Animated Solution for Mathematics - Vector Algebra: Find 3-dimensional vectors ${\vec{v}}_1,{\vec{v}}_2,{\vec{v}}_3$ satisfying ${\vec{v}}_1\cdot {\vec{v}}_1=4,{\vec{v}}_1\cdot {\vec{v}}_2=-2,{\vec{v}}_1\cdot {\vec{v}}_3=6,{\vec{v}}_2\cdot {\vec{v}}_2=2,{\vec{v}}_2\cdot {\vec{v}}_3=-5,{\vec{v}}_3\cdot {\vec{v}}_3=29$.

Imagine you are standing in a vast, empty 3D space. You are given three mysterious vectors, ${\vec{v}}_1,{\vec{v}}_2,{\vec{v}}_3$, and a set of six constraints—their mutual dot products.

If you were to approach this blindly, you might write each vector as $(x_i,y_i,z_i)$, leading to nine unknown variables. That is a recipe for an algebraic nightmare.

But wait—there is a better way. We are not just solving equations; we are uncovering the geometric soul of these vectors. By choosing our coordinate system wisely, we can make the math dance to our tune.

Let us start by placing ${\vec{v}}_1$ exactly along the x-axis. This is our first act of defiance against complexity. By doing this, we force its y and z components to be zero, so ${\vec{v}}_1=x_1\hat{i}$.

Now, look at the first condition: ${\vec{v}}_1\cdot {\vec{v}}_1=4$. The dot product of a vector with itself is just its magnitude squared.

We have our first vector: ${\vec{v}}_1=2\hat{i}$. We have successfully pinned it down!

Now, we turn to ${\vec{v}}_2$. We have already fixed the x-axis, but we still have the freedom to rotate the y and z axes around it. Let us orient the xy-plane so that ${\vec{v}}_2$ lies completely within it.

This is a brilliant move because it means ${\vec{v}}_2$ has no z-component. We can write ${\vec{v}}_2=x_2\hat{i}+y_2\hat{j}$.

We have two unknowns, $x_2$ and $y_2$, and two conditions: ${\vec{v}}_1\cdot {\vec{v}}_2=-2$ and ${\vec{v}}_2\cdot {\vec{v}}_2=2$. Using ${\vec{v}}_1=2\hat{i}$, the first condition becomes:

Now, for the second condition: $x_2^2+y_2^2=2$. Substituting $x_2=-1$, we get:

Choosing the positive direction, we get $y_2=1$. Thus, ${\vec{v}}_2=-\hat{i}+\hat{j}$.

Solving for ${\vec{v}}_3$

Finally, we face ${\vec{v}}_3$. Since we have locked down the x and y axes, ${\vec{v}}_3$ must be a general vector in 3D space: ${\vec{v}}_3=x_3\hat{i}+y_3\hat{j}+z_3\hat{k}$.

We have three remaining conditions: ${\vec{v}}_1\cdot {\vec{v}}_3=6$, ${\vec{v}}_2\cdot {\vec{v}}_3=-5$, and ${\vec{v}}_3\cdot {\vec{v}}_3=29$. First, ${\vec{v}}_1\cdot {\vec{v}}_3=6$:

Next, ${\vec{v}}_2\cdot {\vec{v}}_3=-5$:

Finally, ${\vec{v}}_3\cdot {\vec{v}}_3=x_3^2+y_3^2+z_3^2=29$. Plugging in our values:

This leads to $z_3^2=16$, so $z_3=\pm 4$.

The Elegance of Simplicity

We have arrived at our destination. The vectors are ${\vec{v}}_1=2\hat{i}$, ${\vec{v}}_2=-\hat{i}+\hat{j}$, and ${\vec{v}}_3=3\hat{i}-2\hat{j}\pm 4\hat{k}$.

By choosing our coordinate system with care, we transformed a daunting problem into a series of simple, logical steps. This is the true power of vector algebra—not just calculation, but the art of seeing the geometry beneath the numbers.