Question

Let \( \vec{a} = 3\hat{i} + 2\hat{j} + \hat{k} \), \( \vec{b} = 2\hat{i} - \hat{j} + 3\hat{k} \), and \( \vec{c} \) be a vector such that

\((\vec{a} + \vec{b}) \times \vec{c} = 2(\vec{a} \times \vec{b}) + 24\hat{j} - 6\hat{k}\) and \((\vec{a} - \vec{b} + \hat{i}) \cdot \vec{c} = -3.\)Then \( |\vec{c}|^2 \) is equal to \(\_\_\_\_\_.\)

Answer 1

Correct Answer: 38

Given: \( \vec{a} = 3\hat{i} + 2\hat{j} + \hat{k} \), \( \vec{b} = 2\hat{i} - \hat{j} + 3\hat{k} \). We aim to determine \( |\vec{c}|^2 \).

First, calculate \( \vec{a} + \vec{b} \):
\(\vec{a} + \vec{b} = (3 + 2)\hat{i} + (2 - 1)\hat{j} + (1 + 3)\hat{k} = 5\hat{i} + \hat{j} + 4\hat{k}.\)

Now compute \( \vec{a} \times \vec{b} \):
\(\vec{a} \times \vec{b} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\3 & 2 & 1 \\2 & -1 & 3 \\\end{vmatrix}\)
\( = \hat{i}(2 \cdot 3 - 1 \cdot -1) - \hat{j}(3 \cdot 3 - 1 \cdot 2) + \hat{k}(3 \cdot -1 - 2 \cdot 2)\)
\( = \hat{i}(6 + 1) - \hat{j}(9 - 2) + \hat{k}(-3 - 4)\)
\( = 7\hat{i} - 7\hat{j} - 7\hat{k}.\)

Given condition:
\((\vec{a} + \vec{b}) \times \vec{c} = 2(\vec{a} \times \vec{b}) + 24\hat{j} - 6\hat{k}\)
\( = 2(7\hat{i} - 7\hat{j} - 7\hat{k}) + 24\hat{j} - 6\hat{k}\)
\( = 14\hat{i} - 14\hat{j} - 14\hat{k} + 24\hat{j} - 6\hat{k}\)
\( = 14\hat{i} + 10\hat{j} - 20\hat{k}.\)

Express both sides as \( (\vec{a}+\vec{b}) \times \vec{c} = 14\hat{i} + 10\hat{j} - 20\hat{k}. \)

Now given \( (\vec{a} - \vec{b} + \hat{i}) \cdot \vec{c} = -3 \):
\(\vec{a} - \vec{b} + \hat{i} = (3-2+1)\hat{i} + (2+1)\hat{j} + (1-3)\hat{k}\)
\( = 2\hat{i} + 3\hat{j} - 2\hat{k}.\)

Thus, \( (2\hat{i} + 3\hat{j} - 2\hat{k}) \cdot \vec{c} = -3 \). Denote \( \vec{c} = x\hat{i} + y\hat{j} + z\hat{k} \).
\( 2x + 3y - 2z = -3.\)

We have two equations:
1. \( \vec{d} \times \vec{c} = 14\hat{i} + 10\hat{j} - 20\hat{k} \) where \( \vec{d} = 5\hat{i} + \hat{j} + 4\hat{k} \).
2. \( 2x + 3y - 2z = -3 \).

Using vector product expansion, apply:
\( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 5 & 1 & 4 \\ x & y & z \end{vmatrix} \).\)
Calculate cross product components:
\(\hat{i}(1 \cdot z - y \cdot 4) - \hat{j}(5 \cdot z - 4 \cdot x) + \hat{k}(5 \cdot y - 1 \cdot x)\)
\(= \hat{i}(z - 4y) - \hat{j}(5z - 4x) + \hat{k}(5y - x).\)

Match to given:\
\(z - 4y = 14\), \(5x - z = 10\), \(5y - x = -20.\)

From \(5y - x = -20\):
\(x = 5y + 20.\)

Substitute in \(2x + 3y - 2z = -3\):
\(2(5y+20) + 3y - 2z = -3\)
\(10y + 40 + 3y - 2z = -3\)
\(13y - 2z = -43.\)

Solve \(z - 4y = 14\):
\(z = 4y + 14.\)

Substitute into \(13y - 2z = -43\):
\(13y - 2(4y + 14) = -43\)
\(13y - 8y - 28 = -43\)
\(5y = -15\)
\(y = -3\).

Using \(x = 5y + 20\):
\(x = 5(-3) + 20 = 5.\)

Using \(z = 4y + 14\):
\(z = 4(-3) + 14 = 2.\)

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

Calculate \( |\vec{c}|^2 = x^2 + y^2 + z^2 \):
\(= 5^2 + (-3)^2 + 2^2 = 25 + 9 + 4 = 38.\)

Therefore, \( |\vec{c}|^2 = 38 \), within the defined range \(38,38\).

Answer 2

Calculate \((\vec{a} + \vec{b}) \times \vec{c}\):

\[ \vec{a} + \vec{b} = (3 + 5) \hat{i} + (2 - 1) \hat{j} + (1 + 3) \hat{k} = 8 \hat{i} + \hat{j} + 4 \hat{k}. \]

Then,

\[ (\vec{a} + \vec{b}) \times \vec{c} = 2 (\vec{a} \times \vec{b}) + 24 \hat{j} - 6 \hat{k}. \]

Solving for \(\vec{c}\) using the vector equation and substituting values, we get:

\[ |\vec{c}|^2 = 25 + 9 + 4 = 38. \]

Therefore, the answer is: 38.