Let a = 2i - 3j + 4k, b = 3i + 4j - 5k and a vector c be such that a × (b + c) + b × c = i + 8j + 13

Q: Why did we take the cross product with $\vec{a}$? Couldn't we just solve for $\vec{c}$ directly?

That is a fantastic question! If you try to solve for $\vec{c}$ directly, you will find yourself stuck in a system of equations that is very difficult to untangle because $\vec{c}$ is inside a cross product. By taking the cross product with $\vec{a}$, we are essentially 'projecting' the equation into a form where the Vector Triple Product identity can be applied. This identity allows us to convert cross products into dot products, which are scalars and much easier to handle. It is a strategic move, not just a random one!

Q: Wait, the scalar triple product calculation is confusing. Is there a simpler way?

The scalar triple product $[\vec{a} \quad \vec{d} \quad \vec{b}]$ is just the determinant of a $3 \times 3$ matrix formed by the components of the vectors. While it might look intimidating, it is the most robust and error-proof method. Just be very careful with the signs when expanding the determinant. If you find yourself getting lost, write out the matrix clearly on your paper before you start calculating the minors.

Q: Why do we need to calculate $\vec{b} \cdot \vec{c}$ at the end? Can't we find $\vec{c}$ itself?

You absolutely could find $\vec{c}$ itself, but that would be like using a sledgehammer to crack a nut! The question only asks for the value of $24 - \vec{b} \cdot \vec{c}$. In JEE Advanced problems, time is your most valuable resource. Always look at what the question is asking for before you start calculating. If you can find the final answer without finding the intermediate vector $\vec{c}$, you will save precious minutes.

Question

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

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given two vectors, $\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$ and $\vec{b} = 3\hat{i} + 4\hat{j} - 5\hat{k}$. We are also provided with the vector equation:

$$\vec{a} 	imes (\vec{b} + \vec{c}) + \vec{b} 	imes \vec{c} = \hat{i} + 8\hat{j} + 13\hat{k}$$

Additionally, we have the constraint $\vec{a} \cdot \vec{c} = 13$. Our objective is to determine the value of $24 - \vec{b} \cdot \vec{c}$.

### The Art of Simplification

Let $\vec{d} = \hat{i} + 8\hat{j} + 13\hat{k}$. Using the distributive property of the cross product, the given equation expands to:

$$\vec{a} 	imes \vec{b} + \vec{a} 	imes \vec{c} + \vec{b} 	imes \vec{c} = \vec{d}$$

This expression serves as our primary foundation for the subsequent algebraic manipulations.

### The Master Stroke

To isolate the components involving $\vec{c}$, we utilize the **Vector Triple Product identity**:

$$\vec{x} 	imes (\vec{y} 	imes \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z}$$

We take the cross product of the entire equation with $\vec{a}$ on both sides:

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

### Applying the Triple Product

Applying the identity to each term, we obtain:

1. $\vec{a} 	imes (\vec{a} 	imes \vec{b}) = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}$
2. $\vec{a} 	imes (\vec{a} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{a})\vec{c}$
3. $\vec{a} 	imes (\vec{b} 	imes \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$

### The Calculation

First, we compute the necessary scalar values:
$\vec{a} \cdot \vec{b} = (2)(3) + (-3)(4) + (4)(-5) = 6 - 12 - 20 = -26$
$|\vec{a}|^2 = 2^2 + (-3)^2 + 4^2 = 29$

Substituting these into our expanded equation:

$$(-26)\vec{a} - 29\vec{b} + 13\vec{a} - 29\vec{c} + 13\vec{b} - (-26)\vec{c} = \vec{a} 	imes \vec{d}$$

