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 \[ \vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times \vec{c} = \hat{i} + 8\hat{j} + 13\hat{k}. \] If \( \vec{a} \cdot \vec{c} = 13 \), then \( (24 - \vec{b} \cdot \vec{c}) \) is equal to ______.

Answer 1

Correct Answer: 46

From the given equation:

\[ \vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times \vec{c} = \hat{i} + 8\hat{j} + 13\hat{k}. \]

Expanding using vector algebra:

\[ \vec{a} \times \vec{b} + \vec{a} \times \vec{c} + \vec{b} \times \vec{c} = \hat{i} + 8\hat{j} + 13\hat{k}. \]

It is given:

\[ \vec{a} \times \vec{b} = \hat{i} + 8\hat{j} + 13\hat{k}. \]

So:

\[ \vec{a} \times \vec{c} + \vec{b} \times \vec{c} = \vec{0}. \]

Expanding further:

\[ \vec{b} \times \vec{c} = -\vec{a} \times \vec{c}. \]

Using \( \vec{a} \cdot \vec{c} = 13 \), compute:

\[ \vec{b} \cdot \vec{c} = -\left[\vec{a} \cdot (\hat{i} + 8\hat{j} + 13\hat{k})\right] = -22. \]

From the determinant of \( \vec{b} \cdot \vec{c} \):

\[ 24 - \vec{b} \cdot \vec{c} = 46. \]

Answer 2

We are given two vectors \( \vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k} \), \( \vec{b} = 3\hat{i} + 4\hat{j} - 5\hat{k} \), and a vector \( \vec{c} \). We are provided with a vector equation \( \vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times \vec{c} = \hat{i} + 8\hat{j} + 13\hat{k} \) and a scalar condition \( \vec{a} \cdot \vec{c} = 13 \). The goal is to find the value of the expression \( (24 - \vec{b} \cdot \vec{c}) \).

Concept Used:

The solution involves the use of fundamental properties of vector algebra:

1. Distributive Property of Vector Cross Product: The cross product is distributive over vector addition.

\[ \vec{u} \times (\vec{v} + \vec{w}) = \vec{u} \times \vec{v} + \vec{u} \times \vec{w} \] \[ (\vec{u} + \vec{v}) \times \vec{w} = \vec{u} \times \vec{w} + \vec{v} \times \vec{w} \]

2. Vector Triple Product (BAC-CAB Rule): The expansion of a triple cross product is given by:

\[ \vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w} \]

3. Standard methods for calculating the scalar (dot) and vector (cross) products of vectors in component form.

Step-by-Step Solution:

Step 1: Simplify the given vector equation.

We start with the equation \( \vec{a} \times (\vec{b} + \vec{c}) + \vec{b} \times \vec{c} = \hat{i} + 8\hat{j} + 13\hat{k} \).

Using the distributive property on the first term:

\[ (\vec{a} \times \vec{b} + \vec{a} \times \vec{c}) + \vec{b} \times \vec{c} = \hat{i} + 8\hat{j} + 13\hat{k} \]

Group the terms involving \( \vec{c} \):

\[ \vec{a} \times \vec{b} + (\vec{a} \times \vec{c} + \vec{b} \times \vec{c}) = \hat{i} + 8\hat{j} + 13\hat{k} \]

Using the distributive property again:

\[ \vec{a} \times \vec{b} + (\vec{a} + \vec{b}) \times \vec{c} = \hat{i} + 8\hat{j} + 13\hat{k} \]

Isolate the term containing \( \vec{c} \):

\[ (\vec{a} + \vec{b}) \times \vec{c} = (\hat{i} + 8\hat{j} + 13\hat{k}) - (\vec{a} \times \vec{b}) \]

Step 2: Calculate the constant vectors \( \vec{a} \times \vec{b} \) and \( \vec{a} + \vec{b} \).

\[ \vec{a} + \vec{b} = (2+3)\hat{i} + (-3+4)\hat{j} + (4-5)\hat{k} = 5\hat{i} + \hat{j} - \hat{k} \] \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 4 \\ 3 & 4 & -5 \end{vmatrix} = \hat{i}(15 - 16) - \hat{j}(-10 - 12) + \hat{k}(8 - (-9)) = -\hat{i} + 22\hat{j} + 17\hat{k} \]

Step 3: Substitute these values into the simplified equation from Step 1.

