Question

Let \( \vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \, \vec{b} = 3\hat{i} + \hat{j} - \hat{k} \) and \( \vec{c} \) be three vectors such that \( \vec{c} \) is coplanar with \( \vec{a} \) and \( \vec{b} \). If the vector \( \vec{c} \) is perpendicular to \( \vec{b} \) and \( \vec{a} \cdot \vec{c} = 5 \), then \( |\vec{c}| \) is equal to:

• A. \( \frac{1}{\sqrt{3}} \)
• B. 18
• C. 16
• D. \( \sqrt{\frac{11}{6}} \)

Hint:

For problems involving vectors, always use the dot product to find relationships between the vectors and their magnitudes. Also, make use of the vector equation \( \vec{c} = \lambda \vec{a} + \mu \vec{b} \) for coplanar vectors.

Answer 1

The Correct Option is D

To determine the magnitude of vector \( \vec{c} \), which is coplanar with vectors \( \vec{a} \) and \( \vec{b} \), perpendicular to \( \vec{b} \), and adheres to the condition \( \vec{a} \cdot \vec{c} = 5 \), a sequential problem-solving approach is necessary.

1. As \( \vec{c} \) is coplanar with \( \vec{a} \) and \( \vec{b} \), it can be expressed as a linear combination: \( \vec{c} = x\vec{a} + y\vec{b} \).
2. The condition that \( \vec{c} \) is perpendicular to \( \vec{b} \) yields: \( \vec{b} \cdot \vec{c} = \vec{b} \cdot (x\vec{a} + y\vec{b}) = x(\vec{b} \cdot \vec{a}) + y(\vec{b} \cdot \vec{b}) = 0 \).
3. The necessary dot products are calculated as follows:
   1. \( \vec{b} \cdot \vec{a} = (3\hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) = 3 + 2 - 3 = 2 \).
   2. \( \vec{b} \cdot \vec{b} = 3^2 + 1^2 + (-1)^2 = 9 + 1 + 1 = 11 \).
4. From the previous steps, \( y \) can be solved in terms of \( x \): \( y = -\frac{2}{11}x \).
5. Applying the given condition \( \vec{a} \cdot \vec{c} = 5 \): \( \vec{a} \cdot \vec{c} = (x\vec{a} + y\vec{b}) \cdot \vec{a} = x(\vec{a} \cdot \vec{a}) + y(\vec{b} \cdot \vec{a}) = 5 \).
6. The dot product \( \vec{a} \cdot \vec{a} \) is calculated: \( \vec{a} \cdot \vec{a} = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14 \). Substituting this and the value of \( \vec{b} \cdot \vec{a} \) into the equation yields \( 14x + 2y = 5 \).
7. Substituting \( y = -\frac{2}{11}x \) into \( 14x + 2y = 5 \) results in: \( 14x + 2(-\frac{2}{11}x) = 5 \), which simplifies to \( 14x - \frac{4}{11}x = 5 \). Further simplification gives \( \frac{150x}{11} = 5 \), leading to \( x = \frac{11}{30} \).
8. The value of \( y \) is then calculated: \( y = -\frac{2}{11} \cdot \frac{11}{30} = -\frac{1}{15} \).
9. The vector \( \vec{c} \) is therefore represented as \( \vec{c} = \frac{11}{30} \vec{a} - \frac{1}{15} \vec{b} \).
10. The magnitude of \( \vec{c} \) is computed as follows: \( |\vec{c}| = \sqrt{\left(\frac{11}{30}\right)^2(14) + \left(-\frac{1}{15}\right)^2(11)} = \sqrt{\frac{121}{900} \cdot 14 + \frac{1}{225} \cdot 11} = \sqrt{\frac{1694}{900} + \frac{11}{225}} = \sqrt{\frac{1694 + 44}{900}} = \sqrt{\frac{1738}{900}} = \sqrt{\frac{869}{450}} = \sqrt{\frac{11}{6}} \).
11. Consequently, the magnitude of \( \vec{c} \) is \( \sqrt{\frac{11}{6}} \), confirming the provided correct answer.