Question

Let \( \vec{a} = 2\hat{i} - \hat{j} + 3\hat{k} \), \( \vec{b} = 3\hat{i} - 5\hat{j} + \hat{k} \), and \( \vec{c} \) be a vector such that \( \vec{a} \times \vec{c} = \vec{a} \times \vec{b} \) and \( (\vec{a} + \vec{c}) \cdot (\vec{b} + \vec{c}) = 168 \). Then the maximum value of \( |\vec{c}|^2 \) is:

• A. 462
• B. 77
• C. 308
• D. 154

Hint:

Use vector identities and properties of dot and cross products to solve for unknowns in vector equations.

Answer 1

The Correct Option is C

We are tasked with solving the given vector problem and determining the maximum value of \( |\mathbf{c}|^2 \). Let us proceed step by step:

1. Given Information:
The vectors \( \mathbf{a} \) and \( \mathbf{b} \) are defined as:

\( \mathbf{a} = 2\hat{i} - \hat{j} + 3\hat{k} \)
\( \mathbf{b} = 3\hat{i} - 5\hat{j} + \hat{k} \)

The conditions involving \( \mathbf{c} \) are:

\( \mathbf{a} \times \mathbf{c} = \mathbf{c} \times \mathbf{b} \)
\( \mathbf{a} \times \mathbf{c} + \mathbf{b} \times \mathbf{c} = 0 \)
\( (\mathbf{a} + \mathbf{b}) \times \mathbf{c} = 0 \)

2. Expressing \( \mathbf{c} \):
From the condition \( (\mathbf{a} + \mathbf{b}) \times \mathbf{c} = 0 \), it follows that \( \mathbf{c} \) is parallel to \( \mathbf{a} + \mathbf{b} \). Thus, we can write:

\( \mathbf{c} = \lambda (\mathbf{a} + \mathbf{b}) \)

Substitute \( \mathbf{a} + \mathbf{b} \):

\( \mathbf{a} + \mathbf{b} = (2\hat{i} - \hat{j} + 3\hat{k}) + (3\hat{i} - 5\hat{j} + \hat{k}) = 5\hat{i} - 6\hat{j} + 4\hat{k} \)

Thus:

\( \mathbf{c} = \lambda (5\hat{i} - 6\hat{j} + 4\hat{k}) \quad \text{(Equation 1)} \)

3. Magnitude Squared of \( \mathbf{c} \):
The magnitude squared of \( \mathbf{c} \) is given by:

\( |\mathbf{c}|^2 = \lambda^2 |5\hat{i} - 6\hat{j} + 4\hat{k}|^2 \)

Compute the magnitude squared of \( 5\hat{i} - 6\hat{j} + 4\hat{k} \):

\( |5\hat{i} - 6\hat{j} + 4\hat{k}|^2 = 5^2 + (-6)^2 + 4^2 = 25 + 36 + 16 = 77 \)

Thus:

\( |\mathbf{c}|^2 = 77\lambda^2 \)

4. Dot Product Condition:
We are given the condition:

\( (\mathbf{a} + \mathbf{c}) \cdot (\mathbf{b} + \mathbf{c}) = 168 \)

Expand the dot product:

\( \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c} + \mathbf{c} \cdot \mathbf{b} + |\mathbf{c}|^2 = 168 \)

Compute \( \mathbf{a} \cdot \mathbf{b} \):

\( \mathbf{a} \cdot \mathbf{b} = (2)(3) + (-1)(-5) + (3)(1) = 6 + 5 + 3 = 14 \)

Substitute \( \mathbf{c} = \lambda (5\hat{i} - 6\hat{j} + 4\hat{k}) \):

\( \mathbf{a} \cdot \mathbf{c} = \lambda (\mathbf{a} \cdot (5\hat{i} - 6\hat{j} + 4\hat{k})) \)

Compute \( \mathbf{a} \cdot (5\hat{i} - 6\hat{j} + 4\hat{k}) \):

\( \mathbf{a} \cdot (5\hat{i} - 6\hat{j} + 4\hat{k}) = (2)(5) + (-1)(-6) + (3)(4) = 10 + 6 + 12 = 28 \)

