If b and c are any two non-collinear unit vectors and a is any vector, then (a.b)b + (a.c)c + (a.(b

Q: Wait, why can we just assume $\vec{b} = \hat{i}$ and $\vec{c} = \hat{j}$? Doesn't that make the proof specific to one case?

That is a brilliant question! In vector algebra, if an identity holds for a general basis, it must hold for any specific basis. By choosing a convenient coordinate system, we are not changing the physics; we are just choosing the most comfortable 'lens' through which to view the truth. If the result is $\vec{a}$ in this specific frame, it must be $\vec{a}$ in all frames.

Q: What if $\vec{b}$ and $\vec{c}$ are not unit vectors? Does the formula change?

Yes, absolutely! The formula relies on the fact that the projection of $\vec{a}$ onto $\vec{b}$ is $(\vec{a} \cdot \hat{b})\hat{b}$. If $\vec{b}$ is not a unit vector, you would need to divide by its magnitude squared, i.e., $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\vec{b}$. The given problem is elegant because it assumes unit vectors, simplifying the denominator to one.

Q: Is this related to the Gram-Schmidt process?

You have a sharp eye! Yes, this is essentially the geometric foundation of the Gram-Schmidt process. We are decomposing an arbitrary vector $\vec{a}$ into its components along three mutually perpendicular directions. It is the vector equivalent of writing a number in terms of its digits in a specific base.

Question

If $\vec{b}$ and $\vec{c}$ are any two non-collinear unit vectors and $\vec{a}$ is any vector, then $(\vec{a} \cdot \vec{b})\vec{b} + (\vec{a} \cdot \vec{c})\vec{c} + \frac{\vec{a} \cdot (\vec{b} 	imes \vec{c})}{|\vec{b} 	imes \vec{c}|^2}(\vec{b} 	imes \vec{c}) = .........$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in the center of a vast, three-dimensional room. You have two vectors, $\vec{b}$ and $\vec{c}$, which are non-collinear unit vectors.

They define a plane, but they do not span the entire room. To describe any vector $\vec{a}$ in this space, you need a third direction.

That is where the cross product $\vec{b} 	imes \vec{c}$ comes in. It is the key that unlocks the third dimension, creating a complete basis.

### The Power of the Smart Assumption

In the heat of a JEE Advanced exam, you might look at the expression:

$$(\vec{a} \cdot \vec{b})\vec{b} + (\vec{a} \cdot \vec{c})\vec{c} + \frac{\vec{a} \cdot (\vec{b} 	imes \vec{c})}{|\vec{b} 	imes \vec{c}|^2}(\vec{b} 	imes \vec{c})$$

This expression is a universal truth. It holds for any non-collinear unit vectors.

So, why fight the general case? Let us choose the most beautiful, simple coordinate system possible.

Let $\vec{b} = \hat{i}$ and $\vec{c} = \hat{j}$. Now, the cross product $\vec{b} 	imes \vec{c}$ becomes $\hat{i} 	imes \hat{j} = \hat{k}$.

The magnitude squared $|\vec{b} 	imes \vec{c}|^2$ is simply $|\hat{k}|^2 = 1$. The scary expression has just melted away.

### Decomposing the Vector

Now, let us take an arbitrary vector $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$. We are going to project this vector onto our new basis.

The first term, $(\vec{a} \cdot \vec{b})\vec{b}$, is just the projection of $\vec{a}$ onto the x-axis, which gives us $x\hat{i}$.

The second term, $(\vec{a} \cdot \vec{c})\vec{c}$, is the projection onto the y-axis, giving us $y\hat{j}$.

Finally, the third term, $\frac{\vec{a} \cdot (\vec{b} 	imes \vec{c})}{|\vec{b} 	imes \vec{c}|^2}(\vec{b} 	imes \vec{c})$, is the projection onto the z-axis, which simplifies beautifully to $z\hat{k}$.

### The Grand Unification

When we sum these three projections, $x\hat{i} + y\hat{j} + z\hat{k}$, we are simply reconstructing our original vector $\vec{a}$.

It is a profound realization: any vector can be decomposed into its projections along three mutually perpendicular axes.

The complexity of the original expression was just a mask for this fundamental geometric truth. You have successfully navigated the 3D space and returned to your starting point, **$\vec{a}$**.

Animated Solution for Mathematics - Vector Algebra: If $b$ and $c$ are any two non-collinear unit vectors and $a$ is any vector, then $(a \cdot b) b + (a \cdot c) c + \frac{a \cdot ( b \times c )}{∣ b \times c ∣ ^{2}} (b \times c) = .........$

Visualized Solution

Visualizing the Basis Vectors

The Normal Vector

Smart Substitution

Orthogonal Basis

Arbitrary Vector $a$

Evaluating Term $1$

Simplifying Term $1$

Evaluating Term $2$

Simplifying Term $2$

Evaluating Term $3$

Simplifying Term $3$

Final Addition

The Sigma Insight: Components of a Vector

Analyzing the Setup

The Power of the Smart Assumption

Decomposing the Vector

The Grand Unification

Similar Questions

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 $c =$

If vectors $a_{1} = x \hat{i} - \hat{j} + \hat{k}$ and $a_{2} = \hat{i} + y \hat{j} + z \hat{k}$ are collinear, then a possible unit vector parallel to the vector $x \hat{i} + y \hat{j} + z \hat{k}$ is :

A vector $a = α \hat{i} + 2 \hat{j} + β \hat{k} (α, β \in R)$ lies in the plane of the vectors, $b = \hat{i} + \hat{j}$ and $c = \hat{i} - \hat{j} + 4 \hat{k}$ . If $a$ bisects the angle between $b$ and $c$ , then

The components of a vector $a$ along and perpendicular to a non-zero vector $b$ are ......... and ......... respectively.

Let $α = 4 \hat{i} + 3 \hat{j} + 5 \hat{k}$ and $β = \hat{i} + 2 \hat{j} - 4 \hat{k}$ . Let $β_{1}$ be parallel to $α$ and $β_{2}$ be perpendicular to $α$ . If $β = β_{1} + β_{2}$ , then the value of $5 β_{2} \cdot (\hat{i} + \hat{j} + \hat{k})$ is

If the components of $a = α \hat{i} + β \hat{j} + γ \hat{k}$ along and perpendicular to $b = 3 \hat{i} + \hat{j} - \hat{k}$ respectively, are $\frac{16}{11} (3 \hat{i} + \hat{j} - \hat{k})$ and $\frac{1}{11} (- 4 \hat{i} - 5 \hat{j} - 17 \hat{k})$ , then $α^{2} + β^{2} + γ^{2}$ is equal to :

If a unit vector $a$ makes angles $π /3$ with $\hat{i}$ , $π /4$ with $\hat{j}$ and $θ \in (0, π)$ with $\hat{k}$ , then a value of $θ$ is :-

For any vector $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ , with $10∣ a_{i} ∣ < 1, i = 1, 2, 3$ , consider the following statements: (A) $max {∣ a_{1} ∣, ∣ a_{2} ∣, ∣ a_{3} ∣} \leq ∣ a ∣$ , (B) $∣ a ∣ \leq 3 max {∣ a_{1} ∣, ∣ a_{2} ∣, ∣ a_{3} ∣}$

Let $α, β, γ$ be distinct real numbers. The points with position vectors $α \hat{i} + β \hat{j} + γ \hat{k}$ , $β \hat{i} + γ \hat{j} + α \hat{k}$ , $γ \hat{i} + α \hat{j} + β \hat{k}$

Visualized Solution

Visualizing the Basis Vectors

The Normal Vector

Smart Substitution

Orthogonal Basis

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