Animated Solution for Mathematics - Three Dimensional Geometry: Let $\alpha$ be the angle between the lines whose direction cosines satisfy the equations $l+m-n=0$ and $l^2+m^2-n^2=0$. Then the value of ${\sin }^4\alpha +{\cos }^4\alpha$ is :

A Journey into 3D Space

Welcome, future engineer. Today, we are not just solving a problem; we are exploring the architecture of 3D space. Imagine you are standing at the origin of a coordinate system.

Two lines pass through this origin, piercing the void. We are given two constraints on their direction cosines $(l,m,n)$, and our mission is to find the value of ${\sin }^4\alpha +{\cos }^4\alpha$, where $\alpha$ is the angle between them.

This problem is a classic JEE Advanced challenge because it tests your ability to bridge the gap between algebraic manipulation and geometric intuition.

The Hidden Constraint

We are given two equations: $l+m-n=0$ and $l^2+m^2-n^2=0$. Many students stop here, staring at these two equations, wondering how to find three unknowns.

But here is the secret: in 3D geometry, direction cosines are not free agents. They are bound by the fundamental identity:

This is the 'hidden' third equation that completes our system. Without it, the problem is unsolvable. Always remember, whenever you see direction cosines, this identity is your best friend.

The Algebraic Dance

Let us simplify the system. From the first equation, we have $n=l+m$.

Now, let us substitute this into the second equation:

Expanding this, we get $l^2+m^2-(l^2+m^2+2lm)=0$. Watch closely as the terms cancel out—it is one of those moments in math that feels like a magic trick.

The $l^2$ and $m^2$ terms vanish, leaving us with $-2lm=0$, which simplifies to $lm=0$. This tells us that either $l=0$ or $m=0$. We have just cracked the code!

Finding the Lines

Now, we explore the two cases. If $l=0$, then $n=m$.

Substituting this into our fundamental identity $l^2+m^2+n^2=1$, we get $0^2+m^2+m^2=1$, which means $2m^2=1$, or $m=\pm \frac{1}{\sqrt{2}}$.

Thus, our first line has direction cosines $(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$.

Similarly, if $m=0$, we find $n=l$, leading to $2l^2=1$, so $l=\pm \frac{1}{\sqrt{2}}$. Our second line has direction cosines $(\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}})$. We have successfully pinned down the orientation of these lines in space.

The Final Calculation

With the direction cosines in hand, finding the angle $\alpha$ is straightforward. The cosine of the angle between two lines is given by the absolute value of the dot product of their direction cosines:

Substituting our values, we get:

This means $\alpha =60^{\circ }$.

Finally, we need to evaluate ${\sin }^4\alpha +{\cos }^4\alpha$. Instead of calculating the powers directly, we use the identity:

Since ${\sin }^2\alpha +{\cos }^2\alpha =1$, this becomes $1-2{\sin }^2\alpha {\cos }^2\alpha$. Given $\cos \alpha =\frac{1}{2}$, we have ${\cos }^2\alpha =\frac{1}{4}$ and ${\sin }^2\alpha =\frac{3}{4}$.

Substituting these, we get:

And there you have it! Through logical deduction and the power of fundamental identities, we have arrived at the solution. Keep practicing this blend of algebra and geometry, and you will find that even the most complex JEE problems become elegant stories waiting to be told.