Let a = i - j, b = i + j + k and c be a vector such that a × c + b = 0 and a · c = 4, then |c|² is e

Q: Wait, why can't we just divide by $\vec{a}$ to get $\vec{c}$?

That is a classic trap! In vector algebra, division by a vector is not a defined operation. Unlike scalar algebra, where you can divide by any non-zero number, vectors have direction and magnitude. You cannot 'divide' by a direction. Instead, we use tools like dot products and cross products to isolate variables. Always remember: vectors are not just numbers; they are geometric entities.

Q: How do I know to take the cross product with $\vec{a}$ specifically?

Great question! We look at the structure of the equation $\vec{a} \times \vec{c} = -\vec{b}$. We want to isolate $\vec{c}$. By taking the cross product with $\vec{a}$ again, we create a Vector Triple Product, which allows us to expand the expression into dot products—which are scalars! This effectively 'unpacks' the cross product and lets us solve for $\vec{c}$ linearly.

Q: Is there another way to solve this, perhaps by assuming components for $\vec{c}$?

Yes, you could assume $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$ and solve a system of linear equations using the given conditions. However, that is often tedious and prone to calculation errors. The Vector Triple Product method is the 'JEE Advanced' way—it is elegant, faster, and demonstrates a deeper mastery of vector identities.

Question

Let $\vec{a} = \hat{i} - \hat{j}, \vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{c}$ be a vector such that $\vec{a} 	imes \vec{c} + \vec{b} = \vec{0}$ and $\vec{a} \cdot \vec{c} = 4$, then $|\vec{c}|^2$ is equal to :-

Options:

(A) $\frac{19}{2}$

(B) 8

(C) $\frac{17}{2}$

(D) 9

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey through the elegant landscape of 3D vector space.

We are given two vectors, $\vec{a} = \hat{i} - \hat{j}$ and $\vec{b} = \hat{i} + \hat{j} + \hat{k}$, and an unknown vector $\vec{c}$ that is bound by two constraints:

$$\vec{a} 	imes \vec{c} + \vec{b} = \vec{0} \quad 	ext{and} \quad \vec{a} \cdot \vec{c} = 4$$

Our mission is to find the square of the magnitude of $\vec{c}$. Let's dive in.

### The Strategic Rearrangement

First, let's look at our primary constraint: $\vec{a} 	imes \vec{c} + \vec{b} = \vec{0}$. The immediate instinct might be to try and 'divide' by $\vec{a}$, but as we discussed, that is a mathematical impossibility.

Instead, let's isolate the cross product term. By shifting $\vec{b}$ to the other side, we get:

$$\vec{a} 	imes \vec{c} = -\vec{b}$$

This is our starting point. We have a vector $\vec{c}$ trapped inside a cross product. How do we set it free?

### The Vector Triple Product Magic

This is where the beauty of JEE Advanced physics and math shines. We need to use the **Vector Triple Product** identity.

If we take the cross product of both sides with $\vec{a}$ from the right, we get:

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

Using the property that $-\vec{b} 	imes \vec{a} = \vec{a} 	imes \vec{b}$, our equation becomes:

$$(\vec{a} 	imes \vec{c}) 	imes \vec{a} = \vec{a} 	imes \vec{b}$$

Now, apply the identity $(\vec{u} 	imes \vec{v}) 	imes \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$. Substituting our vectors, we get:

$$(\vec{a} \cdot \vec{a})\vec{c} - (\vec{c} \cdot \vec{a})\vec{a} = \vec{a} 	imes \vec{b}$$

Notice how the cross product has vanished, replaced by dot products—which are just simple scalars.

### The Calculation

We know $\vec{a} = \hat{i} - \hat{j}$, so $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 1^2 + (-1)^2 = 2$. We are also given $\vec{a} \cdot \vec{c} = 4$.

Substituting these into our equation, we get $2\vec{c} - 4\vec{a} = \vec{a} 	imes \vec{b}$. Now, we calculate $\vec{a} 	imes \vec{b}$ using the determinant method:

$$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 0 \ 1 & 1 & 1 \end{vmatrix} = \hat{i}(-1) - \hat{j}(1) + \hat{k}(1 - (-1)) = -\hat{i} - \hat{j} + 2\hat{k}$$

Substituting this back, we have $2\vec{c} = 4\vec{a} + (\vec{a} 	imes \vec{b})$. Plugging in $\vec{a} = \hat{i} - \hat{j}$:

$$2\vec{c} = 4(\hat{i} - \hat{j}) + (-\hat{i} - \hat{j} + 2\hat{k}) = 3\hat{i} - 5\hat{j} + 2\hat{k}$$

Thus, we find the vector $\vec{c}$:

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

### Final Calculation

Finally, we calculate the magnitude squared:

$$|\vec{c}|^2 = \left(\frac{3}{2}ight)^2 + \left(-\frac{5}{2}ight)^2 + 1^2 = \frac{9}{4} + \frac{25}{4} + 1 = \frac{34}{4} + 1 = \frac{17}{2} + 1 = \frac{19}{2}$$

We have arrived at the answer: **$\frac{19}{2}$**. This problem was not just about calculation; it was about recognizing the structure of vector identities. You have successfully navigated the triple product and emerged victorious.

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{i} - \hat{j}, b = \hat{i} + \hat{j} + \hat{k}$ and $c$ be a vector such that $a \times c + b = 0$ and $a \cdot c = 4$ , then $∣ c ∣^{2}$ is equal to :-

Select Answer:

Visualized Solution

Identify Given Vectors $a$ and $b$

Isolate the Cross Product Term

Apply Vector Triple Product Trick

Expand the Vector Triple Product

Calculate $∣ a ∣^{2}$ and Substitute $a \cdot c$

Compute the Cross Product $a \times b$

Evaluate the Determinant

Solve for Vector $c$

Simplify to Find $c$

Calculate $∣ c ∣^{2}$

Final Answer

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Strategic Rearrangement

The Vector Triple Product Magic

The Calculation

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; } Let a=i^−j^​,b=i^+j^​+k^ and c be a vector such that a×c+b=0 and a⋅c=4, then ∣c∣2 is equal to :-

Select Answer:

Visualized Solution

Identify 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

Isolate the Cross Product Term

Apply Vector Triple Product Trick

Expand the Vector Triple Product

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

Compute the Cross Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a×b

Evaluate the Determinant

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

Simplify to Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

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