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 ______.
Correct Answer: 46
Approach Solution - 1
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 \( \vec{b} \cdot \vec{c} \):
\[ 24 - \vec{b} \cdot \vec{c} = 46. \]
Approach Solution -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.
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