Animated Solution for Mathematics - Vector Algebra: If volume of parallelopiped whose coterminous edges are given by $\vec{u}=\hat{i}+\hat{j}+\lambda \hat{k}$, $\vec{v}=\hat{i}+\hat{j}+3\hat{k}$ and $\vec{w}=2\hat{i}+\hat{j}+\hat{k}$ be 1 cu. unit. If $\theta$ be the angle between the edges $\vec{u}$ and $\vec{w}$, then, $\cos \theta$ can be :

Imagine you are standing in a 3D coordinate system. You have three vectors, $\vec{u}$, $\vec{v}$, and $\vec{w}$, all originating from the origin. These vectors define the edges of a parallelepiped—a slanted, 3D box.

The problem asks us to find the cosine of the angle between two of these edges, but there is a mystery: the vector $\vec{u}$ contains an unknown variable, $\lambda$. To solve this, we must first master the geometry of volume.

The Key

The volume of a parallelepiped is not just a random calculation; it is the geometric interpretation of the Scalar Triple Product (STP). The STP, denoted as $[\vec{u}\vec{v}\vec{w}]$, is equivalent to the determinant of the matrix formed by the components of these vectors.

We are told the volume is $1$ cubic unit. Mathematically, this means $|[\vec{u}\vec{v}\vec{w}]|=1$.

This is where many students stumble. The absolute value is crucial. Because the determinant can be negative depending on the orientation of the vectors, we must set the determinant equal to both $1$ and $-1$:

Let us expand this determinant along the first row. We take the first element, $1$, and multiply it by the minor determinant $(1-3)$.

Then, we subtract the second element, $1$, multiplied by its minor $(1-6)$. Finally, we add the third element, $\lambda$, multiplied by its minor $(1-2)$.

This simplifies to the following expression:

Simplifying further, we get $-2+5-\lambda =\pm 1$, which reduces to the elegant equation:

Solving this gives us two distinct possibilities for our unknown. If $3-\lambda =1$, then $\lambda =2$. If $3-\lambda =-1$, then $\lambda =4$. We have successfully constrained our vector $\vec{u}$.

Now that we have our potential values for $\lambda$, we turn our attention to the angle $\theta$ between $\vec{u}$ and $\vec{w}$. The definition of the dot product is our best friend here: $\vec{u}\cdot \vec{w}=|\vec{u}||\vec{w}|\cos \theta$.

Rearranging this, we get:

Let us test the case where $\lambda =4$. Our vector $\vec{u}$ becomes $\hat{i}+\hat{j}+4\hat{k}$, and $\vec{w}$ is $2\hat{i}+\hat{j}+\hat{k}$.

The dot product $\vec{u}\cdot \vec{w}$ is $(1)(2)+(1)(1)+(4)(1)=2+1+4=7$. The magnitude of $\vec{u}$ is $\sqrt{1^2+1^2+4^2}=\sqrt{18}=3\sqrt{2}$. The magnitude of $\vec{w}$ is $\sqrt{2^2+1^2+1^2}=\sqrt{6}$.

Substituting these into our formula, we get:

Since $\sqrt{2}\cdot \sqrt{6}=\sqrt{12}=2\sqrt{3}$, the denominator becomes $3\cdot 2\sqrt{3}=6\sqrt{3}$. Thus, the final result is:

We have arrived at the solution, and it matches our options perfectly. Remember, in JEE Advanced, the math is never just about calculation; it is about understanding the physical reality behind the symbols.