If the vectors $\vec{c}, \vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$ and $\vec{b} = \hat{j}$ are such that $\vec{a}, \vec{c}$ and $\vec{b}$ form a right handed system then $\vec{c}$ is : Options: (A) $z\hat{i} - x\hat{k}$ (B) $\vec{0}$ (C) $y\hat{j}$ (D) $-z\hat{i} + x\hat{k}$

### Analyzing the Setup We are given two vectors in three-dimensional space: $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$ and $\vec{b} = \hat{j}$. These vectors form part of a right-handed system $(\vec{a}, \vec{c}, \vec{b})$. In a right-handed system, the cyclic order $\vec{a} \rightarrow \vec{c} \rightarrow \vec{b} \rightarrow \vec{a}$ must be maintained. This geometric requirement implies the cross-product relationship: $$\vec{b} \times \vec{a} = \vec{c}$$ ### The Master Equation To determine the components of $\vec{c}$, we utilize the determinant form of the cross product. We set up the $3 \times 3$ matrix where the first row consists of the unit vectors, the second row contains the components of $\vec{b}$, and the third row contains the components of $\vec{a}$: $$\vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 0 \\ x & y & z \end{vmatrix}$$ Expanding this determinant along the first row, we perform the following calculation: $$\vec{c} = \hat{i}(1 \cdot z - 0 \cdot y) - \hat{j}(0 \cdot z - 0 \cdot x) + \hat{k}(0 \cdot y - 1 \cdot x)$$ ### Final Calculation Simplifying the expression above, the $\hat{j}$ component vanishes, which is consistent with the requirement that $\vec{c}$ must be perpendicular to $\vec{b}$ (the y-axis). The resulting vector is: $$\vec{c} = z\hat{i} - x\hat{k}$$ This result is **elegant and precise**. By mastering the right-hand rule, you gain the ability to visualize the orientation of vectors within the fabric of physical space.

Wait, why is it $\vec{c} = \vec{b} \times \vec{a}$ and not $\vec{a} \times \vec{b}$?

Great question! In a right-handed system $(\vec{a}, \vec{c}, \vec{b})$, the cyclic order is $\vec{a} \rightarrow \vec{c} \rightarrow \vec{b} \rightarrow \vec{a}$. This means $\vec{a} \times \vec{c} = \vec{b}$, $\vec{c} \times \vec{b} = \vec{a}$, and $\vec{b} \times \vec{a} = \vec{c}$. Since we want to find $\vec{c}$, we use the relation $\vec{c} = \vec{b} \times \vec{a}$.

Why does the $\hat{j}$ component of $\vec{c}$ become zero?

Geometrically, $\vec{c}$ is the cross product of $\vec{b}$ and $\vec{a}$. By definition, the cross product of two vectors is perpendicular to both. Since $\vec{b}$ is along the y-axis, $\vec{c}$ must be perpendicular to the y-axis. Therefore, the $\hat{j}$ component of $\vec{c}$ must be zero.

Does the order of vectors in the right-handed system matter?

Absolutely! The cross product is anti-commutative, meaning $\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$. Changing the order of the vectors in the triplet would change the sign of the resulting vector, so the cyclic order is essential.

JEE Main 2002

LEVELJEE Main

Animated Solution for Mathematics - Vector Algebra: If the vectors $c, a = x \hat{i} + y \hat{j} + z \hat{k}$ and $b = \hat{j}$ are such that $a, c$ and $b$ form a right handed system then $c$ is :

Select Answer:

Visualized Solution

Defining the Given Vectors $a$ and $b$

We are given two vectors in a three-dimensional Cartesian coordinate system.
The first vector is $a = x \hat{i} + y \hat{j} + z \hat{k}$ .
The second vector is $b = \hat{j}$ , which lies entirely along the positive y-axis.

Understanding the Right-Handed System

The problem states that the vectors $a$ , $c$ , and $b$ form a right-handed system.
In a right-handed system, the ordered triplet $(u, v, w)$ satisfies the right-hand rule.
This means the cross product of the first two vectors points in the direction of the third vector.

