Animated Solution for Mathematics - Three Dimensional Geometry: If the two lines $l_1:\frac{x-2}{3}=\frac{y+1}{-2},z=2$ and $l_2:\frac{x-1}{1}=\frac{2y+3}{\alpha }=\frac{z+5}{2}$ are perpendicular, then an angle between the lines $l_2$ and $l_3:\frac{1-x}{3}=\frac{2y-1}{-4}=\frac{z}{4}$ is:

Imagine you are standing in a vast, empty 3D space. Three lines, $l_1$, $l_2$, and $l_3$, are floating before you. To understand the relationship between them, we don't need to know where they are; we only need to know where they are pointing. This is the essence of direction ratios.

A line's equation is like its DNA. In the standard symmetric form,

the denominators $(a,b,c)$ are the direction ratios. However, the coefficients of $x$, $y$, and $z$ must be $1$.

For $l_1$, we see $z=2$, which we rewrite as $\frac{z-2}{0}$, revealing the direction vector ${\vec{b}}_1=(3,-2,0)$.

For $l_2$, we encounter a trap: $\frac{2y+3}{\alpha }$. We must divide by $2$ to get $\frac{y+3/2}{\alpha /2}$, giving us ${\vec{b}}_2=(1,\alpha /2,2)$.

The problem states that $l_1\perp l_2$. This implies their dot product is zero: ${\vec{b}}_1\cdot {\vec{b}}_2=0$.

Substituting our vectors, we get:

This simplifies beautifully to $3-\alpha =0$, or $\alpha =3$.

Now, our vector ${\vec{b}}_2$ becomes $(1,3/2,2)$. To make it cleaner, we multiply by $2$ to get ${\vec{b}}_2=(2,3,4)$. The direction remains the same, just scaled for convenience.

Now we analyze $l_3$. Its equation is $\frac{1-x}{3}=\frac{2y-1}{-4}=\frac{z}{4}$.

We fix the signs and coefficients to obtain:

Our third direction vector is ${\vec{b}}_3=(-3,-2,4)$.

To find the angle $\theta$ between $l_2$ and $l_3$, we use the dot product formula:

The dot product is $(2)(-3)+(3)(-2)+(4)(4)=-6-6+16=4$.

The magnitudes are:

Thus, $\cos \theta =\frac{4}{29}$.

Since the options are in terms of ${\sec }^{-1}$, we use the identity $\sec \theta =\frac{29}{4}$. This leads us to the final result: