Given that a = (1, 1, 1), c = (0, 1, -1), a.b = 3 and a x b = c, then b = .........

Q: Wait, why do we have three equations from the cross product but only two are independent?

That is a brilliant observation! In any cross product $\vec{a} \times \vec{b} = \vec{c}$, the resulting vector $\vec{c}$ is inherently perpendicular to $\vec{a}$. This means the components of $\vec{c}$ are not entirely independent; they are constrained by the geometry of $\vec{a}$. Mathematically, this manifests as a linear dependency in the system of equations derived from the determinant. You only need two of those equations to relate the variables, and the third will be consistent with them.

Q: Why can't we just use the formula $\vec{b} = \frac{\vec{a} \times \vec{c}}{|\vec{a}|^2}$?

That formula is a specific solution for when $\vec{a} \perp \vec{b}$. However, the general solution for $\vec{a} \times \vec{b} = \vec{c}$ is $\vec{b} = \frac{\vec{c} \times \vec{a}}{|\vec{a}|^2} + k\vec{a}$, where $k$ is a scalar. The dot product condition $\vec{a} \cdot \vec{b} = 3$ is exactly what allows us to determine the value of $k$. Without that dot product, $\vec{b}$ would not be uniquely determined!

Q: Is there a unique solution for $\vec{b}$?

Yes, absolutely! The cross product condition fixes the components of $\vec{b}$ that are perpendicular to $\vec{a}$, while the dot product condition fixes the component of $\vec{b}$ that is parallel to $\vec{a}$. Together, they lock the vector $\vec{b}$ into a single, unique position in 3D space.

Question

Given that $\vec{a} = (1, 1, 1)$, $\vec{c} = (0, 1, -1)$, $\vec{a} \cdot \vec{b} = 3$ and $\vec{a} 	imes \vec{b} = \vec{c}$, then $\vec{b} = .........$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a 3D coordinate system, holding two arrows in your hands. One is $\vec{a} = (1, 1, 1)$, and the other is $\vec{c} = (0, 1, -1)$.

Before we even touch the algebra, let us look at the geometry. If we calculate the dot product:

$$\vec{a} \cdot \vec{c} = (1)(0) + (1)(1) + (1)(-1) = 0$$

We discover something profound: these two vectors are perfectly perpendicular. This is not a coincidence; it is the physical soul of the cross product.

When we are told that $\vec{a} 	imes \vec{b} = \vec{c}$, we are being told that $\vec{c}$ is the vector that emerges from the interaction of $\vec{a}$ and $\vec{b}$. By definition, it must be perpendicular to both.

### The Algebraic Blueprint

Since $\vec{b}$ is our mystery, let us give it a name: $\vec{b} = x\hat{i} + y\hat{j} + z\hat{k}$. Our mission is to find the scalars $x, y,$ and $z$.

We have two powerful tools at our disposal: the cross product and the dot product. The cross product $\vec{a} 	imes \vec{b} = \vec{c}$ is best handled using the determinant method. We set up the matrix:

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

Expanding this determinant along the first row, we get:

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

This gives us three component equations:
1. $z - y = 0$
2. $x - z = 1$
3. $y - x = -1$

Notice how these equations are beautifully consistent. The equation $z - y = 0$ tells us $z = y$. Substituting this into $x - z = 1$, we get $x = 1 + z$.

### The Final Anchor

We have expressions for $x$ and $y$ in terms of $z$, but we need a numerical value. This is where the dot product $\vec{a} \cdot \vec{b} = 3$ comes in. It acts as our anchor, pinning down the final degree of freedom.

In component form, this is $(1)(x) + (1)(y) + (1)(z) = 3$, or simply $x + y + z = 3$. Now, we substitute our expressions:

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

This simplifies to $1 + 3z = 3$, which leads us to $3z = 2$, or:

$$z = \frac{2}{3}$$

With $z$ found, the rest is a victory lap. Since $y = z$, then $y = \frac{2}{3}$. Since $x = 1 + z$, then $x = 1 + \frac{2}{3} = \frac{5}{3}$.

We have arrived at our destination. The vector $\vec{b}$ is $\frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$, which we can write elegantly as:

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

You have successfully navigated the interplay of cross and dot products, proving that with a clear geometric vision, even the most complex vector problems become a logical, satisfying journey.

Animated Solution for Mathematics - Vector Algebra: Given that $a = (1, 1, 1)$ , $c = (0, 1, - 1)$ , $a \cdot b = 3$ and $a \times b = c$ , then $b = .........$

Visualized Solution

Visualizing the Given Vectors

The Orthogonality Check

Defining the Unknown Vector $b$

Setting up the Cross Product

Expanding the Determinant

Equating the Components

Using the Dot Product Condition

Solving the System of Equations

Finding $x$ and $y$

Final Vector Assembly

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Algebraic Blueprint

The Final Anchor

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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; } Given that a=(1,1,1), c=(0,1,−1), a⋅b=3 and a×b=c, then b=.........