Animated Solution for Mathematics - Vector Algebra: The magnitude of the projection of the vector $2\hat{i}+3\hat{j}+\hat{k}$ on the vector perpendicular to the plane containing the vectors $\hat{i}+\hat{j}+\hat{k}$ and $\hat{i}+2\hat{j}+3\hat{k}$, is :

Welcome, fellow traveler in the world of JEE Advanced mathematics. Today, we are going to peel back the layers of a classic vector problem. It is not just about crunching numbers; it is about visualizing the hidden geometry of space.

Imagine you are standing in a room, and there is a flat, tilted plane defined by two vectors, $\vec{b}$ and $\vec{c}$. You have a third vector, $\vec{a}$, piercing through the air. The problem asks for the magnitude of the projection of $\vec{a}$ onto a vector perpendicular to that plane.

The Normal Vector

To solve this, we first need to find the direction that is perfectly perpendicular to our plane. In the language of vectors, this is the normal vector, $\vec{n}$.

We know that the cross product of two vectors $\vec{b}$ and $\vec{c}$ gives us a vector that is orthogonal to both. Since both vectors lie on the plane, their cross product, $\vec{n}=\vec{b}\times \vec{c}$, will point directly out of the plane.

Let us set up our determinant to find this vector, given $\vec{b}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}+2\hat{j}+3\hat{k}$:

Expanding this, we calculate the components: $\hat{i}$ component: $(1\times 3-2\times 1)=1$ $\hat{j}$ component: $-(1\times 3-1\times 1)=-2$ * $\hat{k}$ component: $(1\times 2-1\times 1)=1$

Thus, our normal vector is $\vec{n}=\hat{i}-2\hat{j}+\hat{k}$.

The Projection

Now that we have our normal vector, we need to find the projection of $\vec{a}=2\hat{i}+3\hat{j}+\hat{k}$ onto $\vec{n}$. Think of this as finding the 'shadow' of $\vec{a}$ on the line defined by $\vec{n}$.

The formula for the magnitude of the projection is:

First, let us compute the dot product:

Next, we find the magnitude of the normal vector:

Finally, we plug these values into our projection formula:

We can simplify this further by writing $3$ as $\sqrt{9}$, giving us $\sqrt{\frac{9}{6}}$, which simplifies to $\sqrt{\frac{3}{2}}$.

And there you have it! We have successfully navigated the geometry of the plane and found the projection. Remember, every time you face a vector problem, visualize the physical space first; the math will follow naturally.