Equation of a Line in 3D

Vector Equation Problems for JEE Main

Hello students! Welcome to the world of vector equation problems. This topic is super important for JEE Main because it combines your knowledge of vector algebra and 3D geometry to solve tricky problems. Mastering vector equations helps you tackle a wide range of questions, often appearing in different forms and requiring a strong conceptual understanding.

Conceptual Explanation

A vector equation is simply an equation where the unknown is a vector. Our goal is to find this unknown vector $\vec{x}$ that satisfies certain conditions. These conditions are usually given in terms of dot products, cross products, and sometimes, perpendicularity. The key is to use the information given to set up a system of equations and then solve for the components of $\vec{x}$. Think of it like solving for $x$ and $y$ in linear equations, but now we're dealing with vectors!

Let's build some intuition. Imagine you're given that the dot product of $\vec{a}$ and $\vec{x}$ is a constant, say $k_1$. That means you know something about the projection of $\vec{x}$ onto $\vec{a}$. If you also know that the dot product of $\vec{b}$ and $\vec{x}$ is $k_2$, you have even more information about $\vec{x}$'s direction and magnitude. When you combine this with cross-product information, like $\vec{a} \times \vec{x} = \vec{c}$, you're essentially constraining $\vec{x}$ to lie in a very specific region in space.

Important Formulas

1. If $\vec{a} \cdot \vec{x} = k_1$ and $\vec{b} \cdot \vec{x} = k_2$

Explanation: This is the most basic scenario. You have two dot products defined. Let's assume $\vec{x} = x_1\hat{i} + x_2\hat{j} + x_3\hat{k}$, $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$.

Then, the equations become:

$$a_1x_1 + a_2x_2 + a_3x_3 = k_1$$ $$b_1x_1 + b_2x_2 + b_3x_3 = k_2$$

You now have two equations with three unknowns ($x_1$, $x_2$, $x_3$). You'll need more information to solve for all three components. This is where other conditions (like a cross product or magnitude) come in.

Example: Suppose you know $\vec{a} = \hat{i} + \hat{j}$, $\vec{b} = \hat{j} + \hat{k}$, $k_1 = 1$, and $k_2 = 1$. Then $x_1 + x_2 = 1$ and $x_2 + x_3 = 1$. You need another equation (perhaps from $|\vec{x}|$) to solve for $x_1$, $x_2$, and $x_3$ uniquely.

2. If $\vec{a} \times \vec{x} = \vec{c}$

Explanation: This is a classic scenario involving the cross product. The cross product $\vec{a} \times \vec{x}$ results in a vector $\vec{c}$ that is perpendicular to both $\vec{a}$ and $\vec{x}$. This implies that $\vec{x}$ lies in the plane perpendicular to $\vec{c}$. But, crucially, we only get information about two components of $\vec{x}$ since we only know the *direction* of $\vec{x}$ is perpendicular to $\vec{c}$.

This means $\vec{x}$ has a component along $\vec{a}$! Why? $\vec{a} \times \vec{x} = \vec{c}$ tells us $\vec{x}$ has to have a component orthogonal to $\vec{a}$, but it doesn't tell us what the component *parallel* to $\vec{a}$ is. So, we can express $\vec{x}$ as:

$$\vec{x} = \lambda \vec{a} + \vec{p}$$

where $\lambda$ is a scalar and $\vec{p}$ is a vector perpendicular to $\vec{a}$. To solve, you also usually need $\vec{a} \cdot \vec{x} = k$ (some scalar), which provides the extra equation needed.

Derivation: Taking the dot product with $\vec{a}$:

$$\vec{a} \cdot \vec{x} = \vec{a} \cdot (\lambda \vec{a} + \vec{p}) = \lambda |\vec{a}|^2 + \vec{a} \cdot \vec{p} = \lambda |\vec{a}|^2$$

Since $\vec{a} \cdot \vec{p} = 0$ (because $\vec{a}$ and $\vec{p}$ are perpendicular). Therefore,

$$\lambda = \frac{\vec{a} \cdot \vec{x}}{|\vec{a}|^2} = \frac{k}{|\vec{a}|^2}$$

So, we know $\lambda$. Now we can plug $\vec{x} = \lambda \vec{a} + \vec{p}$ back into $\vec{a} \times \vec{x} = \vec{c}$:

$$\vec{a} \times (\lambda \vec{a} + \vec{p}) = \vec{a} \times \lambda \vec{a} + \vec{a} \times \vec{p} = \vec{a} \times \vec{p} = \vec{c}$$

Now $\vec{p} = \frac{\vec{a} \times \vec{c}}{|\vec{a}|^2}$, which gives you $\vec{x} = \lambda \vec{a} + \frac{\vec{a} \times \vec{c}}{|\vec{a}|^2}$

3. For $\vec{r}$ perpendicular to $\vec{a}$: $\vec{r} \cdot \vec{a} = 0$

Explanation: This is the condition for perpendicularity. If two vectors are perpendicular, their dot product is zero. This condition helps establish relationships between the components of the vectors. For instance, if $\vec{x}$ is perpendicular to a known vector $\vec{a}$, you can write an equation relating the components of $\vec{x}$ and $\vec{a}$.

Example: If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{x}$ is perpendicular to $\vec{a}$, then $x_1 + x_2 + x_3 = 0$, where $\vec{x} = x_1\hat{i} + x_2\hat{j} + x_3\hat{k}$.

Tips for Solving Problems

Common Mistakes to Avoid

JEE-Specific Tricks