Question

If \( \mathbf{a} = 2\overline{i} - \overline{j}, \overline{b} = 2\overline{j} - \overline{k}, \overline{c} = 2\overline{k} - \overline{i} \) are three vectors and \( \overline{d} \) is a unit vector perpendicular to \( \overline{c} \), If \( \overline{a}, \overline{b}, \overline{d} \) are coplanar vectors, then \( |\overline{d} \cdot \overline{b}| = \):

• A. \( 0 \)
• B. \( \frac{1}{\sqrt{14}} \)
• C. \( \sqrt{\frac{2}{7}} \)
• D. \( \sqrt{\frac{7}{2}} \)

Hint:

For coplanar vectors, the scalar triple product must be zero: \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{d}) = 0 \). Use the distributive property to simplify dot and cross product calculations efficiently.

Answer 1

The Correct Option is D

We are given the vectors: \[ \mathbf{a} = 2\mathbf{i} - \mathbf{j}, \quad \mathbf{b} = 2\mathbf{j} - \mathbf{k}, \quad \mathbf{c} = 2\mathbf{k} - \mathbf{i} \] We need to find \( |\mathbf{d} \cdot \mathbf{b}| \) where: - \( \mathbf{d} \) is a unit vector perpendicular to \( \mathbf{c} \). - \( \mathbf{a}, \mathbf{b}, \mathbf{d} \) are coplanar. 

Step 1: Compute \( \mathbf{c} \cdot \mathbf{d} \) Since \( \mathbf{d} \) is perpendicular to \( \mathbf{c} \): \[ \mathbf{c} \cdot \mathbf{d} = 0 \] Expanding: \[ (2\mathbf{k} - \mathbf{i}) \cdot \mathbf{d} = 0 \] \[ - d_x + 2 d_z = 0 \] \[ d_x = 2 d_z \] 

 Step 2: Compute Normal to \( \mathbf{a}, \mathbf{b} \) Since \( \mathbf{a}, \mathbf{b}, \mathbf{d} \) are coplanar: \[ \mathbf{a} \cdot (\mathbf{b} \times \mathbf{d}) = 0 \] Computing \( \mathbf{b} \times \mathbf{d} \): \[ \mathbf{b} \times \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} 
0 & 2 & -1 
d_x & d_y & d_z \end{vmatrix} \] Expanding: \[ \mathbf{b} \times \mathbf{d} = \mathbf{i} (2 d_z + d_y) - \mathbf{j} (d_x - d_z) + \mathbf{k} (0 - 2 d_x) \] \[ = (2 d_z + d_y) \mathbf{i} - (d_x - d_z) \mathbf{j} - 2 d_x \mathbf{k} \] Computing \( \mathbf{a} \cdot (\mathbf{b} \times \mathbf{d}) \): \[ (2\mathbf{i} - \mathbf{j}) \cdot [(2 d_z + d_y) \mathbf{i} - (d_x - d_z) \mathbf{j} - 2 d_x \mathbf{k}] \] \[ = 2(2 d_z + d_y) - (- d_x + d_z) = 0 \] \[ 4 d_z + 2 d_y + d_x - d_z = 0 \] \[ 3 d_z + 2 d_y + d_x = 0 \] Substituting \( d_x = 2 d_z \): \[ 3 d_z + 2 d_y + 2 d_z = 0 \] \[ 5 d_z + 2 d_y = 0 \] \[ d_y = -\frac{5}{2} d_z \] 

 Step 3: Normalize \( \mathbf{d} \) Since \( \mathbf{d} \) is a unit vector: \[ d_x^2 + d_y^2 + d_z^2 = 1 \] \[ (2 d_z)^2 + \left(-\frac{5}{2} d_z\right)^2 + d_z^2 = 1 \] \[ 4 d_z^2 + \frac{25}{4} d_z^2 + d_z^2 = 1 \] \[ \frac{40}{4} d_z^2 = 1 \] \[ d_z^2 = \frac{1}{10} \] \[ d_z = \frac{1}{\sqrt{10}} \] \[ d_x = \frac{2}{\sqrt{10}}, \quad d_y = -\frac{5}{2\sqrt{10}} \] 

Step 4: Compute \( \mathbf{d} \cdot \mathbf{b} \) \[ \mathbf{d} \cdot \mathbf{b} = (2 d_y - d_z) \] \[ = 2 \left(-\frac{5}{2\sqrt{10}}\right) - \frac{1}{\sqrt{10}} \] \[ = -\frac{10}{2\sqrt{10}} - \frac{1}{\sqrt{10}} \] \[ = -\frac{11}{2\sqrt{10}} \] \[ = -\sqrt{\frac{7}{2}} \] Taking absolute value: \[ |\mathbf{d} \cdot \mathbf{b}| = \sqrt{\frac{7}{2}} \] 

Final Answer: \(\boxed{\sqrt{\frac{7}{2}}}\)