Animated Solution for Mathematics - Vector Algebra: If $\vec{a}$ and $\vec{b}$ are vectors such that $|\vec{a}+\vec{b}|=\sqrt{29}$ and $\vec{a}\times (2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times \vec{b}$, then a possible value of $(\vec{a}+\vec{b})\cdot (-7\hat{i}+2\hat{j}+3\hat{k})$ is

Unlocking the Hidden Symmetry

Welcome, future engineer! Today, we are going to peel back the layers of a classic JEE Advanced vector problem.

At first glance, it looks like a system of equations that might require tedious component-wise solving. But hold on—the beauty of vector algebra is that it often rewards the elegant thinker over the brute-force calculator. Let us embark on this journey together.

The Setup

We are given a constant vector, let us call it $\vec{v}=2\hat{i}+3\hat{j}+4\hat{k}$. The problem presents us with a fascinating relationship:

I know what you are thinking: 'Should I write $\vec{a}$ as $(a_1,a_2,a_3)$ and $\vec{b}$ as $(b_1,b_2,b_3)$?' Stop right there! If you do that, you will be drowning in variables.

Instead, let us manipulate the equation as a whole. Bring everything to one side:

The Algebraic Dance

Here is where the magic happens. We cannot simply factor out $\vec{v}$ because the cross product is not commutative.

However, we know it is anti-commutative: $\vec{v}\times \vec{b}=-(\vec{b}\times \vec{v})$. Substituting this into our equation, we get:

Now, look at that! We have a common factor of $\vec{v}$ on the right side. Using the distributive property of the cross product, we can write this as:

The Geometric Insight

This is the 'Aha!' moment. When the cross product of two vectors is the zero vector, it tells us something profound about their geometry: they are parallel.

This means the vector $(\vec{a}+\vec{b})$ must be a scalar multiple of $\vec{v}$. We can write this as:

where $k$ is some scalar. Now, let us use the magnitude condition provided: $|\vec{a}+\vec{b}|=\sqrt{29}$.

Substituting our parallel condition, we get $|k\vec{v}|=\sqrt{29}$, which simplifies to $|k||\vec{v}|=\sqrt{29}$.

Calculating the magnitude of $\vec{v}$ is straightforward:

Substituting this back, we find $|k|\sqrt{29}=\sqrt{29}$, which means $|k|=1$. Therefore, $k$ can be either $1$ or $-1$.

The Final Calculation

We are asked to find the possible value of $(\vec{a}+\vec{b})\cdot (-7\hat{i}+2\hat{j}+3\hat{k})$. Let us call the second vector $\vec{w}=-7\hat{i}+2\hat{j}+3\hat{k}$.

Using our substitution $(\vec{a}+\vec{b})=k\vec{v}$, the expression becomes:

Now, calculate the dot product $\vec{v}\cdot \vec{w}$:

Finally, our result is $k(4)$. Since $k=\pm 1$, the possible values are $4$ or $-4$.

Looking at our options, $4$ is the clear winner. You have just navigated a complex vector problem with elegance and precision. Keep this mindset—always look for the geometric structure before diving into the algebra!