vec{b}=-2\hat{i}+\hat{j}+\hat{k}$.

The angle $\theta$ between the lines is given by:

$\cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$

$\cos \theta =\frac{(-1)(-2)+(-2)(1)+(1)(1)}{\sqrt{(-1{)}^2+(-2{)}^2+1^2}\sqrt{(-2{)}^2+1^2+1^2}}$

$\cos \theta =\frac{2-2+1}{\sqrt{6}\sqrt{6}}=\frac{1}{6}$

We need to find $\sin \theta$:

$\sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-{\left(\frac{1}{6}\right)}^2}=\sqrt{1-\frac{1}{36}}=\frac{\sqrt{35}}{6}$

Answer: $\frac{\sqrt{35}}{6}$