Combining like terms yields:

$$-13\vec{a} - 16\vec{b} - 3\vec{c} = \vec{a} 	imes \vec{d}$$

### Final Calculation

To find $\vec{b} \cdot \vec{c}$, we take the dot product of the simplified equation with $\vec{b}$:

$$-13(\vec{a} \cdot \vec{b}) - 16(\vec{b} \cdot \vec{b}) - 3(\vec{b} \cdot \vec{c}) = \vec{b} \cdot (\vec{a} 	imes \vec{d})$$

The term $\vec{b} \cdot (\vec{a} 	imes \vec{d})$ is the scalar triple product $[\vec{b} \, \vec{a} \, \vec{d}]$, which evaluates to $-396$. Given $|\vec{b}|^2 = 3^2 + 4^2 + (-5)^2 = 50$, we substitute:

$$-13(-26) - 16(50) - 3(\vec{b} \cdot \vec{c}) = -396$$
$$338 - 800 - 3(\vec{b} \cdot \vec{c}) = -396$$
$$-462 - 3(\vec{b} \cdot \vec{c}) = -396 \Rightarrow -3(\vec{b} \cdot \vec{c}) = 66 \Rightarrow \vec{b} \cdot \vec{c} = -22$$

Finally, the requested value is $24 - (-22) = \mathbf{46}$.

Animated Solution for Mathematics - Vector Algebra: 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 _______

Enter Numerical Value:

Visualized Solution

Given Vectors and Equation

Simplifying the RHS

The Master Stroke: Cross Product with $a$

Applying Vector Triple Product

Calculating Dot Products

Substituting Known Values

Simplifying the Vector Equation

Isolating $b \cdot c$

Calculating $∣ b ∣^{2}$ and STP

Solving for $b \cdot c$

Final Calculation

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Art of Simplification

The Master Stroke

Applying the Triple Product

The Calculation

Final Calculation

Similar Questions

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 ____.

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 $λ \in R, a = λ \hat{i} + 2 \hat{j} - 3 \hat{k}, b = \hat{i} - λ \hat{j} + 2 \hat{k}$ . If $((a + b) \times (a \times b)) \times (a - b) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$ , then $∣ λ (a + b) \times (a - b) ∣^{2}$ 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 :

Let $a = 2 \hat{i} - \hat{j} + 5 \hat{k}$ and $b = α \hat{i} + β \hat{j} + 2 \hat{k}$ . If $((a \times b) \times \hat{i}) \cdot \hat{k} = \frac{23}{2}$ , then $∣ b \times 2 \hat{j} ∣$ 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

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 :-

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

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 $(2 a + b) \cdot [(a \times b) \times (a - 2 b)]$ .

Enter Numerical Value:

Visualized Solution

Given Vectors and Equation

Simplifying the RHS

The Master Stroke: Cross Product with .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Applying Vector Triple Product

Calculating Dot Products

Substituting Known Values

Simplifying the Vector Equation

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

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

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

Final Calculation

The Sigma Insight: Vector Triple Product

Analyzing the Setup

The Art of Simplification

The Master Stroke

Applying the Triple Product

The Calculation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 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, the value of 3(c⋅a) is equal to ____.

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The Master Stroke: Cross Product with $a$

Isolating $b \cdot c$

Calculating $∣ b ∣^{2}$ and STP

Solving for $b \cdot c$

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 ____.

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 $λ \in R, a = λ \hat{i} + 2 \hat{j} - 3 \hat{k}, b = \hat{i} - λ \hat{j} + 2 \hat{k}$ . If $((a + b) \times (a \times b)) \times (a - b) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$ , then $∣ λ (a + b) \times (a - b) ∣^{2}$ 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 :

Let $a = 2 \hat{i} - \hat{j} + 5 \hat{k}$ and $b = α \hat{i} + β \hat{j} + 2 \hat{k}$ . If $((a \times b) \times \hat{i}) \cdot \hat{k} = \frac{23}{2}$ , then $∣ b \times 2 \hat{j} ∣$ 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

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 :-

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

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 $(2 a + b) \cdot [(a \times b) \times (a - 2 b)]$ .