Let a = i + j + k, b and c = j - k be three vectors such that a × b = c and a · b = 1. If the length

Q: Wait, why do we use Lagrange's Identity here? Can't we just find $\vec{b}$ by solving the system of equations?

That is a brilliant instinct! However, solving for $\vec{b}$ directly is an algebraic nightmare because $\vec{a} \times \vec{b} = \vec{c}$ only gives you two independent equations for the three components of $\vec{b}$. Lagrange's Identity is the elegant shortcut that bypasses the need to find the individual components of $\vec{b}$ entirely.

Q: Why is the numerator of the projection equal to $|\vec{c}|^2$?

Great observation! The numerator is the scalar triple product $\vec{b} \cdot (\vec{a} \times \vec{c})$. By the cyclic property, this is equivalent to $(\vec{b} \times \vec{a}) \cdot \vec{c}$. Since we know $\vec{a} \times \vec{b} = \vec{c}$, it follows that $\vec{b} \times \vec{a} = -\vec{c}$. Thus, the expression becomes $(-\vec{c}) \cdot \vec{c}$, which is simply $-|\vec{c}|^2$.

Q: How do I know if $\vec{a}$ and $\vec{c}$ are perpendicular?

Always check the dot product! If $\vec{a} \cdot \vec{c} = 0$, they are orthogonal. In this problem, $(1)(0) + (1)(1) + (1)(-1) = 0$. This simplifies the cross product magnitude significantly, as $\sin(90^\circ) = 1$.

Q: Is the projection length always positive?

Yes, the length of a projection is defined as the magnitude of the scalar projection. We use the absolute value $|\vec{b} \cdot \vec{n}|$ to ensure that the length is a non-negative scalar, regardless of the direction of the vectors.

Q: Why did the problem ask for $3l^2$ instead of just $l$?

In JEE Advanced, this is a common technique to ensure the final answer is an integer or a simple rational number. It also serves as a hint that $l^2$ will likely involve a fraction with a denominator of 3, which will be cancelled out perfectly.

Question

Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b}$ and $\vec{c} = \hat{j} - \hat{k}$ be three vectors such that $\vec{a} 	imes \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 1$. If the length of projection vector of the vector $\vec{b}$ on the vector $\vec{a} 	imes \vec{c}$ is $l$, then the value of $3l^2$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

You are given two vectors, $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{c} = \hat{j} - \hat{k}$. A third vector, $\vec{b}$, satisfies the conditions $\vec{b} 	imes \vec{a} = \vec{c}$ and $\vec{b} \cdot \vec{a} = 1$.

While one could assign variables $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$ to solve a system of linear equations, we will instead utilize the elegance of vector algebra to find the length of the projection of $\vec{b}$ onto $\vec{a} 	imes \vec{c}$.

### Unlocking the Magnitude of $\vec{b}$

To find the magnitude of $\vec{b}$, we employ **Lagrange's Identity**:

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

We calculate the known components: $||\vec{a}||^2 = 1^2 + 1^2 + 1^2 = 3$ and $||\vec{a} 	imes \vec{b}|| = ||\vec{c}|| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}$.

Substituting these values into the identity, we obtain $(\sqrt{2})^2 + (1)^2 = 3 ||\vec{b}||^2$. This simplifies to $3 = 3 ||\vec{b}||^2$, revealing that $||\vec{b}|| = 1$. Thus, $\vec{b}$ is a **unit vector**.

### The Art of the Scalar Triple Product

Let $\vec{n} = \vec{a} 	imes \vec{c}$ be the vector onto which we project $\vec{b}$. The length of the projection $l$ is given by:

$$l = \frac{|\vec{b} \cdot (\vec{a} 	imes \vec{c})|}{||\vec{a} 	imes \vec{c}||}$$

The numerator is the scalar triple product $[\vec{b} \vec{a} \vec{c}]$. By the cyclic property of the scalar triple product, this is equivalent to $(\vec{b} 	imes \vec{a}) \cdot \vec{c}$.

Given $\vec{a} 	imes \vec{b} = \vec{c}$, it follows that $\vec{b} 	imes \vec{a} = -\vec{c}$. Therefore, the numerator becomes $(-\vec{c}) \cdot \vec{c} = -||\vec{c}||^2 = -2$. Taking the absolute value, the numerator is $2$.

### The Final Convergence

Next, we calculate the denominator $||\vec{a} 	imes \vec{c}||$. We observe that $\vec{a} \cdot \vec{c} = (1)(0) + (1)(1) + (1)(-1) = 0$, meaning $\vec{a}$ and $\vec{c}$ are orthogonal.

Consequently, the magnitude of the cross product is:

$$||\vec{a} 	imes \vec{c}|| = ||\vec{a}|| ||\vec{c}|| \sin(90^{\circ}) = \sqrt{3} \cdot \sqrt{2} \cdot 1 = \sqrt{6}$$

Substituting these into our projection formula, we find $l = \frac{2}{\sqrt{6}}$. Squaring this result yields $l^2 = \frac{4}{6} = \frac{2}{3}$.

The final value requested is $3l^2$, which is $3 \cdot \frac{2}{3}$. The final answer is **2**.

Enter Numerical Value:

Visualized Solution

Given Vectors $a$ and $c$

Magnitudes of $a$ and $c$

Lagrange's Identity

Substituting Values

Solving for $∣ b ∣^{2}$

Defining Projection Length $l$

Simplifying the Numerator

Orthogonality of $a$ and $c$

Calculating Denominator $∣ a \times c ∣$

Finding $l$ and $l^{2}$

Final Result: $3 l^{2}$

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

Unlocking the Magnitude of $b$

The Art of the Scalar Triple Product

The Final Convergence

Similar Questions

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Enter Numerical Value:

Visualized Solution

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

Magnitudes of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Lagrange's Identity

Substituting Values

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣b∣2

Defining Projection Length .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l

Simplifying the Numerator

Orthogonality of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Calculating Denominator .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣a×c∣

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l2

Final Result: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3l2

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

Unlocking the Magnitude of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b