Let a = i - 2j + 3k, b = i + j + k and c be a vector such that a + (b × c) = 0 and b · c = 5. Then,

Q: Wait, why do we cross with $\vec{b}$? Couldn't we just assume $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$ and solve the system of equations?

That is a valid approach, but it is often a trap in JEE Advanced problems! While you could solve for $x, y, z$ using the cross product and dot product constraints, it leads to a system of linear equations that is time-consuming and prone to arithmetic errors. Using the Vector Triple Product identity is the elegant, 'JEE-style' shortcut that directly isolates the vector $\vec{c}$ without needing to solve for individual components.

Q: Is the order of the cross product important in the identity?

Absolutely! The identity $\vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z}$ is specific. If you had written $(\vec{b} \times \vec{b}) \times \vec{c}$, you would get $\vec{0} \times \vec{c} = \vec{0}$, which tells you nothing about $\vec{c}$. Always ensure you are crossing from the correct side to maintain the structure of the identity.

Q: Why did we get a negative sign when we moved $\vec{a}$ to the other side?

Great observation! The original equation is $\vec{a} + (\vec{b} \times \vec{c}) = \vec{0}$. Subtracting $\vec{a}$ from both sides gives $\vec{b} \times \vec{c} = -\vec{a}$. This is just basic algebraic manipulation of vectors. Remember, the cross product is anti-commutative, so $\vec{b} \times \vec{c} = -(\vec{c} \times \vec{b})$. Keeping track of these signs is crucial for the final result!

Question

Let $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}, \vec{b} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{c}$ be a vector such that $\vec{a} + (\vec{b} 	imes \vec{c}) = \vec{0}$ and $\vec{b} \cdot \vec{c} = 5$. Then, the value of $3(\vec{c} \cdot \vec{a})$ is equal to ____.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a 3D coordinate system. You have two vectors, $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} + \hat{j} + \hat{k}$, fixed in space.

There is a mysterious vector $\vec{c}$ defined by two clues: a constraint involving a cross product, $\vec{a} + (\vec{b} 	imes \vec{c}) = \vec{0}$, and a scalar value from a dot product, $\vec{b} \cdot \vec{c} = 5$.

This is a classic JEE Advanced puzzle. It requires understanding the geometric relationship between these vectors rather than simple number crunching.

### The Strategic Rearrangement

First, let's simplify our starting equation: $\vec{a} + (\vec{b} 	imes \vec{c}) = \vec{0}$. By moving $\vec{a}$ to the other side, we obtain:

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

This reveals that the cross product of $\vec{b}$ and $\vec{c}$ is anti-parallel to $\vec{a}$. Since vector division is not defined, we must use the power of the Vector Triple Product to isolate $\vec{c}$.

### The Power of the Triple Product

To isolate $\vec{c}$, we take the cross product of $\vec{b}$ with both sides of our equation:

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

The left side is a perfect candidate for the Vector Triple Product identity: $\vec{x} 	imes (\vec{y} 	imes \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z}$. Applying this, we get:

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

Given $\vec{b} \cdot \vec{c} = 5$ and $|\vec{b}|^2 = \vec{b} \cdot \vec{b}$, the equation simplifies to:

$$5\vec{b} - |\vec{b}|^2\vec{c} = \vec{a} 	imes \vec{b}$$

### The Final Calculation

First, we compute the magnitude squared of $\vec{b}$:
$|\vec{b}|^2 = (1)^2 + (1)^2 + (1)^2 = 3$.

Next, we calculate the cross product $\vec{a} 	imes \vec{b}$ using the determinant:

$$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -2 & 3 \ 1 & 1 & 1 \end{vmatrix} = (-2-3)\hat{i} - (1-3)\hat{j} + (1+2)\hat{k} = -5\hat{i} + 2\hat{j} + 3\hat{k}$$

Substituting these values into our equation $5\vec{b} - 3\vec{c} = \vec{a} 	imes \vec{b}$:

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

Rearranging to solve for $3\vec{c}$:

$$3\vec{c} = 5(\hat{i} + \hat{j} + \hat{k}) - (-5\hat{i} + 2\hat{j} + 3\hat{k}) = 10\hat{i} + 3\hat{j} + 2\hat{k}$$

Finally, we calculate $3(\vec{c} \cdot \vec{a})$, which is equivalent to $(3\vec{c}) \cdot \vec{a}$:

$$(10\hat{i} + 3\hat{j} + 2\hat{k}) \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = (10)(1) + (3)(-2) + (2)(3) = 10 - 6 + 6 = 10$$

The final result is **10**.

