Let a = i + 2j - k, b = i - j and c = i - j - k be three given vectors. If r is a vector such that r

Q: Wait, why do we assume $\vec{r} = \vec{c} + \lambda \vec{a}$? Can't $\vec{r}$ be something else?

That is a fantastic question! The condition $(\vec{r} - \vec{c}) \times \vec{a} = \vec{0}$ implies that the vector $(\vec{r} - \vec{c})$ must be parallel to $\vec{a}$. In vector algebra, if two vectors have a cross product of zero, they must be scalar multiples of each other. Thus, $(\vec{r} - \vec{c}) = \lambda \vec{a}$ is the most general form for any vector $\vec{r}$ satisfying that condition. It essentially defines a line in 3D space passing through the point represented by $\vec{c}$ and extending in the direction of $\vec{a}$.

Q: Why does $\vec{c} \cdot \vec{a}$ become zero? Did I make a calculation error?

Not at all! It is actually a beautiful, intentional simplification. When you calculate $\vec{c} \cdot \vec{a} = (1)(1) + (-1)(2) + (-1)(-1)$, you get $1 - 2 + 1$, which is indeed zero. This means that $\vec{c}$ and $\vec{a}$ are orthogonal. In JEE problems, such elegant cancellations are often a sign that you are on the right track!

Q: Why can't we just solve for $\vec{r}$ directly as a system of linear equations?

You could try, but it would be much more tedious. The cross product condition $\vec{r} \times \vec{a} = \vec{c} \times \vec{a}$ does not uniquely determine $\vec{r}$ because it only constrains the component of $\vec{r}$ perpendicular to $\vec{a}$. It leaves the component parallel to $\vec{a}$ free, which is why we use the parameter $\lambda$. Using the parametric form is the most efficient and elegant way to handle this.

Question

Let $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}, \vec{b} = \hat{i} - \hat{j}$ and $\vec{c} = \hat{i} - \hat{j} - \hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r} 	imes \vec{a} = \vec{c} 	imes \vec{a}$ and $\vec{r} \cdot \vec{b} = 0$, then $\vec{r} \cdot \vec{a}$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given three vectors in three-dimensional space:
$\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$
$\vec{b} = \hat{i} - \hat{j}$
$\vec{c} = \hat{i} - \hat{j} - \hat{k}$

We seek a vector $\vec{r}$ that satisfies two specific geometric constraints:
1. $\vec{r} 	imes \vec{a} = \vec{c} 	imes \vec{a}$
2. $\vec{r} \cdot \vec{b} = 0$

### The Cross Product Mystery

We begin with the first condition: $\vec{r} 	imes \vec{a} = \vec{c} 	imes \vec{a}$. By rearranging the terms, we obtain:
$$\vec{r} 	imes \vec{a} - \vec{c} 	imes \vec{a} = \vec{0}$$

Applying the distributive property of the cross product, this simplifies to:
$$(\vec{r} - \vec{c}) 	imes \vec{a} = \vec{0}$$

Since the cross product of two vectors is zero if and only if they are parallel, the vector $(\vec{r} - \vec{c})$ must be parallel to $\vec{a}$. This allows us to express $\vec{r}$ in terms of a scalar parameter $\lambda$:
$$\vec{r} - \vec{c} = \lambda \vec{a} \implies \vec{r} = \vec{c} + \lambda \vec{a}$$

### The Orthogonality Constraint

Now, we apply the second condition, $\vec{r} \cdot \vec{b} = 0$, which implies that $\vec{r}$ is orthogonal to $\vec{b}$. Substituting our parametric form for $\vec{r}$ into this equation yields:
$$(\vec{c} + \lambda \vec{a}) \cdot \vec{b} = 0$$

Expanding the dot product, we get the master equation:
$$\vec{c} \cdot \vec{b} + \lambda (\vec{a} \cdot \vec{b}) = 0$$

Calculating the individual dot products:
$\vec{c} \cdot \vec{b} = (1)(1) + (-1)(-1) + (-1)(0) = 1 + 1 + 0 = 2$
$\vec{a} \cdot \vec{b} = (1)(1) + (2)(-1) + (-1)(0) = 1 - 2 + 0 = -1$

Substituting these values back into the master equation:
$$2 + \lambda(-1) = 0 \implies \lambda = 2$$

### Final Calculation

With $\lambda = 2$ determined, we find the value of $\vec{r} \cdot \vec{a}$ by substituting $\vec{r} = \vec{c} + 2\vec{a}$:
$$\vec{r} \cdot \vec{a} = (\vec{c} + 2\vec{a}) \cdot \vec{a} = \vec{c} \cdot \vec{a} + 2(\vec{a} \cdot \vec{a})$$

First, we compute $\vec{c} \cdot \vec{a}$:
$$\vec{c} \cdot \vec{a} = (1)(1) + (-1)(2) + (-1)(-1) = 1 - 2 + 1 = 0$$

Next, we compute the magnitude squared of $\vec{a}$:
$$|\vec{a}|^2 = \vec{a} \cdot \vec{a} = (1)^2 + (2)^2 + (-1)^2 = 1 + 4 + 1 = 6$$

Combining these results:
$$\vec{r} \cdot \vec{a} = 0 + 2(6) = 12$$

The final result is **12**.

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{i} + 2 \hat{j} - \hat{k}, b = \hat{i} - \hat{j}$ and $c = \hat{i} - \hat{j} - \hat{k}$ be three given vectors. If $r$ is a vector such that $r \times a = c \times a$ and $r \cdot b = 0$ , then $r \cdot a$ is equal to

Enter Numerical Value:

Visualized Solution

Visualizing the Given Vectors

Analyzing the Cross Product Condition

Defining Vector $r$ Parametrically

Using the Orthogonality Condition

Calculating $c \cdot b$

Calculating $a \cdot b$

Solving for $λ$

Setting up the Final Expression

Calculating $c \cdot a$

Calculating $∣ a ∣^{2}$

Final Result and Takeaway

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Cross Product Mystery

The Orthogonality Constraint

Final Calculation

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Enter Numerical Value:

Visualized Solution

Visualizing the Given Vectors

Analyzing the Cross Product Condition

Defining Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r Parametrically

Using the Orthogonality Condition

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c⋅b

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a⋅b

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

Setting up the Final Expression

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

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