Animated Solution for Physics - Kinematics: Three particles $P,Q$ and $R$ are moving along the vectors $\mathbf{A}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$, $\mathbf{B}=\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{C}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}$, respectively. They strike on a point and start to move in different directions. Now, particle $P$ is moving normal to the plane which contain vectors $\mathbf{A}$ and $\mathbf{B}$. Similarly, particle $Q$ is moving normal to the plane which contain vectors $\mathbf{A}$ and $\mathbf{C}$. The angle between the direction of motion of $P$ and $Q$ is ${\cos }^{-1}\left(\frac{1}{\sqrt{x}}\right)$. Then, the value of $x$ is ......... .

Welcome to a fascinating journey into the world of three-dimensional geometry and vector algebra! In this problem, we are not just dealing with abstract numbers; we are visualizing particles moving through space, interacting with planes, and defining their trajectories based on strict mathematical rules.

Imagine you are standing in a room. The floor is the $xy$-plane, the walls are the $xz$ and $yz$ planes. We have three vectors, $\mathbf{A}$, $\mathbf{B}$, and $\mathbf{C}$, pointing to different corners of this room. Our mission is to find the angle between the paths of two particles, $P$ and $Q$, which are moving perpendicular to specific planes formed by these vectors.

Understanding the Vectors

Let's first understand our given vectors. We have $\mathbf{A}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$, which lies perfectly flat on the $xy$-plane, pointing diagonally. Then we have $\mathbf{B}=\hat{\mathbf{j}}+\hat{\mathbf{k}}$, which lies on the $yz$-plane. Finally, $\mathbf{C}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}$, which also lies on the $xy$-plane but points in a different direction.

These vectors are the building blocks of our planes. A plane in 3D space can be uniquely defined by any two non-collinear vectors that lie within it.

Finding the Normals

The problem states a crucial condition: particle $P$ moves normal (or perpendicular) to the plane containing vectors $\mathbf{A}$ and $\mathbf{B}$. How do we mathematically find a direction that is perpendicular to two given vectors?

This is where the cross product shines! The cross product of two vectors inherently produces a third vector that is perfectly orthogonal to both original vectors, and thus orthogonal to the entire plane they span. Therefore, the direction of particle $P$ is given by the cross product $\mathbf{A}\times \mathbf{B}$.

Let's roll up our sleeves and calculate this cross product. We set up a determinant with our unit vectors $\hat{\mathbf{i}}$, $\hat{\mathbf{j}}$, and $\hat{\mathbf{k}}$ in the top row, followed by the components of $\mathbf{A}$ and $\mathbf{B}$.

Expanding this determinant, we get:

This new vector $\mathbf{P}$ represents the exact direction in which our first particle is moving.

Similarly, particle $Q$ moves normal to the plane containing vectors $\mathbf{A}$ and $\mathbf{C}$. We apply the exact same logic here. The direction of $Q$ will be along the cross product $\mathbf{A}\times \mathbf{C}$.

Notice something interesting here? Both vectors $\mathbf{A}$ and $\mathbf{C}$ have a $z$-component of zero. They both lie entirely in the $xy$-plane. Therefore, any vector perpendicular to both of them must point straight up or down along the $z$-axis. Let's verify this mathematically:

As expected, the direction vector for $Q$ points purely along the $z$-axis.

Bridging the Angle

Now we have the direction vectors for both particles: $\mathbf{P}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{Q}=2\hat{\mathbf{k}}$. The ultimate goal is to find the angle $\theta$ between them.

To find the angle between any two vectors in 3D space, we use the dot product. The fundamental definition of the dot product is:

By rearranging this formula, we can isolate $\cos \theta$:

Bringing It All Together

Let's substitute our values into this elegant formula. First, we calculate the dot product $\mathbf{P}\cdot \mathbf{Q}$:

Next, we find the magnitudes of both vectors. The magnitude of $\mathbf{P}$ is:

The magnitude of $\mathbf{Q}$ is simply:

Now, we plug these back into our cosine equation:

The $2$ in the numerator and denominator cancel out beautifully, leaving us with:

The problem states that the angle between the direction of motion of $P$ and $Q$ is ${\cos }^{-1}\left(\frac{1}{\sqrt{x}}\right)$. By directly comparing this given expression with our derived result, it is crystal clear that:

And there we have it! A seemingly complex 3D geometry problem elegantly dismantled using the fundamental tools of vector algebra.

Before we conclude, consider this: what if the question had asked for the angle between the two planes themselves, rather than the particles? The beautiful truth of geometry is that the angle between two planes is exactly equal to the angle between their normal vectors. So, by finding the angle between the paths of particles $P$ and $Q$, we have simultaneously found the angle between the plane containing $\mathbf{A}$ and $\mathbf{B}$ and the plane containing $\mathbf{A}$ and $\mathbf{C}$. Keep exploring, and never stop questioning the geometry around you!