The angle between the lines `vecr=(2hati-5hatj+hatk)+lamda(3hati+2hatj+6hatk)andvecr=(7hati-6hatk)+mu(hati+2hatj+2hatk)` is

Options

A. `cos^(-1)((19)/(21))`
B. `cos^(-1)((1)/(21))`
C. `cos^(-1)((9)/(19))`
D. `cos^(-1)((9)/(21))`

Answer

To find the angle between the two lines given in vector form, we can follow these steps:

**Step 1: Identify the direction vectors of the lines.**

The first line is given by:
\[ \vec{r} = (2\hat{i} - 5\hat{j} + \hat{k}) + \lambda(3\hat{i} + 2\hat{j} + 6\hat{k}) \]
The direction vector \( \vec{b_1} \) for the first line is:
\[ \vec{b_1} = 3\hat{i} + 2\hat{j} + 6\hat{k} \]

The second line is given by:
\[ \vec{r} = (7\hat{i} - 6\hat{k}) + \mu(\hat{i} + 2\hat{j} + 2\hat{k}) \]
The direction vector \( \vec{b_2} \) for the second line is:
\[ \vec{b_2} = \hat{i} + 2\hat{j} + 2\hat{k} \]

**Step 2: Use the formula for the cosine of the angle between two vectors.**

The cosine of the angle \( \theta \) between two vectors \( \vec{b_1} \) and \( \vec{b_2} \) can be calculated using the formula:
\[ \cos \theta = \frac{\vec{b_1} \cdot \vec{b_2}}{|\vec{b_1}| |\vec{b_2}|} \]

**Step 3: Calculate the dot product \( \vec{b_1} \cdot \vec{b_2} \).**

Calculating the dot product:
\[ \vec{b_1} \cdot \vec{b_2} = (3)(1) + (2)(2) + (6)(2) = 3 + 4 + 12 = 19 \]

**Step 4: Calculate the magnitudes of \( \vec{b_1} \) and \( \vec{b_2} \).**

Calculating the magnitude of \( \vec{b_1} \):
\[ |\vec{b_1}| = \sqrt{3^2 + 2^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7 \]

Calculating the magnitude of \( \vec{b_2} \):
\[ |\vec{b_2}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

**Step 5: Substitute the values into the cosine formula.**

Now substituting the values into the cosine formula:
\[ \cos \theta = \frac{19}{7 \cdot 3} = \frac{19}{21} \]

**Step 6: Find the angle \( \theta \).**

To find \( \theta \), we take the inverse cosine:
\[ \theta = \cos^{-1}\left(\frac{19}{21}\right) \]

Thus, the angle between the lines is \( \theta = \cos^{-1}\left(\frac{19}{21}\right) \).

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