The vector r satisfying the conditions that I. it is perrpendicular to `3hati+2hatj+2hatk and 18hati-22hatj-5hatk` II. It makes an obtuse angle with Y-axis III. `|r|=14`.

Options

A. `2(-2hati-3hatj+6hatk)`
B. `2(2hati-3hatj+6hatk)`
C. `4hati+6hatj-12hatk`
D. none of these

Answer

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

1. It is perpendicular to the vectors \( \mathbf{a} = 3\hat{i} + 2\hat{j} + 2\hat{k} \) and \( \mathbf{b} = 18\hat{i} - 22\hat{j} - 5\hat{k} \).
2. It makes an obtuse angle with the Y-axis.
3. The magnitude of \( \mathbf{r} \) is \( | \mathbf{r} | = 14 \).

### Step 1: Find the Cross Product of the Two Vectors

Since \( \mathbf{r} \) is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \), we can find \( \mathbf{r} \) using the cross product \( \mathbf{r} = \mathbf{a} \times \mathbf{b} \).

\[ \mathbf{r} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & 2 \\ 18 & -22 & -5 \end{vmatrix} \]

Calculating the determinant:

\[ \mathbf{r} = \hat{i} \begin{vmatrix} 2 & 2 \\ -22 & -5 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 2 \\ 18 & -5 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & 2 \\ 18 & -22 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 2 & 2 \\ -22 & -5 \end{vmatrix} = (2)(-5) - (2)(-22) = -10 + 44 = 34 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 3 & 2 \\ 18 & -5 \end{vmatrix} = (3)(-5) - (2)(18) = -15 - 36 = -51 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 3 & 2 \\ 18 & -22 \end{vmatrix} = (3)(-22) - (2)(18) = -66 - 36 = -102 \]

Putting it all together:

\[ \mathbf{r} = 34\hat{i} + 51\hat{j} - 102\hat{k} \]

### Step 2: Express \( \mathbf{r} \) in Terms of a Scalar \( \lambda \)

Since \( \mathbf{r} \) can be expressed as a scalar multiple of the vector we found, we write:

\[ \mathbf{r} = \lambda (34\hat{i} + 51\hat{j} - 102\hat{k}) \]

### Step 3: Use the Magnitude Condition

We know that \( |\mathbf{r}| = 14 \):

\[ |\mathbf{r}| = |\lambda| \sqrt{34^2 + 51^2 + (-102)^2} = 14 \]

Calculating \( \sqrt{34^2 + 51^2 + (-102)^2} \):

\[ 34^2 = 1156, \quad 51^2 = 2601, \quad (-102)^2 = 10404 \]

\[ 34^2 + 51^2 + (-102)^2 = 1156 + 2601 + 10404 = 14261 \]

Thus,

\[ |\lambda| \sqrt{14261} = 14 \]

\[ |\lambda| = \frac{14}{\sqrt{14261}} \]

### Step 4: Determine the Sign of \( \lambda \)

Since \( \mathbf{r} \) must make an obtuse angle with the Y-axis, we need the \( j \)-component of \( \mathbf{r} \) to be negative. The \( j \)-component of \( \mathbf{r} \) is \( \lambda \cdot 51 \). Therefore, we need:

\[ \lambda < 0 \]

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

From the magnitude condition:

\[ |\lambda| = \frac{14}{\sqrt{14261}} \implies \lambda = -\frac{14}{\sqrt{14261}} \]

### Step 6: Substitute \( \lambda \) Back into \( \mathbf{r} \)

Now substitute \( \lambda \) back into the expression for \( \mathbf{r} \):

\[ \mathbf{r} = -\frac{14}{\sqrt{14261}} (34\hat{i} + 51\hat{j} - 102\hat{k}) \]

### Final Result

Thus, the vector \( \mathbf{r} \) is:

\[ \mathbf{r} = -\frac{476}{\sqrt{14261}} \hat{i} - \frac{714}{\sqrt{14261}} \hat{j} + \frac{1428}{\sqrt{14261}} \hat{k} \]