Animated Solution for Mathematics - Vector Algebra: Let $\vec{u}=\hat{i}+\hat{j},\vec{v}=\hat{i}-\hat{j}$ and $\vec{w}=\hat{i}+2\hat{j}+3\hat{k}$. If $\hat{n}$ is a unit vector such that $\vec{u}\cdot \hat{n}=0$ and $\vec{v}\cdot \hat{n}=0$, then $|\vec{w}\cdot \hat{n}|$ is equal to

A Journey into 3D Space

Welcome, future engineer. Today, we are not just solving a vector problem; we are learning to navigate the architecture of 3D space.

When you look at vectors $\vec{u}=\hat{i}+\hat{j}$ and $\vec{v}=\hat{i}-\hat{j}$, I want you to see them as two lines drawn on a flat sheet of paper—the $xy$-plane. They are trapped in that two-dimensional world.

But the problem introduces a third vector, $\vec{w}=\hat{i}+2\hat{j}+3\hat{k}$, which breaks free into the $z$-dimension. Our goal is to find the relationship between this 3D vector $\vec{w}$ and a mysterious unit vector $\hat{n}$ that is perpendicular to our original 2D plane.

The Search for the Normal

We are told that $\vec{u}\cdot \hat{n}=0$ and $\vec{v}\cdot \hat{n}=0$. In the language of physics and geometry, a dot product of zero is a signal. It screams orthogonality.

It tells us that $\hat{n}$ is standing perfectly upright, perpendicular to the plane formed by $\vec{u}$ and $\vec{v}$.

How do we find a vector that is perpendicular to two others? We reach for the most powerful tool in our vector arsenal: the cross product.

The cross product $\vec{u}\times \vec{v}$ is specifically designed to generate a vector that is orthogonal to the plane containing $\vec{u}$ and $\vec{v}$.

The Determinant Dance

Let us compute this cross product. We set up our determinant, the standard machinery for this operation:

I know determinants can look intimidating, but look at the structure. The third column is all zeros, except for the $\hat{k}$ component. This is a gift!

We expand along the third column:

Just like that, the complexity collapses. We have found that the vector perpendicular to our plane is $-2\hat{k}$.

Normalizing the Vector

We are almost there. The problem asks for a unit vector $\hat{n}$. A vector is only a unit vector if its magnitude is 1.

Our current vector, $-2\hat{k}$, has a magnitude of 2. To fix this, we divide by the magnitude:

Why the $\pm$? Because a line perpendicular to a plane can point in two directions—up or down. Both are valid normals. This is the beauty of the math; it accounts for all possibilities.

The Final Projection

Now, we bring in our third vector, $\vec{w}=\hat{i}+2\hat{j}+3\hat{k}$. We need to find the absolute value of the dot product $|\vec{w}\cdot \hat{n}|$.

This is essentially asking: "How much of $\vec{w}$ lies along the direction of the normal vector $\hat{n}$?"

Substituting our values:

Using the properties of dot products, we know that $\hat{i}\cdot \hat{k}=0$ and $\hat{j}\cdot \hat{k}=0$. Only the $\hat{k}$ component survives the interaction:

Finally, we take the absolute value as requested:

And there it is. The answer is 3. You have successfully navigated the 3D space, used the cross product to find a normal, and projected a vector onto that normal.