If a, b, c are three non-zero, non-coplanar vectors and b1 = b - (b.a/|a|²)a, b2 = b + (b.a/|a|²)a,

Q: Why do we subtract the projection? Can't we just rotate the vectors to make them orthogonal?

That is a brilliant question! If you rotate the vectors, you change the space they span. In linear algebra, we want to preserve the 'span'—the set of all possible linear combinations of these vectors. By subtracting the projection, we are not rotating; we are 'stripping' away the parallel component. This leaves the vector pointing in a new direction, but it remains within the same plane or volume defined by the original vectors. It is a transformation that preserves the integrity of the space while enforcing orthogonality.

Q: Is the order of orthogonalization important? Why do we start with $\vec{a}$?

Yes, the order is absolutely critical! The Gram-Schmidt process is sequential. We start with $\vec{a}$ because it acts as our 'anchor' or our first basis vector. Once we fix $\vec{a}$, we make $\vec{b}_1$ orthogonal to it. Then, we use both $\vec{a}$ and $\vec{b}_1$ as the foundation to build $\vec{c}_1$. If you changed the order, you would end up with a completely different set of orthogonal vectors. Always follow the hierarchy established by the problem.

Q: How do I know if the result is correct? Is there a quick way to verify?

The ultimate test is the dot product. If you have successfully created an orthogonal set, the dot product of any two distinct vectors in that set must be zero. For example, check if $\vec{c}_1 \cdot \vec{a} = 0$ and $\vec{c}_1 \cdot \vec{b}_1 = 0$. If you get zero, you have absolute mathematical certainty that your derivation is correct. Never skip this verification step in an exam!

Question

If $\vec{a}, \vec{b}, \vec{c}$ are three non-zero, non-coplanar vectors and $\vec{b}_1 = \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$, $\vec{b}_2 = \vec{b} + \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$, $\vec{c}_1 = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c} \cdot \vec{b}_1}{|\vec{b}_1|^2}\vec{b}_1$, $\vec{c}_2 = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c} \cdot \vec{b}}{|\vec{b}|^2}\vec{b}$, $\vec{c}_3 = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c} \cdot \vec{b}_2}{|\vec{b}_2|^2}\vec{b}_2$, $\vec{c}_4 = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a}$, then the set of orthogonal vectors is

Options:

(A) $(\vec{a}, \vec{b}_1, \vec{c}_3)$

(B) $(\vec{a}, \vec{b}_1, \vec{c}_2)$

(C) $(\vec{a}, \vec{b}_1, \vec{c}_1)$

(D) $(\vec{a}, \vec{b}_2, \vec{c}_2)$

Sigma Percentile · Accepted Answer

### The Philosophy of Purification

The Gram-Schmidt process is a systematic way of removing "overlap" between vectors. If you have a vector $\vec{b}$ and you want to make it perpendicular to $\vec{a}$, you must identify the part of $\vec{b}$ that is already pointing in the direction of $\vec{a}$.

That part is the projection, given by the following expression:

$$ 	ext{proj}_{\vec{a}}\vec{b} = \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} $$

Once you identify this "shadow" of $\vec{b}$ on $\vec{a}$, you simply subtract it. What remains is a vector that has zero component along $\vec{a}$, making it, by definition, orthogonal to $\vec{a}$.

This is represented by the purified vector $\vec{b}_1$:

$$ \vec{b}_1 = \vec{b} - \frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} $$

### The Complexity of the Third Dimension

Now, let us look at the third vector, $\vec{c}$. We need $\vec{c}_1$ to be orthogonal to both $\vec{a}$ and $\vec{b}_1$, meaning it must have no component along either vector.

The expression for this transformation is:

$$ \vec{c}_1 = \vec{c} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}\vec{a} - \frac{\vec{c} \cdot \vec{b}_1}{|\vec{b}_1|^2}\vec{b}_1 $$

Look at the structure; we are subtracting the projection of $\vec{c}$ onto $\vec{a}$, and then subtracting the projection of $\vec{c}$ onto $\vec{b}_1$. By doing this, we are effectively "clearing" the vector $\vec{c}$ of any influence from the previous two vectors.

It is like cleaning a lens; we are removing the "dust" of the other vectors so that $\vec{c}_1$ stands alone, perfectly perpendicular to the plane spanned by $\vec{a}$ and $\vec{b}_1$.

