Find th esine of angles between vectors `vec(a)= 2hat(i) - 6hat(j) - 3hat(k), and vec(b) = 4hat(i) + 3hat(j)- hat(k)`?

Options

A. `(1)/(sqrt26)`
B. `(5)/(sqrt26)`
C. `(5)/(26)`
D. `(1)/(26)`

Answer

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 \(\vec{a} \times \vec{b}\)

We can set up the determinant to find the cross product:

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

### Step 2: Expand the determinant

Using the determinant formula, we expand it as follows:

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

Calculating each of these 2x2 determinants:

1. For \(\hat{i}\):
\[ (-6)(-1) - (-3)(3) = 6 + 9 = 15 \]

2. For \(\hat{j}\):
\[ (2)(-1) - (-3)(4) = -2 + 12 = 10 \]

3. For \(\hat{k}\):
\[ (2)(3) - (-6)(4) = 6 + 24 = 30 \]

Thus, we have:

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

### Step 3: 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 4: Calculate the magnitude of \(\vec{a}\)

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

### Step 5: Calculate the magnitude of \(\vec{b}\)

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

### Step 6: Substitute into the sine formula

Now we can substitute these values 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:

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

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