A unit vector is represented as `(0.8hat(i)+bhat(j)+0.4hat(k))`. Hence the value of 'b' must be

Options

A. 0.4
B. `sqrt0.6`
C. 0.2
D. `sqrt0.2`

Answer

To find the value of 'b' in the unit vector represented as \( (0.8 \hat{i} + b \hat{j} + 0.4 \hat{k}) \), we need to use the property of unit vectors. A unit vector has a magnitude of 1.

### Step-by-Step Solution:

1. **Understand the Definition of a Unit Vector**:
A unit vector \( \vec{A} \) can be expressed in the form \( \vec{A} = x \hat{i} + y \hat{j} + z \hat{k} \) where the magnitude is given by:
\[ |\vec{A}| = \sqrt{x^2 + y^2 + z^2} = 1 \]

2. **Identify Components**:
In our case, the components are:
- \( x = 0.8 \)
- \( y = b \)
- \( z = 0.4 \)

3. **Set Up the Magnitude Equation**:
According to the unit vector property, we can write:
\[ \sqrt{(0.8)^2 + b^2 + (0.4)^2} = 1 \]

4. **Square Both Sides**:
To eliminate the square root, we square both sides:
\[ (0.8)^2 + b^2 + (0.4)^2 = 1 \]

5. **Calculate the Squares**:
Calculate \( (0.8)^2 \) and \( (0.4)^2 \):
\[ (0.8)^2 = 0.64 \quad \text{and} \quad (0.4)^2 = 0.16 \]

6. **Substitute Back into the Equation**:
Substitute these values into the equation:
\[ 0.64 + b^2 + 0.16 = 1 \]

7. **Combine Like Terms**:
Combine \( 0.64 \) and \( 0.16 \):
\[ 0.8 + b^2 = 1 \]

8. **Isolate \( b^2 \)**:
Rearranging gives:
\[ b^2 = 1 - 0.8 \]
\[ b^2 = 0.2 \]

9. **Solve for \( b \)**:
Taking the square root of both sides gives:
\[ b = \sqrt{0.2} \]
Since \( b \) can be positive or negative, we have:
\[ b = \pm \sqrt{0.2} \]

10. **Final Value of \( b \)**:
The approximate value of \( \sqrt{0.2} \) is about \( 0.447 \). Therefore:
\[ b \approx \pm 0.447 \]