The sine of the angle between vectors `vec(a)=2hat(i)-6hat(j)-3hat(k) and vec(b)=4hat(i)+3hat(j)-hat(k)`

Options

A. `(1)/(sqrt(26))`
B. `(5)/(sqrt(26))`
C. `5/26`
D. `1/26`

Explanation

To find the sine of the angle between the vectors \(\vec{a} = 2\hat{i} - 6\hat{j} - 3\hat{k}\) and \(\vec{b} = 4\hat{i} + 3\hat{j} - \hat{k}\), we can use the formula:

\[ \sin \theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}| |\vec{b}|} \]

### Step 1: Calculate the cross product \(\vec{a} \times \vec{b}\)

To find the cross product, we can use the determinant of a matrix formed by the unit vectors and the components of the vectors:

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -6 & -3 \\ 4 & 3 & -1 \end{vmatrix} \]

Calculating this determinant, we have:

\[ \vec{a} \times \vec{b} = \hat{i} \begin{vmatrix} -6 & -3 \\ 3 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -3 \\ 4 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & -6 \\ 4 & 3 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \(\begin{vmatrix} -6 & -3 \\ 3 & -1 \end{vmatrix} = (-6)(-1) - (-3)(3) = 6 + 9 = 15\)

2. \(\begin{vmatrix} 2 & -3 \\ 4 & -1 \end{vmatrix} = (2)(-1) - (-3)(4) = -2 + 12 = 10\)

3. \(\begin{vmatrix} 2 & -6 \\ 4 & 3 \end{vmatrix} = (2)(3) - (-6)(4) = 6 + 24 = 30\)

Putting it all together:

\[ \vec{a} \times \vec{b} = 15\hat{i} - 10\hat{j} + 30\hat{k} \]

### Step 2: Calculate the magnitude of \(\vec{a} \times \vec{b}\)

The magnitude is given by:

\[ |\vec{a} \times \vec{b}| = \sqrt{(15)^2 + (-10)^2 + (30)^2} = \sqrt{225 + 100 + 900} = \sqrt{1225} = 35 \]

### Step 3: Calculate the magnitudes of \(\vec{a}\) and \(\vec{b}\)

For \(\vec{a}\):

\[ |\vec{a}| = \sqrt{(2)^2 + (-6)^2 + (-3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7 \]

For \(\vec{b}\):

\[ |\vec{b}| = \sqrt{(4)^2 + (3)^2 + (-1)^2} = \sqrt{16 + 9 + 1} = \sqrt{26} \]

### Step 4: Substitute into the sine formula

Now substituting into the sine formula:

\[ \sin \theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{35}{7 \cdot \sqrt{26}} = \frac{35}{7\sqrt{26}} = \frac{5}{\sqrt{26}} \]

### Final Answer

Thus, the sine of the angle between the vectors is:

\[ \sin \theta = \frac{5}{\sqrt{26}} \]