Let `a =i-j,b=j-k,c=k-i`. If d is a unit vector such that` a.d =0 = [vecb,vecc,vecd]` then d equals

Options

A. `pm(k+j-2k)//sqrt(6)`
B. `pm(i+j-k)//sqrt(3)`
C. `pm(i+j+k)//sqrt(3)`
D. `pmk`

Answer

To solve the problem step by step, we will follow the given conditions and properties of vectors.

### Step 1: Define the vectors
We have the following vectors:
- \( \vec{a} = \hat{i} - \hat{j} \)
- \( \vec{b} = \hat{j} - \hat{k} \)
- \( \vec{c} = \hat{k} - \hat{i} \)

### Step 2: Understand the conditions
We are given that:
1. \( \vec{a} \cdot \vec{d} = 0 \) (which means \( \vec{d} \) is perpendicular to \( \vec{a} \))
2. The scalar triple product \( [\vec{b}, \vec{c}, \vec{d}] = 0 \) (which means \( \vec{b}, \vec{c}, \vec{d} \) are coplanar)

### Step 3: Express \( \vec{d} \) in terms of \( \vec{b} \) and \( \vec{c} \)
Since \( \vec{b}, \vec{c}, \vec{d} \) are coplanar, we can express \( \vec{d} \) as:
\[ \vec{d} = \vec{b} + \lambda \vec{c} \]
where \( \lambda \) is a scalar.

### Step 4: Substitute the vectors
Substituting the expressions for \( \vec{b} \) and \( \vec{c} \):
\[ \vec{d} = (\hat{j} - \hat{k}) + \lambda (\hat{k} - \hat{i}) \]
This simplifies to:
\[ \vec{d} = -\lambda \hat{i} + \hat{j} + (\lambda - 1) \hat{k} \]

### Step 5: Use the condition \( \vec{a} \cdot \vec{d} = 0 \)
Now we compute the dot product:
\[ \vec{a} \cdot \vec{d} = (\hat{i} - \hat{j}) \cdot (-\lambda \hat{i} + \hat{j} + (\lambda - 1) \hat{k}) \]
Calculating this gives:
\[ -\lambda (\hat{i} \cdot \hat{i}) + (-1)(\hat{j} \cdot \hat{j}) + 0 = -\lambda - 1 \]
Setting this equal to zero:
\[ -\lambda - 1 = 0 \]
Thus, we find:
\[ \lambda = -1 \]

### Step 6: Substitute \( \lambda \) back into \( \vec{d} \)
Substituting \( \lambda = -1 \) into the expression for \( \vec{d} \):
\[ \vec{d} = -(-1) \hat{i} + \hat{j} + (-1 - 1) \hat{k} = \hat{i} + \hat{j} - 2\hat{k} \]

### Step 7: Find the unit vector
To find the unit vector \( \hat{d} \):
\[ \hat{d} = \frac{\vec{d}}{|\vec{d}|} \]
First, we calculate the magnitude of \( \vec{d} \):
\[ |\vec{d}| = \sqrt{(1)^2 + (1)^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6} \]
Thus, the unit vector is:
\[ \hat{d} = \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} - 2\hat{k}) \]

### Final Answer
The unit vector \( \hat{d} \) is:
\[ \hat{d} = \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} - 2\hat{k}) \]

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