If a = i + 2k, b = i + j + k, c = 7i - 3j + 4k, r × b + b × c = 0 and r · a = 0 then r · c is equal

Q: Wait, why can't we just solve for the components of $\vec{r}$ directly from the cross product?

That is a fantastic question! The cross product $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ does not define $\vec{r}$ uniquely. It only defines the component of $\vec{r}$ perpendicular to $\vec{b}$. Geometrically, this equation describes a line in space, not a single point. That is why we introduce the scalar parameter $\lambda$ to represent all possible vectors $\vec{r}$ that satisfy the condition.

Q: Why did we use $\lambda$ instead of just solving for components?

Using the scalar $\lambda$ is a powerful technique in vector algebra. It allows us to express the unknown vector $\vec{r}$ in terms of known vectors $\vec{c}$ and $\vec{b}$ and a single unknown variable. This reduces a complex 3D vector problem into a simple 1D algebraic equation, which is much easier to solve and less prone to calculation errors.

Q: Is there a geometric way to think about $\vec{r} \cdot \vec{a} = 0$?

Absolutely! The dot product being zero is the mathematical definition of orthogonality. Geometrically, it means that the vector $\vec{r}$ lies on a plane that is perpendicular to vector $\vec{a}$. By satisfying both the cross product condition (which puts $\vec{r}$ on a line) and the dot product condition (which puts $\vec{r}$ on a plane), we find the unique point where that line intersects that plane.

Question

If $\vec{a} = \hat{i} + 2\hat{k}, \vec{b} = \hat{i} + \hat{j} + \hat{k}, \vec{c} = 7\hat{i} - 3\hat{j} + 4\hat{k}$, $\vec{r} 	imes \vec{b} + \vec{b} 	imes \vec{c} = \vec{0}$ and $\vec{r} \cdot \vec{a} = 0$ then $\vec{r} \cdot \vec{c}$ is equal to

Options:

(A) 34

(B) 12

(C) 36

(D) 30

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a 3D coordinate system, looking at three fixed vectors: $\vec{a}$, $\vec{b}$, and $\vec{c}$. You are tasked with finding a mysterious vector $\vec{r}$.

At first glance, the conditions given—a cross product equation and a dot product equation—might seem like a jumble of symbols. We are not just solving for a variable; we are navigating the geometric relationships between these vectors.

### Decoding the Cross Product

We begin with the condition $\vec{r} 	imes \vec{b} + \vec{b} 	imes \vec{c} = \vec{0}$.

Applying the anti-commutative property of the cross product, $\vec{b} 	imes \vec{c} = -\vec{c} 	imes \vec{b}$, our equation transforms into:

$$\vec{r} 	imes \vec{b} - \vec{c} 	imes \vec{b} = \vec{0}$$

This simplifies to the elegant form:

$$(\vec{r} - \vec{c}) 	imes \vec{b} = \vec{0}$$

In the language of vectors, when the cross product of two vectors is the zero vector, it implies that the vectors are parallel. This is our breakthrough, indicating that $(\vec{r} - \vec{c})$ must be collinear with $\vec{b}$.

We express this relationship using a scalar parameter $\lambda$:

$$\vec{r} - \vec{c} = \lambda \vec{b}$$

### The Parametric Path

By rearranging our equation, we find the general form of our mystery vector:

$$\vec{r} = \vec{c} + \lambda \vec{b}$$

Geometrically, this tells us that to reach the tip of $\vec{r}$, we start at the origin, travel along $\vec{c}$, and then slide along a line parallel to $\vec{b}$. The only thing missing is the value of $\lambda$.

We use the second condition, $\vec{r} \cdot \vec{a} = 0$, which states that $\vec{r}$ is perpendicular to $\vec{a}$. Substituting our expression for $\vec{r}$ into this dot product yields:

$$(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 0$$

### The Elegant Finish

Expanding the dot product, we obtain:

$$\vec{c} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 0$$

We compute the necessary dot products:

$$\vec{c} \cdot \vec{a} = (7)(1) + (-3)(0) + (4)(2) = 15$$

$$\vec{b} \cdot \vec{a} = (1)(1) + (1)(0) + (1)(2) = 3$$

Plugging these into our equation, we get $15 + 3\lambda = 0$, which results in $\lambda = -5$. Consequently, our vector is defined as $\vec{r} = \vec{c} - 5\vec{b}$.

Finally, we calculate $\vec{r} \cdot \vec{c}$ by substituting our definition:

$$\vec{r} \cdot \vec{c} = (\vec{c} - 5\vec{b}) \cdot \vec{c} = |\vec{c}|^2 - 5(\vec{b} \cdot \vec{c})$$

Given $|\vec{c}|^2 = 74$ and $\vec{b} \cdot \vec{c} = 8$, we arrive at the final result:

$$74 - 5(8) = 34$$

We have successfully navigated the geometry and arrived at the solution. The final answer is **34**.

Animated Solution for Mathematics - Vector Algebra: If $a = \hat{i} + 2 \hat{k}, b = \hat{i} + \hat{j} + \hat{k}, c = 7 \hat{i} - 3 \hat{j} + 4 \hat{k}$ , $r \times b + b \times c = 0$ and $r \cdot a = 0$ then $r \cdot c$ is equal to

Select Answer:

Visualized Solution

Given Vectors

Analyzing the Cross Product

Factoring the Equation

Geometric Interpretation

Defining Vector $r$

The Orthogonality Condition

Substituting $r$

Expanding the Dot Product

Calculating $c \cdot a$

Calculating $b \cdot a$

Solving for $λ$

Setting up the Final Target

Expanding the Final Target

Calculating Components

Final Evaluation

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

Decoding the Cross Product

The Parametric Path

The Elegant Finish

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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=i^+2k^,b=i^+j^​+k^,c=7i^−3j^​+4k^, r×b+b×c=0 and r⋅a=0 then r⋅c is equal to