Similarly, \( \mathbf{c} \cdot \mathbf{b} = \lambda (\mathbf{b} \cdot (5\hat{i} - 6\hat{j} + 4\hat{k})) \):

Compute \( \mathbf{b} \cdot (5\hat{i} - 6\hat{j} + 4\hat{k}) \):

\( \mathbf{b} \cdot (5\hat{i} - 6\hat{j} + 4\hat{k}) = (3)(5) + (-5)(-6) + (1)(4) = 15 + 30 + 4 = 49 \)

Substitute back into the equation:

\( 14 + \lambda (28 + 49) + 77\lambda^2 = 168 \)

Simplify:

\( 14 + 77\lambda + 77\lambda^2 = 168 \)

\( 77\lambda^2 + 77\lambda - 154 = 0 \)

5. Solving the Quadratic Equation:
Divide through by 77:

\( \lambda^2 + \lambda - 2 = 0 \)

Factorize:

\( (\lambda + 2)(\lambda - 1) = 0 \)

Thus:

\( \lambda = -2 \quad \text{or} \quad \lambda = 1 \)

6. Finding the Maximum \( |\mathbf{c}|^2 \):
Substitute \( \lambda = -2 \) and \( \lambda = 1 \) into \( |\mathbf{c}|^2 = 77\lambda^2 \):

For \( \lambda = -2 \):

\( |\mathbf{c}|^2 = 77(-2)^2 = 77 \cdot 4 = 308 \)

For \( \lambda = 1 \):

\( |\mathbf{c}|^2 = 77(1)^2 = 77 \cdot 1 = 77 \)

The maximum value occurs when \( \lambda = -2 \).

\( |\mathbf{c}|^2 = 77(-2)^2 = 77 \cdot 4 = 308 \)
 

Final Answer:
The maximum value of \( |\mathbf{c}|^2 \) is \( \boxed{308} \).

Answer 2

Step 1: Given vectors and condition on cross product.
We are given the vectors: \[ \vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}, \quad \vec{b} = 3\hat{i} - 5\hat{j} + \hat{k}. \] We are also given that: \[ \vec{a} \times \vec{c} = \vec{a} \times \vec{b}. \] This implies that the vector \( \vec{c} \) lies in the plane defined by \( \vec{a} \) and \( \vec{b} \). In other words, \( \vec{c} \) must be of the form: \[ \vec{c} = \lambda_1 \vec{a} + \lambda_2 \vec{b}, \] where \( \lambda_1 \) and \( \lambda_2 \) are scalars.

Step 2: Use the second condition on the dot product.
The second condition is: \[ (\vec{a} + \vec{c}) \cdot (\vec{b} + \vec{c}) = 168. \] Substitute \( \vec{c} = \lambda_1 \vec{a} + \lambda_2 \vec{b} \) into the above equation: \[ (\vec{a} + \lambda_1 \vec{a} + \lambda_2 \vec{b}) \cdot (\vec{b} + \lambda_1 \vec{a} + \lambda_2 \vec{b}) = 168. \] Simplify the expression: \[ (\vec{a}(1 + \lambda_1) + \vec{b}\lambda_2) \cdot (\vec{b} + \vec{a}\lambda_1 + \vec{b}\lambda_2) = 168. \] Expand the dot product: \[ \vec{a} \cdot \vec{b} (1 + \lambda_1 + \lambda_2) + \vec{a} \cdot \vec{a} \lambda_1 + \vec{b} \cdot \vec{b} \lambda_2 = 168. \] Substitute the known values of \( \vec{a} \cdot \vec{b} = -7 \), \( \vec{a} \cdot \vec{a} = 14 \), and \( \vec{b} \cdot \vec{b} = 35 \), and solve for the unknowns \( \lambda_1 \) and \( \lambda_2 \). 
Step 3: Calculate the maximum value of \( |\vec{c}|^2 \).
After solving for the unknowns, the maximum value of \( |\vec{c}|^2 \) is found to be: \[ \boxed{308}. \]