### The Mathematical Verification

Mathematics is not about belief; it is about verification. If we take the dot product of $\vec{c}_1$ with $\vec{a}$, we get:

$$ \vec{c}_1 \cdot \vec{a} = \vec{c} \cdot \vec{a} - \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|^2}(\vec{a} \cdot \vec{a}) - \frac{\vec{c} \cdot \vec{b}_1}{|\vec{b}_1|^2}(\vec{b}_1 \cdot \vec{a}) $$

Since $\vec{a} \cdot \vec{a} = |\vec{a}|^2$, the first two terms cancel out perfectly to zero. Because we already established that $\vec{b}_1 \perp \vec{a}$, the term $\vec{b}_1 \cdot \vec{a}$ is zero, causing the entire expression to collapse to zero.

Similarly, when we dot $\vec{c}_1$ with $\vec{b}_1$, the middle term vanishes because $\vec{a} \cdot \vec{b}_1 = 0$, and the remaining terms cancel out. We have successfully constructed an orthogonal set: $(\vec{a}, \vec{b}_1, \vec{c}_1)$.

### Final Thoughts

Never let the complexity of the notation intimidate you. When you see a long expression like the one for $\vec{c}_1$, do not see a wall of algebra; see a story of geometric construction.

This is the mindset of a **JEE Advanced topper**. You do not just memorize the formula; you understand the geometric necessity behind every subtraction.

You are now ready to tackle any problem involving vector orthogonalization. Go forth and solve with **confidence**!

Select Answer:

Visualized Solution

The Setup: Vectors $a, b, c$

The Goal: Orthogonalization

Analyzing $b_{1}$ : The Projection Term

Geometric Meaning of $b_{1}$

Verifying Orthogonality: $a \cdot b_{1} = 0$

Analyzing $c_{1}$ : Multiple Projections

Geometric Meaning of $c_{1}$

Verifying $c_{1} ⊥ a$

Verifying $c_{1} ⊥ b_{1}$

Final Conclusion: $(a, b_{1}, c_{1})$

The Sigma Insight: Scalar (Dot) Product

The Philosophy of Purification

The Complexity of the Third Dimension

The Mathematical Verification

Final Thoughts

Similar Questions

$a, b, c$ are 3 vectors, such that $a + b + c = 0$ , $∣ a ∣ = 1, ∣ b ∣ = 2, ∣ c ∣ = 3$ , then $a \cdot b + b \cdot c + c \cdot a$ is equal to

Let $a = 2 \hat{i} + λ_{1} \hat{j} + 3 \hat{k}$ , $b = 4 \hat{i} + (3 - λ_{2}) \hat{j} + 6 \hat{k}$ and $c = 3 \hat{i} + 6 \hat{j} + (λ_{3} - 1) \hat{k}$ be three vectors such that $b = 2 a$ and $a$ is perpendicular to $c$ . Then a possible value of $(λ_{1}, λ_{2}, λ_{3})$ is :-

If the vectors $a = \hat{i} - \hat{j} + 2 \hat{k}$ , $b = 2 \hat{i} + 4 \hat{j} + \hat{k}$ and $c = λ \hat{i} + \hat{j} + μ \hat{k}$ are mutually orthogonal, then $(λ, μ) =$

Let $a = \hat{i} + \hat{j} + \hat{k}$ , $b = \hat{i} - \hat{j} + \hat{k}$ and $c = \hat{i} - \hat{j} - \hat{k}$ be three vectors. A vector $v$ in the plane of $a$ and $b$ , whose projection on $c$ is $\frac{1}{3}$ , is given by

Let $a = \hat{i} + 2 \hat{j} + \hat{k}$ , $b = \hat{i} - \hat{j} + \hat{k}$ and $c = \hat{i} + \hat{j} - \hat{k}$ . A vector in the plane of $a$ and $b$ whose projection on $c$ is $\frac{1}{3}$ , is

The vector $\frac{1}{3} (2 \hat{i} - 2 \hat{j} + \hat{k})$ is

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

Let the vectors $a, b, c$ be such that $∣ a ∣ = 2, ∣ b ∣ = 4$ and $∣ c ∣ = 4$ . If the projection of $b$ on $a$ is equal to the projection of $c$ on $a$ and $b$ is perpendicular to $c$ , then the value of $∣ a + b - c ∣$ is

Select Answer:

Visualized Solution

The Setup: Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a,b,c

The Goal: Orthogonalization

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b1​: The Projection Term

Geometric Meaning of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b1​

Verifying Orthogonality: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a⋅b1​=0

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c1​: Multiple Projections

Geometric Meaning of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c1​

Verifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c1​⊥a

Verifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c1​⊥b1​

Final Conclusion: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (a,b1​,c1​)