The vector(s) which is/are coplanar with vectors i + j + 2k and i + 2j + k, and perpendicular to the

Q: Wait, why can't I just take the cross product of $\vec{a}$ and $\vec{b}$ to find the vector?

That is a brilliant instinct, but be careful! The cross product $\vec{a} \times \vec{b}$ gives you a vector that is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$. The problem asks for a vector that *lies* in that plane. Using the cross product would give you the normal to the plane, not a vector within it. We need the linear combination approach to stay 'inside' the plane.

Q: Why are there two correct options in the answer key? Did I do something wrong?

Not at all! You are actually doing it perfectly. Remember that a vector is defined by its direction and magnitude. When we solve for $\lambda$ and $\mu$, we find a relationship between them, but not their absolute values. This means the vector $\vec{r}$ can be any scalar multiple of the result. Both options provided are just different scalar multiples of the same directional vector. You have successfully captured the entire family of solutions!

Q: Why do we set the dot product to zero? Is there a physical intuition for this?

Think of the dot product as a measure of 'alignment.' When the dot product is zero, it means there is absolutely no projection of one vector onto the other—they are perfectly orthogonal. In physics, this is like saying the work done by a force perpendicular to displacement is zero. It is the ultimate mathematical filter for perpendicularity.

Question

The vector(s) which is/are coplanar with vectors $\hat{i} + \hat{j} + 2\hat{k}$ and $\hat{i} + 2\hat{j} + \hat{k}$, and perpendicular to the vector $\hat{i} + \hat{j} + \hat{k}$ is/are

Options:

(A) $\hat{j} - \hat{k}$

(B) $-\hat{i} + \hat{j}$

(C) $\hat{i} - \hat{j}$

(D) $-\hat{j} + \hat{k}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are not just solving a problem; we are navigating the elegant geometry of three-dimensional space. Imagine you are standing in a room with two vectors, $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.

These two vectors, anchored at the origin, define a flat, infinite sheet—a plane—cutting through your room. Our goal is to find a mystery vector, $\vec{r}$, that lives on this sheet.

To 'trap' a vector on a plane, we use the concept of a linear combination. By scaling $\vec{a}$ and $\vec{b}$ by constants $\lambda$ and $\mu$, we can reach any point on that plane:

$$\vec{r} = \lambda\vec{a} + \mu\vec{b}$$

This equation is your construction kit. Substituting the components of $\vec{a}$ and $\vec{b}$, we define the general vector $\vec{r}$ as:

$$\vec{r} = \lambda(\hat{i} + \hat{j} + 2\hat{k}) + \mu(\hat{i} + 2\hat{j} + \hat{k})$$

### The Filter: Applying the Perpendicularity Constraint

We have a general form for $\vec{r}$, but it is too broad. The problem provides a specific constraint: our vector must be perpendicular to a third vector, $\vec{c} = \hat{i} + \hat{j} + \hat{k}$.

In vector geometry, 'perpendicular' implies that the **dot product must be zero**. This filter will strip away all vectors on the plane that do not satisfy this condition.

First, let's group the components of our general vector $\vec{r}$ to make the algebra cleaner:

$$\vec{r} = (\lambda + \mu)\hat{i} + (\lambda + 2\mu)\hat{j} + (2\lambda + \mu)\hat{k}$$

Now, we apply the condition $\vec{r} \cdot \vec{c} = 0$:

$$((\lambda + \mu)\hat{i} + (\lambda + 2\mu)\hat{j} + (2\lambda + \mu)\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$$

### The Elegant Cancellation

When we take the dot product, we multiply the corresponding components. Since the components of $\vec{c}$ are all unity, the calculation simplifies to a summation:

$$(\lambda + \mu)(1) + (\lambda + 2\mu)(1) + (2\lambda + \mu)(1) = 0$$

Combining the terms, we have $\lambda + \lambda + 2\lambda = 4\lambda$ and $\mu + 2\mu + \mu = 4\mu$. The equation reduces to:

$$4\lambda + 4\mu = 0 \implies \lambda = -\mu$$

This is our golden key. The relationship $\lambda = -\mu$ allows us to express the vector in terms of a single parameter.

### The Final Reveal

Substituting $\lambda = -\mu$ back into our expression for $\vec{r}$, we get:

$$\vec{r} = (-\mu + \mu)\hat{i} + (-\mu + 2\mu)\hat{j} + (-2\mu + \mu)\hat{k}$$

The $\hat{i}$ component vanishes entirely. We are left with the final form:

$$\vec{r} = \mu(\hat{j} - \hat{k})$$

This represents the family of all vectors satisfying the given conditions. For any non-zero scalar $\mu$, the resulting vector lies on the plane and is perpendicular to $\vec{c}$. The general solution is **$\vec{r} = \mu(\hat{j} - \hat{k})$**.

Animated Solution for Mathematics - Vector Algebra: The vector(s) which is/are coplanar with vectors $\hat{i} + \hat{j} + 2 \hat{k}$ and $\hat{i} + 2 \hat{j} + \hat{k}$ , and perpendicular to the vector $\hat{i} + \hat{j} + \hat{k}$ is/are

Select Answer:

Visualized Solution

Visualizing the Vectors

The Coplanar Condition

Substituting the Vectors

Grouping the Components

The Perpendicularity Constraint

The Dot Product Condition

Setting up the Dot Product

Evaluating the Dot Product

Solving for the Scalars

Substituting Back into $r$

The Final Vector Form

Matching the Options

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Filter

The Elegant Cancellation

The Final Reveal

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