Animated Solution for Mathematics - Vector Algebra: Let the volume of a parallelepiped 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 :

Welcome, future engineer. Today, we are not just solving a problem; we are exploring the architecture of three-dimensional space. Imagine you are standing in a room, and you have three vectors, $\vec{u}$, $\vec{v}$, and $\vec{w}$, all originating from the corner of the floor.

If you extend these vectors, they define a slanted, box-like shape—a parallelepiped. This is the fundamental building block of vector geometry. We are given the components of these vectors, but there is a mystery: the variable $\lambda$ hidden in the $z$-component of $\vec{u}$.

Our mission is to find $\lambda$ using the volume, and then determine the angle between two of these edges.

How do we measure the volume of this 3D shape? We use the scalar triple product, often called the box product. Mathematically, the volume $V$ of a parallelepiped formed by vectors $\vec{u}$, $\vec{v}$, and $\vec{w}$ is given by the absolute value of their scalar triple product: $V=|[\vec{u}\vec{v}\vec{w}]|$.

Why the absolute value? Because the determinant of these vectors can be negative depending on their orientation in space. But volume is a physical reality; it cannot be negative.

So, we write $V=|[\vec{u}\vec{v}\vec{w}]|=1$. This is our anchor. This is the equation that will reveal the secret of $\lambda$.

Now, let us construct our determinant. We place the components of $\vec{u}$, $\vec{v}$, and $\vec{w}$ into a $3\times 3$ matrix:

Take a deep breath. Expanding a determinant is a rhythmic process. We expand along the first row.

First, we take the element $1$ and multiply it by the minor determinant of the remaining $2\times 2$ matrix: $(1\cdot 1-3\cdot 1)=-2$.

Next, we subtract the second element, $1$, and multiply it by its minor: $(1\cdot 1-3\cdot 2)=1-6=-5$.

Finally, we add the third element, $\lambda$, multiplied by its minor: $(1\cdot 1-1\cdot 2)=1-2=-1$.

Putting it all together, we get:

This is the beauty of algebra. A complex-looking determinant collapses into a simple linear expression: $3-\lambda$.

We know the volume is $1$. Therefore, $|3-\lambda |=1$. This is where many students stumble. An absolute value equation splits into two distinct paths.

Case 1: $3-\lambda =1$, which implies $\lambda =2$.

Case 2: $3-\lambda =-1$, which implies $\lambda =4$.

We have two possible vectors for $\vec{u}$. The problem asks for $\cos \theta$ between $\vec{u}$ and $\vec{w}$. Usually, in competitive exams, one of the cases will lead to an answer that matches the given options. Let us test $\lambda =4$ first.

With $\lambda =4$, our vector $\vec{u}$ becomes $\hat{i}+\hat{j}+4\hat{k}$. Our vector $\vec{w}$ is fixed as $2\hat{i}+\hat{j}+\hat{k}$.

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

First, the dot product:

Next, the magnitudes:

Now, we combine them:

Since $\sqrt{12}=2\sqrt{3}$, the denominator becomes $3\cdot 2\sqrt{3}=6\sqrt{3}$.

Thus, the final result is:

Look at that! It matches our options perfectly. You have navigated the geometry, mastered the determinant, handled the absolute value, and calculated the projection. You didn't just solve a problem; you mastered a concept. Keep this momentum going!