Animated Solution for Mathematics - Vector Algebra: If $\vec{a},\vec{b},\vec{c}$ and $\vec{d}$ are unit vectors such that $(\vec{a}\times \vec{b})\cdot (\vec{c}\times \vec{d})=1$ and $\vec{a}\cdot \vec{c}=\frac{1}{2}$, then

A Journey into Coplanarity

Imagine you are standing in a three-dimensional space, holding four unit vectors: $\vec{a},\vec{b},\vec{c},$ and $\vec{d}$. You are given a seemingly simple, yet incredibly powerful condition:

At first glance, this looks like a standard vector algebra problem, but it is actually a gateway into a beautiful geometric reality. Let us break this down together.

We define two new vectors: $\vec{p}=\vec{a}\times \vec{b}$ and $\vec{q}=\vec{c}\times \vec{d}$. We know that for any two unit vectors, the magnitude of their cross product is given by:

Since the sine function is bounded between $0$ and $1$, the maximum magnitude of $\vec{p}$ is $1$. Similarly, the maximum magnitude of $\vec{q}$ is $1$.

Now, look at our condition: $\vec{p}\cdot \vec{q}=1$. The dot product is defined as $|\vec{p}||\vec{q}|\cos \varphi$, where $\varphi$ is the angle between $\vec{p}$ and $\vec{q}$.

Since $|\vec{p}|\le 1$ and $|\vec{q}|\le 1$, the only way their product can be $1$ is if $|\vec{p}|=1$, $|\vec{q}|=1$, and $\cos \varphi =1$. This is the "Aha!" moment.

If $|\vec{p}|=1$, then $\sin {\theta }_{ab}=1$, which forces ${\theta }_{ab}=90^{\circ }$. Thus, $\vec{a}\perp \vec{b}$. By the same logic, $\vec{c}\perp \vec{d}$.

Furthermore, $\cos \varphi =1$ implies $\varphi =0^{\circ }$, meaning $\vec{p}$ and $\vec{q}$ are parallel. Since they have the same magnitude and direction, $\vec{a}\times \vec{b}=\vec{c}\times \vec{d}$.

Because $\vec{a}\times \vec{b}$ and $\vec{c}\times \vec{d}$ are the same vector, let us call this vector $\vec{n}$. This vector $\vec{n}$ is perpendicular to $\vec{a}$ and $\vec{b}$, and it is also perpendicular to $\vec{c}$ and $\vec{d}$.

This means all four vectors lie in the same plane—the plane perpendicular to $\vec{n}$. We have just unlocked the secret of the problem: all four vectors are coplanar.

Now that we know they are coplanar, let us simplify our lives by placing them in the $xy$-plane. We can set $\vec{a}=\hat{i}$ (along the x-axis). Since $\vec{a}\perp \vec{b}$, we can set $\vec{b}=\hat{j}$ (along the y-axis).

We are also given $\vec{a}\cdot \vec{c}=\frac{1}{2}$. Since these are unit vectors, $\cos {\theta }_{ac}=\frac{1}{2}$, which means the angle between $\vec{a}$ and $\vec{c}$ is $60^{\circ }$. Thus:

Finally, since $\vec{c}\perp \vec{d}$ and $\vec{c}\times \vec{d}=\vec{a}\times \vec{b}=\hat{k}$, vector $\vec{d}$ is simply $\vec{c}$ rotated by $90^{\circ }$ counter-clockwise. The angle of $\vec{d}$ is $60^{\circ }+90^{\circ }=150^{\circ }$.

Therefore:

With our vectors defined as $\vec{a}=\hat{i}$, $\vec{b}=\hat{j}$, $\vec{c}=\frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j}$, and $\vec{d}=-\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}$, we can easily evaluate the options.

Options 1 and 2 are false because the vectors are coplanar. Option 4 is false because $\vec{a}$ and $\vec{d}$ are not parallel.

Looking at Option 3, $\vec{b}=\hat{j}$ and $\vec{d}=-\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}$. They are clearly not parallel. Thus, Option 3 is the correct answer. You have successfully navigated the geometry of vectors!