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

Step 1: Since \( \vec{c} \) is coplanar with \( \vec{a} \) and \( \vec{b} \), we can express \( \vec{c} \) as a linear combination of \( \vec{a} \) and \( \vec{b} \): \[ \vec{c} = \lambda \vec{a} + \mu \vec{b} \] where \( \lambda \) and \( \mu \) are constants. 
Step 2: Since \( \vec{c} \) is perpendicular to \( \vec{b} \), we use the dot product: \[ \vec{c} \cdot \vec{b} = 0 \] Substitute \( \vec{c} = \lambda \vec{a} + \mu \vec{b} \) into the equation: \[ (\lambda \vec{a} + \mu \vec{b}) \cdot \vec{b} = 0 \] This simplifies to: \[ \lambda \vec{a} \cdot \vec{b} + \mu \vec{b} \cdot \vec{b} = 0 \] Now calculate the dot products: \[ \vec{a} \cdot \vec{b} = (1)(3) + (2)(1) + (3)(-1) = 3 + 2 - 3 = 2 \] \[ \vec{b} \cdot \vec{b} = (3)^2 + (1)^2 + (-1)^2 = 9 + 1 + 1 = 11 \] Substitute these values into the equation: \[ \lambda (2) + \mu (11) = 0 \quad \Rightarrow \quad 2\lambda + 11\mu = 0 \quad \cdots (1) \] 
Step 3: We are also given that \( \vec{a} \cdot \vec{c} = 5 \), so: \[ \vec{a} \cdot (\lambda \vec{a} + \mu \vec{b}) = 5 \] This expands to: \[ \lambda (\vec{a} \cdot \vec{a}) + \mu (\vec{a} \cdot \vec{b}) = 5 \] We know \( \vec{a} \cdot \vec{a} = 1^2 + 2^2 + 3^2 = 14 \), and from earlier \( \vec{a} \cdot \vec{b} = 2 \), so: \[ \lambda (14) + \mu (2) = 5 \quad \Rightarrow \quad 14\lambda + 2\mu = 5 \quad \cdots (2) \] 
Step 4: Now, solve the system of equations (1) and (2): From (1), we have: \[ 2\lambda = -11\mu \quad \Rightarrow \quad \lambda = -\frac{11\mu}{2} \] Substitute this into (2): \[ 14\left(-\frac{11\mu}{2}\right) + 2\mu = 5 \] Simplifying: \[ -77\mu + 2\mu = 5 \quad \Rightarrow \quad -75\mu = 5 \quad \Rightarrow \quad \mu = -\frac{1}{15} \] Substitute \( \mu = -\frac{1}{15} \) into the expression for \( \lambda \): \[ \lambda = -\frac{11(-\frac{1}{15})}{2} = \frac{11}{30} \] 
Step 5: Now, calculate the magnitude of \( \vec{c} \): \[ |\vec{c}|^2 = \lambda^2 |\vec{a}|^2 + \mu^2 |\vec{b}|^2 + 2\lambda\mu (\vec{a} \cdot \vec{b}) \] Substitute \( \lambda = \frac{11}{30} \), \( \mu = -\frac{1}{15} \), \( |\vec{a}|^2 = 14 \), \( |\vec{b}|^2 = 11 \), and \( \vec{a} \cdot \vec{b} = 2 \): \[ |\vec{c}|^2 = \left(\frac{11}{30}\right)^2(14) + \left(-\frac{1}{15}\right)^2(11) + 2\left(\frac{11}{30}\right)\left(-\frac{1}{15}\right)(2) \] Simplifying: \[ |\vec{c}|^2 = \frac{121}{900} \times 14 + \frac{1}{225} \times 11 + \frac{-22}{450} \] \[ |\vec{c}|^2 = \frac{1694}{900} + \frac{11}{225} - \frac{22}{450} \] After simplifying, we find: \[ |\vec{c}|^2 = \frac{11}{6} \] Thus, \[ |\vec{c}| = \sqrt{\frac{11}{6}} \]