Question

Calculate the values of \(a\) and \(b\) if vectors \(a\hat{i}+b\hat{j} = \hat{n}\) and \((\hat{i}+\hat{j})\) are perpendicular to each other

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

Hint:

Perpendicular means zero dot product; unit vector means unit magnitude.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
\(\hat n = a\hat i + b\hat j\) is a unit vector (the hat shows it) perpendicular to \(\hat i + \hat j\).

Step 2: Two conditions:
Perpendicular: \((a\hat i + b\hat j)\cdot(\hat i+\hat j) = a + b = 0\), so \(b = -a\).
Unit magnitude: \(a^2 + b^2 = 1\), so \(2a^2 = 1\) and \(a = \pm\frac{1}{\sqrt2}\).

Step 3: Pick the option:
\(a = \frac{1}{\sqrt2}\), \(b = -\frac{1}{\sqrt2}\), which is option D.
Option A: \(a+b = 1\ne0\). Option B: the zero vector is not a unit vector. Option C: \(a + b\ne0\), since \(\sqrt2 - \frac{1}{\sqrt2}\ne0\).

Final Answer:
The values are \(a=\frac{1}{\sqrt2}\) and \(b=-\frac{1}{\sqrt2}\), option (D). \[ \boxed{a=\frac{1}{\sqrt2},\ b=-\frac{1}{\sqrt2}} \]