If ` a=-i+j+k and b =2i+0j+k`, then the vector c satisfying the conditions (i) that it is coplanar with a and b (ii) that it is `bot` to b and (iii) that `a.c = 7`, is

Options

A. `-(3)/(2) i+(5)/(2)j+3k`
B. `-3i+5j+6k`
C. `-6i+0j+k`
D. `-i+2j+2k`

Answer

To solve the problem step by step, we need to find the vector \( \mathbf{c} \) that satisfies the following conditions:

1. It is coplanar with vectors \( \mathbf{a} \) and \( \mathbf{b} \).
2. It is perpendicular to vector \( \mathbf{b} \).
3. The dot product \( \mathbf{a} \cdot \mathbf{c} = 7 \).

Given:
\[ \mathbf{a} = -\mathbf{i} + \mathbf{j} + \mathbf{k} \]
\[ \mathbf{b} = 2\mathbf{i} + 0\mathbf{j} + \mathbf{k} \]

### Step 1: Express \( \mathbf{c} \) in terms of \( \mathbf{a} \) and \( \mathbf{b} \)

Since \( \mathbf{c} \) is coplanar with \( \mathbf{a} \) and \( \mathbf{b} \), we can express \( \mathbf{c} \) as a linear combination of \( \mathbf{a} \) and \( \mathbf{b} \):
\[ \mathbf{c} = \lambda \mathbf{a} + \mu \mathbf{b} \]
Substituting the values of \( \mathbf{a} \) and \( \mathbf{b} \):
\[ \mathbf{c} = \lambda (-\mathbf{i} + \mathbf{j} + \mathbf{k}) + \mu (2\mathbf{i} + 0\mathbf{j} + \mathbf{k}) \]
\[ \mathbf{c} = (-\lambda + 2\mu)\mathbf{i} + \lambda\mathbf{j} + (\lambda + \mu)\mathbf{k} \]

### Step 2: Apply the condition that \( \mathbf{c} \) is perpendicular to \( \mathbf{b} \)

For \( \mathbf{c} \) to be perpendicular to \( \mathbf{b} \), the dot product \( \mathbf{b} \cdot \mathbf{c} \) must equal zero:
\[ \mathbf{b} \cdot \mathbf{c} = (2\mathbf{i} + 0\mathbf{j} + \mathbf{k}) \cdot ((-\lambda + 2\mu)\mathbf{i} + \lambda\mathbf{j} + (\lambda + \mu)\mathbf{k}) = 0 \]
Calculating the dot product:
\[ 2(-\lambda + 2\mu) + 0\cdot\lambda + 1(\lambda + \mu) = 0 \]
\[ -2\lambda + 4\mu + \lambda + \mu = 0 \]
\[ -2\lambda + 5\mu = 0 \quad \Rightarrow \quad 2\lambda = 5\mu \quad \Rightarrow \quad \lambda = \frac{5}{2}\mu \]

### Step 3: Apply the condition \( \mathbf{a} \cdot \mathbf{c} = 7 \)

Now, we need to compute the dot product \( \mathbf{a} \cdot \mathbf{c} \):
\[ \mathbf{a} \cdot \mathbf{c} = (-\mathbf{i} + \mathbf{j} + \mathbf{k}) \cdot ((-\lambda + 2\mu)\mathbf{i} + \lambda\mathbf{j} + (\lambda + \mu)\mathbf{k}) \]
Calculating the dot product:
\[ -1(-\lambda + 2\mu) + 1\lambda + 1(\lambda + \mu) = 7 \]
\[ \lambda - 2\mu + \lambda + \lambda + \mu = 7 \]
\[ 3\lambda - \mu = 7 \]

### Step 4: Substitute \( \lambda \) in terms of \( \mu \)

Substituting \( \lambda = \frac{5}{2}\mu \) into the equation:
\[ 3\left(\frac{5}{2}\mu\right) - \mu = 7 \]
\[ \frac{15}{2}\mu - \mu = 7 \]
\[ \frac{15}{2}\mu - \frac{2}{2}\mu = 7 \]
\[ \frac{13}{2}\mu = 7 \quad \Rightarrow \quad \mu = \frac{14}{13} \]

### Step 5: Find \( \lambda \)

Now substituting \( \mu \) back to find \( \lambda \):
\[ \lambda = \frac{5}{2} \cdot \frac{14}{13} = \frac{70}{26} = \frac{35}{13} \]

### Step 6: Substitute \( \lambda \) and \( \mu \) back into \( \mathbf{c} \)

Now substitute \( \lambda \) and \( \mu \) into the expression for \( \mathbf{c} \):
\[ \mathbf{c} = \left(-\frac{35}{13} + 2 \cdot \frac{14}{13}\right)\mathbf{i} + \frac{35}{13}\mathbf{j} + \left(\frac{35}{13} + \frac{14}{13}\right)\mathbf{k} \]
Calculating each component:
\[ \mathbf{c} = \left(-\frac{35}{13} + \frac{28}{13}\right)\mathbf{i} + \frac{35}{13}\mathbf{j} + \left(\frac{49}{13}\right)\mathbf{k} \]
\[ \mathbf{c} = \left(-\frac{7}{13}\right)\mathbf{i} + \left(\frac{35}{13}\right)\mathbf{j} + \left(\frac{49}{13}\right)\mathbf{k} \]

### Final Result

Thus, the vector \( \mathbf{c} \) is:
\[ \mathbf{c} = -\frac{7}{13}\mathbf{i} + \frac{35}{13}\mathbf{j} + \frac{49}{13}\mathbf{k} \]