Animated Solution for Mathematics - Three Dimensional Geometry: An angle between the lines whose direction cosines are given by the equations, $l+3m+5n=0$ and $5lm-2mn+6nl=0$, is :-

We are given two constraints involving the direction ratios $(l,m,n)$ of lines: 1. $l+3m+5n=0$ 2. $5lm-2mn+6nl=0$

The first equation represents a plane passing through the origin, while the second represents a cone. Finding the intersection of these constraints allows us to determine the specific directions of the lines lying on both surfaces.

We begin by isolating $l$ from the linear constraint:

Next, we substitute this expression into the quadratic equation $5lm-2mn+6nl=0$:

Expanding the terms, we obtain:

Combining like terms yields:

Dividing the entire equation by $-15$, we simplify the expression to:

We factorize the quadratic equation $m^2+3mn+2n^2=0$ as follows:

This gives us two distinct cases for the relationship between $m$ and $n$:

Case 1: $m=-n$ Substituting into $l=-(3m+5n)$: The ratio $l:m:n$ is $-2n:-n:n$, which simplifies to $-2:-1:1$. Thus, $\overrightarrow{v_1}=(-2,-1,1)$.

Case 2: $m=-2n$ Substituting into $l=-(3m+5n)$: The ratio $l:m:n$ is $n:-2n:n$, which simplifies to $1:-2:1$. Thus, $\overrightarrow{v_2}=(1,-2,1)$.

To find the angle $\theta$ between the two lines, we use the dot product formula:

The numerator is:

The magnitudes are:

Substituting these into the formula:

The final angle is: