If a,b,c be non-zero vectors such that a is perpendicular to b and c and `|a|=1,|b|=2,|c|=1,b*c=1` and there is a non-zero vector d coplanar with a+b and 2b-c and `d*a=1`, then minimum value of |d| is

Options

A. `(2)/(sqrt(13))`
B. `(3)/(sqrt(13))`
C. `(4)/(sqrt(5))`
D. `(4)/(sqrt(13))`.

Answer

To find the minimum value of the vector \( |d| \), we start with the given conditions and work through the problem step by step.

### Step 1: Understand the Given Information
We have three vectors \( a, b, c \) such that:
- \( a \) is perpendicular to both \( b \) and \( c \) (i.e., \( a \cdot b = 0 \) and \( a \cdot c = 0 \)).
- The magnitudes are given as \( |a| = 1 \), \( |b| = 2 \), \( |c| = 1 \).
- The dot product \( b \cdot c = 1 \).
- There is a vector \( d \) that is coplanar with \( a + b \) and \( 2b - c \).
- \( d \cdot a = 1 \).

### Step 2: Establish the Coplanarity Condition
Since \( d \) is coplanar with \( a + b \) and \( 2b - c \), we can express \( d \) as a linear combination of these vectors:
\[ d = \lambda (a + b) + \mu (2b - c) \]
for some scalars \( \lambda \) and \( \mu \).

### Step 3: Substitute and Simplify
Expanding \( d \):
\[ d = \lambda a + \lambda b + 2\mu b - \mu c = \lambda a + (\lambda + 2\mu)b - \mu c \]

### Step 4: Use the Condition \( d \cdot a = 1 \)
Taking the dot product of \( d \) with \( a \):
\[ d \cdot a = \lambda (a \cdot a) + (\lambda + 2\mu)(b \cdot a) - \mu (c \cdot a) \]
Since \( a \cdot a = |a|^2 = 1 \), \( b \cdot a = 0 \), and \( c \cdot a = 0 \), we have:
\[ d \cdot a = \lambda \cdot 1 + 0 - 0 = \lambda \]
Given \( d \cdot a = 1 \), we find:
\[ \lambda = 1 \]

### Step 5: Substitute \( \lambda \) Back into \( d \)
Now substituting \( \lambda = 1 \) into the expression for \( d \):
\[ d = a + b + 2\mu b - \mu c = a + (1 + 2\mu)b - \mu c \]

### Step 6: Calculate \( |d|^2 \)
To find \( |d|^2 \):
\[ |d|^2 = (a + (1 + 2\mu)b - \mu c) \cdot (a + (1 + 2\mu)b - \mu c) \]
Expanding this:
\[ |d|^2 = |a|^2 + (1 + 2\mu)^2 |b|^2 + \mu^2 |c|^2 + 2(a \cdot (1 + 2\mu)b) - 2\mu(a \cdot c) - 2\mu(1 + 2\mu)(b \cdot c) \]
Substituting the known values:
- \( |a|^2 = 1 \)
- \( |b|^2 = 4 \)
- \( |c|^2 = 1 \)
- \( a \cdot b = 0 \)
- \( a \cdot c = 0 \)
- \( b \cdot c = 1 \)

Thus, we have:
\[ |d|^2 = 1 + (1 + 2\mu)^2 \cdot 4 + \mu^2 \cdot 1 - 2\mu(1 + 2\mu) \cdot 1 \]
\[ = 1 + 4(1 + 4\mu + 4\mu^2) + \mu^2 - 2\mu - 4\mu^2 \]
\[ = 1 + 4 + 16\mu + 16\mu^2 + \mu^2 - 2\mu - 4\mu^2 \]
\[ = 5 + 14\mu + 13\mu^2 \]

### Step 7: Minimize \( |d|^2 \)
To minimize \( |d|^2 = 13\mu^2 + 14\mu + 5 \), we can use the vertex formula for a quadratic:
\[ \mu = -\frac{b}{2a} = -\frac{14}{2 \cdot 13} = -\frac{14}{26} = -\frac{7}{13} \]

### Step 8: Substitute \( \mu \) Back to Find Minimum \( |d|^2 \)
Substituting \( \mu = -\frac{7}{13} \):
\[ |d|^2 = 13\left(-\frac{7}{13}\right)^2 + 14\left(-\frac{7}{13}\right) + 5 \]
\[ = 13 \cdot \frac{49}{169} - \frac{98}{13} + 5 \]
\[ = \frac{637}{169} - \frac{1274}{169} + \frac{845}{169} \]
\[ = \frac{637 - 1274 + 845}{169} = \frac{208}{169} \]

### Step 9: Final Calculation of \( |d| \)
Thus, the minimum value of \( |d| \) is:
\[ |d| = \sqrt{\frac{208}{169}} = \frac{\sqrt{208}}{13} = \frac{4\sqrt{13}}{13} \]

### Final Answer
The minimum value of \( |d| \) is \( \frac{4}{\sqrt{13}} \).