Let \( \vec{V} = (\hat{i} + 8\hat{j} + 13\hat{k}) - (\vec{a} \times \vec{b}) \).

\[ \vec{V} = (\hat{i} + 8\hat{j} + 13\hat{k}) - (-\hat{i} + 22\hat{j} + 17\hat{k}) = 2\hat{i} - 14\hat{j} - 4\hat{k} \]

The simplified equation becomes: \( (\vec{a} + \vec{b}) \times \vec{c} = \vec{V} \).

Step 4: Use the vector triple product to find an expression for \( \vec{c} \).

Take the cross product of the equation with \( \vec{a} \):

\[ \vec{a} \times ((\vec{a} + \vec{b}) \times \vec{c}) = \vec{a} \times \vec{V} \]

Applying the BAC-CAB rule to the left side:

\[ (\vec{a} \cdot \vec{c})(\vec{a} + \vec{b}) - (\vec{a} \cdot (\vec{a} + \vec{b}))\vec{c} = \vec{a} \times \vec{V} \]

Step 5: Calculate the necessary scalar and vector products.

We are given \( \vec{a} \cdot \vec{c} = 13 \).

Calculate \( \vec{a} \cdot (\vec{a} + \vec{b}) \):

\[ \vec{a} \cdot (\vec{a} + \vec{b}) = (2\hat{i} - 3\hat{j} + 4\hat{k}) \cdot (5\hat{i} + \hat{j} - \hat{k}) = (2)(5) + (-3)(1) + (4)(-1) = 10 - 3 - 4 = 3 \]

Calculate \( \vec{a} \times \vec{V} \):

\[ \vec{a} \times \vec{V} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 4 \\ 2 & -14 & -4 \end{vmatrix} = \hat{i}(12 - (-56)) - \hat{j}(-8 - 8) + \hat{k}(-28 - (-6)) = 68\hat{i} + 16\hat{j} - 22\hat{k} \]

Step 6: Substitute these values back into the equation from Step 4 to solve for \( \vec{c} \).

\[ (13)(\vec{a} + \vec{b}) - (3)\vec{c} = \vec{a} \times \vec{V} \] \[ 13(5\hat{i} + \hat{j} - \hat{k}) - 3\vec{c} = 68\hat{i} + 16\hat{j} - 22\hat{k} \] \[ 65\hat{i} + 13\hat{j} - 13\hat{k} - 3\vec{c} = 68\hat{i} + 16\hat{j} - 22\hat{k} \]

Rearranging to solve for \( \vec{c} \):

\[ -3\vec{c} = (68 - 65)\hat{i} + (16 - 13)\hat{j} + (-22 + 13)\hat{k} \] \[ -3\vec{c} = 3\hat{i} + 3\hat{j} - 9\hat{k} \] \[ \vec{c} = -\hat{i} - \hat{j} + 3\hat{k} \]

Final Computation & Result:

Step 7: Calculate the scalar product \( \vec{b} \cdot \vec{c} \).

We have \( \vec{b} = 3\hat{i} + 4\hat{j} - 5\hat{k} \) and \( \vec{c} = -\hat{i} - \hat{j} + 3\hat{k} \).

\[ \vec{b} \cdot \vec{c} = (3)(-1) + (4)(-1) + (-5)(3) \] \[ \vec{b} \cdot \vec{c} = -3 - 4 - 15 = -22 \]

Step 8: Compute the final required value.

\[ 24 - (\vec{b} \cdot \vec{c}) = 24 - (-22) = 24 + 22 = 46 \]

The value of \( (24 - \vec{b} \cdot \vec{c}) \) is 46.