The sine of the angle between the lines `r=2i+2j-k+(i+j+k)t` and the plane `r.(3i-4j+5k)=q` is

Options

A. `2sqrt(6)//15`
B. `2sqrt(3)//15`
C. `sqrt(201)//15`
D. none of these

Answer

To find the sine of the angle between the line given by the equation \( \mathbf{r} = 2\mathbf{i} + 2\mathbf{j} - \mathbf{k} + (\mathbf{i} + \mathbf{j} + \mathbf{k})t \) and the plane defined by \( \mathbf{r} \cdot (3\mathbf{i} - 4\mathbf{j} + 5\mathbf{k}) = q \), we can follow these steps:

### Step 1: Identify the direction vector of the line and the normal vector of the plane.
The direction vector \( \mathbf{b} \) of the line can be extracted from the equation of the line:
\[ \mathbf{b} = \mathbf{i} + \mathbf{j} + \mathbf{k} \]
The normal vector \( \mathbf{n} \) of the plane can be identified from the equation of the plane:
\[ \mathbf{n} = 3\mathbf{i} - 4\mathbf{j} + 5\mathbf{k} \]

### Step 2: Calculate the dot product \( \mathbf{b} \cdot \mathbf{n} \).
We compute the dot product:
\[ \mathbf{b} \cdot \mathbf{n} = (1)(3) + (1)(-4) + (1)(5) = 3 - 4 + 5 = 4 \]

### Step 3: Calculate the magnitudes of \( \mathbf{b} \) and \( \mathbf{n} \).
The magnitude of \( \mathbf{b} \) is:
\[ |\mathbf{b}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]
The magnitude of \( \mathbf{n} \) is:
\[ |\mathbf{n}| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \]

### Step 4: Use the formula for sine of the angle between the line and the plane.
The sine of the angle \( \theta \) between the line and the plane can be found using the formula:
\[ \sin \theta = \frac{|\mathbf{b} \cdot \mathbf{n}|}{|\mathbf{b}| |\mathbf{n}|} \]
Substituting the values we calculated:
\[ \sin \theta = \frac{4}{\sqrt{3} \cdot 5\sqrt{2}} = \frac{4}{5\sqrt{6}} \]

### Step 5: Simplify the expression for sine.
To express this in a more standard form:
\[ \sin \theta = \frac{4\sqrt{6}}{30} = \frac{2\sqrt{6}}{15} \]

### Final Answer:
Thus, the sine of the angle between the lines and the plane is:
\[ \sin \theta = \frac{2\sqrt{6}}{15} \]