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{i} - 2 \hat{j} + 3 \hat{k}, b = \hat{i} + \hat{j} + \hat{k}$ and $c$ be a vector such that $a + (b \times c) = 0$ and $b \cdot c = 5$ . Then, the value of $3 (c \cdot a)$ is equal to ____.

Enter Numerical Value:

Visualized Solution

Identify Given Vectors $a$ and $b$

Analyze the Cross Product Relation

Apply Vector Triple Product (VTP)

Simplify Using Given Dot Product

Calculate $∣ b ∣^{2}$

Calculate $a \times b$

Form the Equation for $3 c$

Solve for $3 c$

Final Dot Product Calculation

Conclusion and Key Takeaway

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Strategic Rearrangement

The Power of the Triple Product

The Final Calculation

Similar Questions

If $a = \frac{1}{10} (3 \hat{i} + \hat{k})$ and $b = \frac{1}{7} (2 \hat{i} + 3 \hat{j} - 6 \hat{k})$ , then the value of $(2 a - b) \cdot [(a \times b) \times (a + 2 b)]$ is

Let $a = \hat{i} + \hat{j} + \hat{k}$ and $b = \hat{j} - \hat{k}$ . If $c$ is a vector such that $a \times c = b$ and $a \cdot c = 3$ , then $a \cdot (b \times c)$ is equal to :

Let $a = 2 \hat{i} + \hat{j} - 2 \hat{k}$ and $b = \hat{i} + \hat{j}$ . Let $c$ be a vector such that $∣ c - a ∣ = 3$ , $∣ (a \times b) \times c ∣ = 3$ and the angle between $c$ and $a \times b$ be $3 0^{\circ}$ . Then $a \cdot c$ is equal to:

Let $a = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}, b = 3 \hat{i} + 4 \hat{j} - 5 \hat{k}$ and a vector $c$ be such that $a \times (b + c) + b \times c = \hat{i} + 8 \hat{j} + 13 \hat{k}$ . If $a \cdot c = 13$ , then $(24 - b \cdot c)$ is equal to _______

If $a$ and $b$ are vectors in space given by $a = \frac{i ^ - 2 j ^}{5}$ and $b = \frac{2 i ^ + j ^ + 3 k ^}{14}$ , then find the value of $(2 a + b) \cdot [(a \times b) \times (a - 2 b)]$ .

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

Let $a = - 5 \hat{i} + \hat{j} - 3 \hat{k}$ , $b = \hat{i} + 2 \hat{j} - 4 \hat{k}$ and $c = (((a \times b) \times \hat{i}) \times \hat{i}) \times \hat{i}$ . Then $c \cdot (- \hat{i} + \hat{j} + \hat{k})$ is equal to

If $a \cdot b = 1, b \cdot c = 2$ and $c \cdot a = 3$ , then the value of $[a \times (b \times c), b \times (c \times a), c \times (b \times a)]$ is:

Let $a, b$ and $c$ be three vectors such that $∣ a ∣ = 3, ∣ b ∣ = 5, b . c = 10$ and the angle between $b$ and $c$ is $\frac{π}{3}$ . If $a$ is perpendicular to vector $b \times c$ , then $∣ a \times (b \times c) ∣$ is equal to ________

Let $a = - \hat{i} + \hat{j} + 2 \hat{k}$ and $b = \hat{i} - \hat{j} - 3 \hat{k}, c = a \times b$ and $d = c \times a$ . Then $(a - b) \cdot d$ is equal to :

Enter Numerical Value:

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

Analyze the Cross Product Relation

Apply Vector Triple Product (VTP)

Simplify Using Given Dot Product

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

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

Form the Equation for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3c

Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3c