A vector `vecc` of magnitude `sqrt(7)` which is perpendicular to the vector `veca=2hatj-hatk` and `vecb=-hati+2hatj-3hatk` and makes an obtuse angle with the y-axis is given by

Options

A. (a)`((-4)/(sqrt(3)),(-1)/(sqrt(3)),(-2)/(sqrt(3)))`
B. (b)`((-4)/(sqrt(3)),(-1)/(sqrt(3)),(2)/(sqrt(3)))`
C. (c)`((4)/(sqrt(3)),(-1)/(sqrt(3)),(2)/(sqrt(3)))`
D. (d)Both (2) & (3)

Answer

To solve the problem, we need to find a vector \( \vec{c} \) of magnitude \( \sqrt{7} \) that is perpendicular to the vectors \( \vec{a} = 2\hat{j} - \hat{k} \) and \( \vec{b} = -\hat{i} + 2\hat{j} - 3\hat{k} \), and makes an obtuse angle with the y-axis.

### Step 1: Find the cross product of \( \vec{a} \) and \( \vec{b} \)

The vector \( \vec{c} \) must be perpendicular to both \( \vec{a} \) and \( \vec{b} \). We can find a vector that is perpendicular to both by calculating the cross product \( \vec{a} \times \vec{b} \).

\[ \vec{a} = 0\hat{i} + 2\hat{j} - 1\hat{k} \]
\[ \vec{b} = -1\hat{i} + 2\hat{j} - 3\hat{k} \]

The cross product \( \vec{a} \times \vec{b} \) can be calculated using the determinant of a matrix:

\[ \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & -1 \\ -1 & 2 & -3 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{c} = \hat{i} \begin{vmatrix} 2 & -1 \\ 2 & -3 \end{vmatrix} - \hat{j} \begin{vmatrix} 0 & -1 \\ -1 & -3 \end{vmatrix} + \hat{k} \begin{vmatrix} 0 & 2 \\ -1 & 2 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 2 & -1 \\ 2 & -3 \end{vmatrix} = (2)(-3) - (2)(-1) = -6 + 2 = -4 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 0 & -1 \\ -1 & -3 \end{vmatrix} = (0)(-3) - (-1)(-1) = 0 - 1 = -1 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 0 & 2 \\ -1 & 2 \end{vmatrix} = (0)(2) - (2)(-1) = 0 + 2 = 2 \]

Putting it all together:

\[ \vec{c} = -4\hat{i} + 1\hat{j} + 2\hat{k} \]

### Step 2: Normalize \( \vec{c} \) to find the direction

Now we need to find the unit vector in the direction of \( \vec{c} \):

\[ |\vec{c}| = \sqrt{(-4)^2 + 1^2 + 2^2} = \sqrt{16 + 1 + 4} = \sqrt{21} \]

The unit vector \( \hat{c} \) in the direction of \( \vec{c} \) is:

\[ \hat{c} = \frac{-4\hat{i} + 1\hat{j} + 2\hat{k}}{\sqrt{21}} \]

### Step 3: Scale \( \hat{c} \) to have a magnitude of \( \sqrt{7} \)

To get \( \vec{c} \) with a magnitude of \( \sqrt{7} \), we scale the unit vector:

\[ \vec{c} = \sqrt{7} \cdot \hat{c} = \sqrt{7} \cdot \frac{-4\hat{i} + 1\hat{j} + 2\hat{k}}{\sqrt{21}} = \frac{-4\sqrt{7}}{\sqrt{21}}\hat{i} + \frac{\sqrt{7}}{\sqrt{21}}\hat{j} + \frac{2\sqrt{7}}{\sqrt{21}}\hat{k} \]

### Step 4: Check the angle with the y-axis

To ensure that \( \vec{c} \) makes an obtuse angle with the y-axis, we check the dot product with \( \hat{j} \):

\[ \vec{c} \cdot \hat{j} = \frac{\sqrt{7}}{\sqrt{21}} \]

For an obtuse angle, this dot product must be negative. Since \( \frac{\sqrt{7}}{\sqrt{21}} > 0 \), we need to take the negative of \( \vec{c} \):

\[ \vec{c} = \frac{4\sqrt{7}}{\sqrt{21}}\hat{i} - \frac{\sqrt{7}}{\sqrt{21}}\hat{j} - \frac{2\sqrt{7}}{\sqrt{21}}\hat{k} \]

### Final Answer

Thus, the vector \( \vec{c} \) that meets all the conditions is:

\[ \vec{c} = \frac{4\sqrt{7}}{\sqrt{21}}\hat{i} - \frac{\sqrt{7}}{\sqrt{21}}\hat{j} - \frac{2\sqrt{7}}{\sqrt{21}}\hat{k} \]