Let the vectors a, b, c be such that |a| = 2, |b| = 4 and |c| = 4. If the projection of b on a is eq

Q: Wait, why do we square the expression? Can't we just find the components of $\vec{a}, \vec{b},$ and $\vec{c}$?

Great question! In JEE Advanced, you rarely have enough information to find the individual components of vectors. Squaring the magnitude is a powerful algebraic 'trick' that converts vector geometry into scalar dot products, which are much easier to handle. It bypasses the need to know the specific orientation of the vectors.

Q: Why does the projection equality imply $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}$?

Remember the definition: the projection of $\vec{b}$ on $\vec{a}$ is $\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}$. If these projections are equal, then $\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|} = \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|}$. Since $|\vec{a}| = 2$ (which is non-zero), we can safely multiply both sides by $|\vec{a}|$ to get $\vec{b} \cdot \vec{a} = \vec{c} \cdot \vec{a}$.

Q: Is it possible that $\vec{a} + \vec{b} - \vec{c}$ results in a negative value?

No! By definition, the magnitude of a vector, denoted by $|\vec{v}|$, is the length of the vector. Length is a scalar quantity and must always be non-negative. When we take the square root of the squared magnitude, we only consider the positive root.

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing in a vast, three-dimensional space with three vectors: $\vec{a}$, $\vec{b}$, and $\vec{c}$. You know their lengths, but you do not know their directions.

This is the classic JEE Advanced setup—a problem that seems to demand specific coordinates but actually rewards those who understand the deep, underlying symmetry of vector algebra.

### The Shadow of Reality

The problem begins with the concept of projection. Think of a projection as a shadow; if you shine a light directly above vector $\vec{b}$ onto vector $\vec{a}$, the length of the shadow is defined as:

$$\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}$$

The problem states that the projection of $\vec{b}$ on $\vec{a}$ is equal to the projection of $\vec{c}$ on $\vec{a}$. This gives us the equation:

$$\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|} = \frac{\vec{c} \cdot \vec{a}}{|\vec{a}|}$$

Since $|\vec{a}| = 2$ (which is non-zero), we can multiply both sides by $|\vec{a}|$ to reveal a beautiful truth:

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

This is our first major breakthrough. We have translated a geometric statement about shadows into a powerful algebraic equality.

### The Power of the Square

Now, we face the target expression: $|\vec{a} + \vec{b} - \vec{c}|$. Calculating the magnitude of a sum of vectors directly is often messy and prone to error.

The 'JEE Master' approach is to square the magnitude, because the square of a magnitude is simply the dot product of the vector with itself: $|\vec{v}|^2 = \vec{v} \cdot \vec{v}$.

Let us expand $|\vec{a} + \vec{b} - \vec{c}|^2$. Using the distributive property of the dot product, we get:

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

### The Moment of Cancellation

This is where the magic happens. We have two pieces of information: $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}$ and the fact that $\vec{b}$ is perpendicular to $\vec{c}$, meaning $\vec{b} \cdot \vec{c} = 0$.

Plugging these into our expanded equation, the term $2(\vec{a} \cdot \vec{b})$ and $-2(\vec{a} \cdot \vec{c})$ cancel each other out perfectly. Furthermore, the term $-2(\vec{b} \cdot \vec{c})$ becomes zero.

The entire complex expression collapses into a simple sum of squares:

$$|\vec{a} + \vec{b} - \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2$$

### Final Calculation

We are given the magnitudes: $|\vec{a}| = 2$, $|\vec{b}| = 4$, and $|\vec{c}| = 4$. Substituting these values, we get:

$$|\vec{a} + \vec{b} - \vec{c}|^2 = (2)^2 + (4)^2 + (4)^2$$

$$|\vec{a} + \vec{b} - \vec{c}|^2 = 4 + 16 + 16 = 36$$

Taking the square root, we find the final result:

**$|\vec{a} + \vec{b} - \vec{c}| = 6$**

Animated Solution for Mathematics - Vector Algebra: 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

Enter Numerical Value:

Visualized Solution

Given Magnitudes: $∣ a ∣, ∣ b ∣, ∣ c ∣$

Projection of $b$ and $c$ on $a$

Equating Projections

Simplifying to $a \cdot b = a \cdot c$

Condition $b ⊥ c$

Evaluating $∣ a + b - c ∣^{2}$

Expanding the Square

Substituting Dot Products

Canceling Terms

Substituting Magnitudes

Calculating the Sum

Final Answer for $∣ a + b - c ∣$

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Shadow of Reality

The Power of the Square

The Moment of Cancellation

Final Calculation

Similar Questions

Let $u, v, w$ be such that $∣ u ∣ = 1, ∣ v ∣ = 2, ∣ w ∣ = 3$ . If the projection $v$ along $u$ is equal to that of $w$ along $u$ and $v, w$ are perpendicular to each other then $∣ u - v + w ∣$ equals

Let $x$ be a vector in the plane containing vectors $a = 2 \hat{i} - \hat{j} + \hat{k}$ and $b = \hat{i} + 2 \hat{j} - \hat{k}$ . If the vector $x$ is perpendicular to $(3 \hat{i} + 2 \hat{j} - \hat{k})$ and its projection on $a$ is $\frac{17 6}{2}$ , then the value of $∣ x ∣^{2}$ is equal to

Let $a = 2 \hat{i} + \hat{j} - \hat{k}$ and $b = \hat{i} + 2 \hat{j} + \hat{k}$ be two vectors. Consider a vector $c = α a + β b, α, β \in R$ . If the projection of $c$ on the vector $(a + b)$ is $32$ , then the minimum value of $(c - (a \times b)) \cdot c$ equals

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

Let $a = 2 \hat{i} - \hat{j} + \hat{k}$ , $b = \hat{i} + 2 \hat{j} - \hat{k}$ and $c = \hat{i} + \hat{j} - 2 \hat{k}$ be three vectors. A vector in the plane of $b$ and $c$ , whose projection on $a$ is of magnitude $2/3$ , is:

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

If $∣ a ∣ = 5, ∣ b ∣ = 4, ∣ c ∣ = 3$ thus what will be the value of $∣ a \cdot b + b \cdot c + c \cdot a ∣$ , given that $a + b + c = 0$

Let $a, b$ and $c$ be three unit vectors such that $∣ a - b ∣^{2} + ∣ a - c ∣^{2} = 8$ . Then $∣ a + 2 b ∣^{2} + ∣ a + 2 c ∣^{2}$ is equal to

Let $A, B, C$ be vectors of length 3, 4, 5 respectively. Let $A$ be perpendicular to $B + C$ , $B$ to $C + A$ and $C$ to $A + B$ . Then the length of vector $A + B + C$ is .........