Question

The unit vector \( (ai + bj) \) is perpendicular to \( (i + j) \). The value of 'b' is

• A. \( +\frac{1}{\sqrt{3}} \)
• B. \( -\frac{1}{\sqrt{3}} \)
• C. \( +\frac{1}{2} \)
• D. \( -\frac{1}{\sqrt{2}} \)

Hint:

When two vectors are perpendicular, their dot product is zero. For unit vectors, ensure that the sum of the squares of the components equals 1.

Answer 1

The Correct Option is D

Step 1: Understanding the condition.
We are told that the unit vector \( (ai + bj) \) is perpendicular to \( (i + j) \). The condition for perpendicular vectors is that their dot product is zero.

Step 2: Dot product calculation.
The dot product of \( (ai + bj) \) and \( (i + j) \) is given by: \[ (ai + bj) \cdot (i + j) = a \cdot 1 + b \cdot 1 = a + b \] For the vectors to be perpendicular, we must have \( a + b = 0 \). Thus, \( b = -a \).

Step 3: Normalizing the vector.
Since the vector \( (ai + bj) \) is a unit vector, we have: \[ a^2 + b^2 = 1 \] Substituting \( b = -a \), we get: \[ a^2 + (-a)^2 = 1 \quad \Rightarrow \quad 2a^2 = 1 \quad \Rightarrow \quad a = \pm \frac{1}{\sqrt{2}} \] Thus, \( b = \mp \frac{1}{\sqrt{2}} \). The correct answer is \( b = -\frac{1}{\sqrt{2}} \).