Let a = 2i + j + k, b = i + 2j - k and a unit vector c be coplanar. If c is perpendicular to a, then

Q: Why can we write $\vec{c} = \lambda \vec{a} + \mu \vec{b}$? Is this always true for coplanar vectors?

Yes, absolutely! By definition, if a vector $\vec{c}$ is coplanar with two non-collinear vectors $\vec{a}$ and $\vec{b}$, it must lie in the plane defined by them. Any vector in that plane can be uniquely represented as a linear combination of the two basis vectors spanning that plane. It is the fundamental language of vector geometry.

Q: Wait, why did we use $\vec{c} \cdot \vec{a} = 0$? Could we have used $\vec{c} \cdot \vec{b} = 0$ instead?

Great observation! The problem specifically states $\vec{c}$ is perpendicular to $\vec{a}$, so we must use $\vec{c} \cdot \vec{a} = 0$. If the problem had stated $\vec{c}$ was perpendicular to $\vec{b}$, then we would use $\vec{c} \cdot \vec{b} = 0$. Always stick to the constraints given in the problem statement!

Q: I got $\mu = -2\lambda$. Does it matter if I solve for $\lambda$ in terms of $\mu$ or vice versa?

It does not matter at all! As long as your algebraic substitution is consistent, the final vector $\vec{c}$ will be the same. The scalars $\lambda$ and $\mu$ are just placeholders for the ratio between the components. The direction of the resulting vector remains invariant.

Q: Why do we need the unit vector condition $|\vec{c}| = 1$ at the end?

The linear combination $\lambda \vec{a} + \mu \vec{b}$ gives us the *direction* of $\vec{c}$, but not its length. Since the problem defines $\vec{c}$ as a unit vector, we must normalize it. Without this step, we would have a vector pointing in the right direction but with the wrong magnitude.

Q: Is there a faster way to do this using cross products?

Yes! You could find a vector perpendicular to the plane using $\vec{n} = \vec{a} \times \vec{b}$, and then find $\vec{c}$ by taking the cross product of $\vec{n}$ and $\vec{a}$. However, the linear combination method is often more intuitive for beginners and less prone to sign errors in cross-product calculations.

Question

Let $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}, \vec{b} = \hat{i} + 2\hat{j} - \hat{k}$ and a unit vector $\vec{c}$ be coplanar. If $\vec{c}$ is perpendicular to $\vec{a}$, then $\vec{c} =$

Options:

(A) $\frac{1}{\sqrt{2}}(-\hat{j} + \hat{k})$

(B) $\frac{1}{\sqrt{3}}(-\hat{i} - \hat{j} - \hat{k})$

(C) $\frac{1}{\sqrt{5}}(\hat{i} - 2\hat{j})$

(D) $\frac{1}{\sqrt{3}}(\hat{i} - \hat{j} - \hat{k})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional space. You have two vectors, $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} - \hat{k}$, anchored at the origin.

These two vectors define a flat surface—a plane—stretching out into infinity. Our goal is to find a special vector, $\vec{c}$, that lies on this plane and is also perpendicular to $\vec{a}$.

### The Power of Linear Combination

Because $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$, it must be a linear combination of them. We express this as:

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

Think of $\lambda$ and $\mu$ as the 'knobs' we turn to adjust the length and direction of our vector within the plane. By substituting the components of $\vec{a}$ and $\vec{b}$, we obtain:

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

When we group the components, the structure of our vector $\vec{c}$ emerges:

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

### The Constraint of Perpendicularity

Now, we introduce the constraint: $\vec{c}$ must be perpendicular to $\vec{a}$. In the language of vectors, this means their dot product must vanish:

$$\vec{c} \cdot \vec{a} = 0$$

We take our expression for $\vec{c}$ and compute the dot product with $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$:

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

Expanding this yields $4\lambda + 2\mu + \lambda + 2\mu + \lambda - \mu = 0$, which simplifies to $6\lambda + 3\mu = 0$. This reveals a hidden symmetry: $\mu = -2\lambda$.

### Normalizing the Vector

Substituting $\mu = -2\lambda$ back into our expression for $\vec{c}$, we find:

$$\vec{c} = (2\lambda - 2\lambda)\hat{i} + (\lambda - 4\lambda)\hat{j} + (\lambda + 2\lambda)\hat{k} = -3\lambda\hat{j} + 3\lambda\hat{k}$$

We now require the vector to be a unit vector, meaning $|\vec{c}| = 1$. We calculate the magnitude:

$$\sqrt{0^2 + (-3\lambda)^2 + (3\lambda)^2} = 1$$

This simplifies to $|3\lambda|\sqrt{2} = 1$, which implies $3\lambda = \pm \frac{1}{\sqrt{2}}$. Substituting this back, we arrive at our final result:

$$\vec{c} = \pm \frac{1}{\sqrt{2}}(-\hat{j} + \hat{k})$$

This is the elegance of vector algebra. Through the rigor of dot products and normalization, we have pinned down the precise mathematical identity of the vector.

Animated Solution for Mathematics - Vector Algebra: Let $a = 2 \hat{i} + \hat{j} + \hat{k}, b = \hat{i} + 2 \hat{j} - \hat{k}$ and a unit vector $c$ be coplanar. If $c$ is perpendicular to $a$ , then $c =$

Select Answer:

Visualized Solution

Visualizing the Vectors

The Coplanarity Equation

Substituting Vectors

Grouping Components

Applying Perpendicularity

Setting up the Dot Product

Solving for $μ$

Refining Vector $c$

The Unit Vector Condition

Setting up the Magnitude

Solving for $λ$

Final Result

The Sigma Insight: Components of a Vector

Analyzing the Setup

The Power of Linear Combination

The Constraint of Perpendicularity

Normalizing the Vector

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=2i^+j^​+k^,b=i^+2j^​−k^ and a unit vector c be coplanar. If c is perpendicular to a, then c=