Establishing the Cyclic Relation

For the triplet $(a, c, b)$ to be right-handed, we follow the cyclic order: $a \to c \to b \to a$ .
Using the cyclic property, the cross product of $b$ and $a$ gives the vector $c$ .
Therefore, we can write the relation: $c = b \times a$ .

Setting up the Cross Product Determinant

We can compute the cross product $b \times a$ using a $3 \times 3$ determinant.
The first row contains the unit vectors: $\hat{i}$ , $\hat{j}$ , and $\hat{k}$ .
The second row contains the components of $b$ : $(0, 1, 0)$ .
The third row contains the components of $a$ : $(x, y, z)$ .

Computing the $\hat{i}$ Component

Let's expand the determinant along the first row.
For the $\hat{i}$ component, we evaluate the $2 \times 2$ minor: $1 y 0 z$ .
This gives: $\hat{i} \cdot (1 \cdot z - 0 \cdot y) = z \hat{i}$ .

Computing the $\hat{j}$ Component

Next, we evaluate the $\hat{j}$ component with an alternating negative sign.
The minor for $\hat{j}$ is: $0 x 0 z$ .
This gives: $- \hat{j} \cdot (0 \cdot z - 0 \cdot x) = 0 \hat{j}$ .

Computing the $\hat{k}$ Component

Finally, we evaluate the $\hat{k}$ component.
The minor for $\hat{k}$ is: $0 x 1 y$ .
This gives: $\hat{k} \cdot (0 \cdot y - 1 \cdot x) = - x \hat{k}$ .

Combining Components for the Final Vector

Combining all three components, we get: $c = z \hat{i} + 0 \hat{j} - x \hat{k}$ .
Simplifying this expression gives: $c = z \hat{i} - x \hat{k}$ .
This matches Option 1 perfectly.

00:00 / 00:00

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

We are given two vectors in three-dimensional space:

a = x \hat{i} + y \hat{j} + z \hat{k}

and

b = \hat{j}

. These vectors form part of a right-handed system

(a, c, b)

In a right-handed system, the cyclic order

a \to c \to b \to a

must be maintained. This geometric requirement implies the cross-product relationship:

b \times a = c

The Master Equation

To determine the components of

c

, we utilize the determinant form of the cross product. We set up the

3 \times 3

matrix where the first row consists of the unit vectors, the second row contains the components of

b

, and the third row contains the components of

a

c = \hat{i} 0 x \hat{j} 1 y \hat{k} 0 z

Expanding this determinant along the first row, we perform the following calculation:

c = \hat{i} (1 \cdot z - 0 \cdot y) - \hat{j} (0 \cdot z - 0 \cdot x) + \hat{k} (0 \cdot y - 1 \cdot x)

Final Calculation

Simplifying the expression above, the

\hat{j}

component vanishes, which is consistent with the requirement that

c

must be perpendicular to

b

(the y-axis).

The resulting vector is:

c = z \hat{i} - x \hat{k}

This result is elegant and precise. By mastering the right-hand rule, you gain the ability to visualize the orientation of vectors within the fabric of physical space.

Animated Solution for Mathematics - Vector Algebra: If the vectors $c, a = x \hat{i} + y \hat{j} + z \hat{k}$ and $b = \hat{j}$ are such that $a, c$ and $b$ form a right handed system then $c$ is :

Select Answer:

Visualized Solution

Defining the Given Vectors $a$ and $b$

Understanding the Right-Handed System

Establishing the Cyclic Relation

Setting up the Cross Product Determinant

Computing the $\hat{i}$ Component

Computing the $\hat{j}$ Component

Computing the $\hat{k}$ Component

Combining Components for the Final Vector

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Master Equation

Final Calculation

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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; } If the vectors c,a=xi^+yj^​+zk^ and b=j^​ are such that a,c and b form a right handed system then c is :

Select Answer:

Visualized Solution

Defining the 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; } b

Understanding the Right-Handed System

Establishing the Cyclic Relation

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Combining Components for the Final Vector

The Sigma Insight: Vector (Cross) Product

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The Master Equation

Final Calculation

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Animated Solution for Mathematics - Vector Algebra: If the vectors $c, a = x \hat{i} + y \hat{j} + z \hat{k}$ and $b = \hat{j}$ are such that $a, c$ and $b$ form a right handed system then $c$ is :

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