Animated Solution for Mathematics - Vector Algebra: If the vectors $\vec{a}=\hat{i}-\hat{j}+2\hat{k}$, $\vec{b}=2\hat{i}+4\hat{j}+\hat{k}$ and $\vec{c}=\lambda \hat{i}+\hat{j}+\mu \hat{k}$ are mutually orthogonal, then $(\lambda ,\mu )=$

Imagine you are standing in the center of a vast, empty room. You have three vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, floating in front of you. They are locked in a rigid, beautiful dance; the problem states they are mutually orthogonal.

In the language of physics, this means they form a perfect, right-angled corner, much like the $x$, $y$, and $z$ axes of a standard coordinate system. When vectors are mutually orthogonal, they are independent and define the very space they inhabit.

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To translate this geometric perfection into algebra, we use the dot product. For any two vectors $\vec{u}$ and $\vec{v}$, the dot product is defined as:

If the angle $\theta$ is $90^{\circ }$, then $\cos (90^{\circ })=0$, and the entire product vanishes. This is our master key. We are given:

$\vec{a}=\hat{i}-\hat{j}+2\hat{k}$ $\vec{b}=2\hat{i}+4\hat{j}+\hat{k}$ $\vec{c}=\lambda \hat{i}+\hat{j}+\mu \hat{k}$

Because they are mutually orthogonal, we know that $\vec{a}\cdot \vec{b}=0$, $\vec{b}\cdot \vec{c}=0$, and $\vec{c}\cdot \vec{a}=0$.

Before we dive into the unknowns, let us verify the consistency of our given vectors $\vec{a}$ and $\vec{b}$. It is a habit of elite problem solvers to check the ground they stand on.

Calculating the dot product:

The condition holds! The geometry is consistent, and we are ready to proceed.

We focus on the unknown vector $\vec{c}$ with variables $\lambda$ and $\mu$. To solve for two variables, we need two independent equations derived from the orthogonality conditions $\vec{b}\cdot \vec{c}=0$ and $\vec{a}\cdot \vec{c}=0$.

First, for $\vec{b}\cdot \vec{c}=0$:

Next, for $\vec{a}\cdot \vec{c}=0$:

We are left with a simple system of linear equations:

To eliminate $\lambda$, multiply the second equation by $2$:

Now, subtract the first equation from this new one:

Substituting $\mu =2$ back into $\lambda +2\mu =1$:

We have found our values: $\lambda =-3$ and $\mu =2$. The vector $\vec{c}$ is $-3\hat{i}+\hat{j}+2\hat{k}$.

We took an abstract geometric condition—mutual orthogonality—and used the dot product to slice through the complexity, turning it into a simple algebraic system. This is the essence of physics: translating the physical world into the language of mathematics, solving it, and translating it back.