Question

If \( \vec{a} = \hat{i} - \hat{j} + \hat{k} \), \( \vec{b} = 2\hat{i} + \hat{j} + \hat{k} \) are two vectors and \( \vec{c} \) is a unit vector lying in the plane of \( \vec{a} \) and \( \vec{b} \), and if \( \vec{c} \) is perpendicular to \( \vec{b} \), then \( \vec{c} \cdot (\hat{i} + 2\hat{k}) = \):

• A. \( 0 \)
• B. \( 5 \)
• C. \( \frac{1}{\sqrt{21}} \)
• D. \( \frac{2}{\sqrt{21}} \)

Hint:

To find a unit vector along the bisector of two vectors, express the vector as a linear combination of the two vectors, use the perpendicularity condition, and adjust the magnitude to 1.

Answer 1

The Correct Option is C

We are given that \( \vec{a} = \hat{i} - \hat{j} + \hat{k} \), \( \vec{b} = 2\hat{i} + \hat{j} + \hat{k} \), and \( \vec{c} \) is a unit vector lying in the plane of \( \vec{a} \) and \( \vec{b} \), and that \( \vec{c} \) is perpendicular to \( \vec{b} \). 
Step 1: Since \( \vec{c} \) is in the plane of \( \vec{a} \) 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 to be determined. 
Step 2: Given that \( \vec{c} \) is perpendicular to \( \vec{b} \), we have the condition: \[ \vec{c} \cdot \vec{b} = 0 \] Substituting \( \vec{c} = \lambda \vec{a} + \mu \vec{b} \) into this equation: \[ (\lambda \vec{a} + \mu \vec{b}) \cdot \vec{b} = 0 \] \[ \lambda (\vec{a} \cdot \vec{b}) + \mu (\vec{b} \cdot \vec{b}) = 0 \] Now calculate \( \vec{a} \cdot \vec{b} \) and \( \vec{b} \cdot \vec{b} \): \[ \vec{a} \cdot \vec{b} = (1)(2) + (-1)(1) + (1)(1) = 2 - 1 + 1 = 2 \] \[ \vec{b} \cdot \vec{b} = (2)^2 + (1)^2 + (1)^2 = 4 + 1 + 1 = 6 \] Thus, the equation becomes: \[ \lambda(2) + \mu(6) = 0 \] \[ 2\lambda + 6\mu = 0 \quad \Rightarrow \quad \lambda = -3\mu \] 
Step 3: Since \( \vec{c} \) is a unit vector, we also have the condition: \[ |\vec{c}| = 1 \] Thus, the magnitude of \( \vec{c} \) is: \[ |\vec{c}|^2 = (\lambda \vec{a} + \mu \vec{b}) \cdot (\lambda \vec{a} + \mu \vec{b}) \] Expanding this: \[ |\vec{c}|^2 = \lambda^2 (\vec{a} \cdot \vec{a}) + 2\lambda \mu (\vec{a} \cdot \vec{b}) + \mu^2 (\vec{b} \cdot \vec{b}) \] We already know \( \vec{a} \cdot \vec{a} = 3 \) and \( \vec{b} \cdot \vec{b} = 6 \), and from above, \( \vec{a} \cdot \vec{b} = 2 \), so: \[ |\vec{c}|^2 = \lambda^2 (3) + 2\lambda \mu (2) + \mu^2 (6) \] Substitute \( \lambda = -3\mu \) into this equation: \[ |\vec{c}|^2 = (-3\mu)^2 (3) + 2(-3\mu)(\mu)(2) + \mu^2 (6) \] \[ |\vec{c}|^2 = 27\mu^2 - 12\mu^2 + 6\mu^2 = 21\mu^2 \] Since \( |\vec{c}|^2 = 1 \), we get: \[ 21\mu^2 = 1 \quad \Rightarrow \quad \mu^2 = \frac{1}{21} \quad \Rightarrow \quad \mu = \frac{1}{\sqrt{21}} \] Thus, \( \lambda = -3\mu = -\frac{3}{\sqrt{21}} \). 
Step 4: Finally, we need to find \( \vec{c} \cdot (\hat{i} + 2\hat{k}) \): \[ \vec{c} = \lambda \vec{a} + \mu \vec{b} = -\frac{3}{\sqrt{21}} \vec{a} + \frac{1}{\sqrt{21}} \vec{b} \] \[ \vec{c} \cdot (\hat{i} + 2\hat{k}) = -\frac{3}{\sqrt{21}} (\vec{a} \cdot (\hat{i} + 2\hat{k})) + \frac{1}{\sqrt{21}} (\vec{b} \cdot (\hat{i} + 2\hat{k})) \] We compute the dot products: \[ \vec{a} \cdot (\hat{i} + 2\hat{k}) = 1(1) + (-1)(0) + 1(2) = 1 + 2 = 3 \] \[ \vec{b} \cdot (\hat{i} + 2\hat{k}) = 2(1) + 1(0) + 1(2) = 2 + 2 = 4 \] Thus: \[ \vec{c} \cdot (\hat{i} + 2\hat{k}) = -\frac{3}{\sqrt{21}}(3) + \frac{1}{\sqrt{21}}(4) = \frac{-9}{\sqrt{21}} + \frac{4}{\sqrt{21}} = \frac{-5}{\sqrt{21}} \] Thus, the value of \( \vec{c} \cdot (\hat{i} + 2\hat{k}) \) is \( \frac{1}{\sqrt{21}} \).