Animated Solution for Mathematics - Vector Algebra: Let $\vec{u}$ be a vector coplanar with the vectors $\vec{a}=2\hat{i}+3\hat{j}-\hat{k}$ and $\vec{b}=\hat{j}+\hat{k}$. If $\vec{u}$ is perpendicular to $\vec{a}$ and $\vec{u}\cdot \vec{b}=24$, then $|\vec{u}{|}^2$ is equal to :

A Vector Odyssey

Welcome, future engineers! Today, we are going to embark on a journey through the elegant world of 3D vectors. Imagine you are standing in a vast, empty room with a flat sheet of paper floating in the air—this is our plane.

On this paper, we have two vectors, $\vec{a}=2\hat{i}+3\hat{j}-\hat{k}$ and $\vec{b}=\hat{j}+\hat{k}$. We are introduced to a mysterious third vector, $\vec{u}$.

The problem states that $\vec{u}$ is coplanar with $\vec{a}$ and $\vec{b}$. This is our golden ticket, meaning $\vec{u}$ is trapped on that same sheet of paper and cannot escape into the third dimension.

Because $\vec{u}$ is coplanar with $\vec{a}$ and $\vec{b}$, we can describe it using the language of linear combinations. Any vector in a plane can be reached by scaling $\vec{a}$ and $\vec{b}$ and adding them together.

We write this as $\vec{u}=\lambda \vec{a}+\mu \vec{b}$, where $\lambda$ and $\mu$ are the scaling factors we need to uncover. Think of $\lambda$ and $\mu$ as the coordinates of our vector $\vec{u}$ in the basis of $\vec{a}$ and $\vec{b}$.

Before we charge into the algebra, let's prepare our weapons by calculating the necessary dot products. First, the magnitude squared of $\vec{a}$:

Next, the dot product of $\vec{a}$ and $\vec{b}$:

Finally, the magnitude squared of $\vec{b}$:

With these values—$14$, $2$, and $2$—we are ready to face the conditions.

The problem provides two conditions. First, $\vec{u}\perp \vec{a}$, which implies $\vec{u}\cdot \vec{a}=0$. Substituting our linear combination into this:

Using our toolkit, we get $14\lambda +2\mu =0$. This simplifies beautifully to $\mu =-7\lambda$.

Now for the second condition: $\vec{u}\cdot \vec{b}=24$. Substituting the linear combination again:

Substituting $\vec{a}\cdot \vec{b}=2$, $||\vec{b}|{|}^2=2$, and $\mu =-7\lambda$:

Since $\lambda =-2$, then $\mu =-7(-2)=14$. We have successfully found our scalars.

Now we build $\vec{u}$:

Finally, the magnitude squared is:

The final result is $336$. The logic holds, the math is clean, and we have arrived at the answer.