If A = (1, 1, 1), C = (0, 1, -1) are given vectors, then a vector B satisfying the equations A x B =

Q: Wait, can't I just divide $\vec{C}$ by $\vec{A}$ to get $\vec{B}$? Why are we doing all this algebra?

Ah, the classic trap! Remember, vector division is not defined. You cannot 'divide' by a vector. The cross product is a many-to-one operation; many different vectors $\vec{B}$ could produce the same $\vec{C}$ when crossed with $\vec{A}$. That is why we need the dot product—it acts as the 'anchor' to pick the one unique solution out of the infinite possibilities.

Q: Why did the cross product give us relations instead of the actual values?

Because the cross product only constrains the part of $\vec{B}$ that is perpendicular to $\vec{A}$. It tells us nothing about the component of $\vec{B}$ that is parallel to $\vec{A}$. That 'parallel' component is exactly what the dot product $\vec{A} \cdot \vec{B} = 3$ reveals. They are two halves of the same puzzle.

Q: Is there a geometric way to solve this without components?

Yes, using the vector triple product identity $\vec{A} \times (\vec{A} \times \vec{B}) = \vec{A}(\vec{A} \cdot \vec{B}) - \vec{B}(\vec{A} \cdot \vec{A})$. This is elegant, but for JEE Advanced, the component method is often safer because it is less prone to sign errors and works even if you forget the identity under exam pressure. Stick to the components until you are a master!

Question

If $\vec{A} = (1, 1, 1)$, $\vec{C} = (0, 1, -1)$ are given vectors, then a vector $\vec{B}$ satisfying the equations $\vec{A} 	imes \vec{B} = \vec{C}$ and $\vec{A} \cdot \vec{B} = 3$ is .........

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are not just solving a vector equation; we are performing a surgical operation on 3D space.

We are given two vectors, $\vec{A} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{C} = \hat{j} - \hat{k}$, and we are tasked with finding a mysterious third vector, $\vec{B}$.

Think of $\vec{B}$ as a point in a room, and our two equations are like two laser beams intersecting to pinpoint its exact location.

### The Geometric Constraint

Our first clue is the cross product: $\vec{A} 	imes \vec{B} = \vec{C}$. Geometrically, this is a powerful statement. It tells us that $\vec{C}$ is perpendicular to the plane containing both $\vec{A}$ and $\vec{B}$.

However, a cross product alone is never enough to fully define a vector. It leaves the 'length' of $\vec{B}$ along the direction of $\vec{A}$ ambiguous.

To solve this, we assume $\vec{B} = x\hat{i} + y\hat{j} + z\hat{k}$. We set up the determinant to relate the components:

$$
\begin{vmatrix} 
\hat{i} & \hat{j} & \hat{k} \ 
1 & 1 & 1 \ 
x & y & z 
\end{vmatrix} = 0\hat{i} + 1\hat{j} - 1\hat{k}
$$

### The Algebraic Expansion

Now, let us expand this determinant with precision. Expanding along the first row, we obtain:

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

Equating the components, we derive a system of linear relations:

$$z - y = 0 \quad \Rightarrow \quad y = z$$
$$x - z = 1 \quad \Rightarrow \quad x = z + 1$$
$$y - x = -1$$

We have successfully expressed all components of $\vec{B}$ in terms of a single variable, $z$.

### The Dot Product Anchor

We have the relationships, but we still lack the absolute values. This is where the second condition, $\vec{A} \cdot \vec{B} = 3$, acts as the 'anchor' that fixes the magnitude.

Substituting our components into the dot product, we get:

$$1(x) + 1(y) + 1(z) = 3 \quad \Rightarrow \quad x + y + z = 3$$

Now, we substitute our relations ($x = z + 1$ and $y = z$) into this equation:

$$(z + 1) + z + z = 3$$

$$3z + 1 = 3 \quad \Rightarrow \quad 3z = 2 \quad \Rightarrow \quad z = \frac{2}{3}$$

### Final Calculation

With $z = \frac{2}{3}$ in hand, the rest falls into place. Since $y = z$, we have $y = \frac{2}{3}$.

Since $x = z + 1$, we have $x = \frac{2}{3} + 1 = \frac{5}{3}$.

Our vector $\vec{B}$ is revealed to be:

$$\vec{B} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$$

The cross product provided the 'shape' of the vector, while the dot product provided its 'size'. This is the essence of vector algebra—balancing constraints to find the truth.

Animated Solution for Mathematics - Vector Algebra: If $A = (1, 1, 1)$ , $C = (0, 1, - 1)$ are given vectors, then a vector $B$ satisfying the equations $A \times B = C$ and $A \cdot B = 3$ is .........

Visualized Solution

Visualizing the Given Vectors

Defining the Unknown Vector $B$

Setting up the Cross Product

Expanding the Determinant

Equating the Components

Applying the Dot Product Condition

Substituting Relations into the Equation

Solving for the Variable $z$

Finding the Remaining Components

The Final Vector $B$

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Geometric Constraint

The Algebraic Expansion

The Dot Product Anchor

Final Calculation

Similar Questions

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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 A=(1,1,1), C=(0,1,−1) are given vectors, then a vector B satisfying the equations A×B=C and A⋅B=3 is .........