Question

A unit vector $\vec{b}$ is coplanar with $\hat{i} + \hat{j} + 2\hat{k}$ and $\hat{i} + 2\hat{j} + \hat{k}$ and is perpendicular to $\hat{i} + \hat{j} + \hat{k}$. Then $\vec{b} \cdot \hat{i}$ equals

• A. $0$
• B. $1$
• C. $\frac{3}{2}$
• D. $2$
• E. $4$

Hint:

Perpendicular condition simplifies equations dramatically—use it early.

Answer 1

The Correct Option is A

Concept:
• Coplanar vectors: $\vec{b} = \lambda \vec{u} + \mu \vec{v}$
• Perpendicular: $\vec{b} \cdot \vec{n} = 0$
• Unit vector: magnitude = 1

Step 1: Write vectors.
\[ \vec{u} = (1,1,2), \vec{v} = (1,2,1) \] \[ \vec{b} = \lambda(1,1,2) + \mu(1,2,1) \] \[ = (\lambda+\mu, \lambda+2\mu, 2\lambda+\mu) \]

Step 2: Use perpendicular condition.
\[ \vec{n} = (1,1,1) \] \[ (\lambda+\mu) + (\lambda+2\mu) + (2\lambda+\mu) = 0 \] \[ 4\lambda + 4\mu = 0 \Rightarrow \lambda + \mu = 0 \]

Step 3: Substitute.
\[ \mu = -\lambda \] \[ \vec{b} = (0, -\lambda, \lambda) \]

Step 4: Unit vector condition.
\[ |\vec{b}|^2 = \lambda^2 + \lambda^2 = 2\lambda^2 = 1 \] \[ \lambda = \frac{1}{\sqrt{2}} \]

Step 5: Compute $\vec{b\cdot \hat{i}$.}
\[ \vec{b} = (0, -\tfrac{1}{\sqrt{2}}, \tfrac{1}{\sqrt{2}}) \] \[ \vec{b}\cdot \hat{i} = 0 \] \[ \boxed{0} \]