Question

The angle between the lines \(\vec{r} = 3\hat{i} - 2\hat{j} + 1\hat{k} + \mu (4\hat{i} + 6\hat{j} + 12\hat{k})\) and \(\vec{r} = 7\hat{i} - 3\hat{j} + 9\hat{k} + \lambda (5\hat{i} + 8\hat{j} - 4\hat{k})\) is:

• A. \(\cos^{-1} \frac{10}{\sqrt{7}\sqrt{105}}\)
• B. \(\cos^{-1} \frac{5}{72}\)
• C. \(\cos^{-1} \frac{2}{35}\)
• D. \(\cos^{-1} \frac{7}{98}\)

Answer 1

The Correct Option is A

To find the angle between the lines given by their vector equations, we utilize the formula for the cosine of the angle \(\theta\) between two vectors \(\vec{a}\) and \(\vec{b}\), which is given by:
\[\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\]

For the first line, the vector direction is \(\vec{a} = 4\hat{i} + 6\hat{j} + 12\hat{k}\).

For the second line, the vector direction is \(\vec{b} = 5\hat{i} + 8\hat{j} - 4\hat{k}\).

We first find the dot product \(\vec{a} \cdot \vec{b}\):

\(\vec{a} \cdot \vec{b} = (4)(5) + (6)(8) + (12)(-4) = 20 + 48 - 48 = 20\)

Next, we calculate the magnitudes of \(\vec{a}\) and \(\vec{b}\):

\(|\vec{a}| = \sqrt{4^2 + 6^2 + 12^2} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14\)

\(|\vec{b}| = \sqrt{5^2 + 8^2 + (-4)^2} = \sqrt{25 + 64 + 16} = \sqrt{105}\)

Substitute these values into the cosine formula:

\(\cos\theta = \frac{20}{14 \cdot \sqrt{105}}\) 

\(\cos\theta = \frac{20}{14\sqrt{105}}\) 

Simplifying the fraction:\(\cos\theta = \frac{10}{7\sqrt{105}}\)

Thus, the angle \(\theta\) is given by:

\(\theta = \cos^{-1}\left(\frac{10}{\sqrt{7}\sqrt{105}}\right)\)

Answer 2

The angle between two lines is determined by the angle between their direction vectors. For the given lines, the direction vectors are:

\(\vec{d_1} = 4\hat{i} + 6\hat{j} + 12\hat{k}\), \(\vec{d_2} = 5\hat{i} + 8\hat{j} - 4\hat{k}\).

The formula for the cosine of the angle between two vectors is:

\[ \cos \theta = \frac{\vec{d_1} \cdot \vec{d_2}}{\|\vec{d_1}\| \|\vec{d_2}\|}. \]

Compute the dot product \(\vec{d_1} \cdot \vec{d_2}\):

\[ \vec{d_1} \cdot \vec{d_2} = (4)(5) + (6)(8) + (12)(-4). \]

\[ \vec{d_1} \cdot \vec{d_2} = 20 + 48 - 48 = 20. \]

Compute the magnitudes of \(\vec{d_1}\) and \(\vec{d_2}\). The magnitude of \(\vec{d_1}\) is:

\[ \|\vec{d_1}\| = \sqrt{(4)^2 + (6)^2 + (12)^2} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14. \]

The magnitude of \(\vec{d_2}\) is:

\[ \|\vec{d_2}\| = \sqrt{(5)^2 + (8)^2 + (-4)^2} = \sqrt{25 + 64 + 16} = \sqrt{105}. \]

Substitute into the cosine formula:

\[ \cos \theta = \frac{\vec{d_1} \times \vec{d_2}}{\|\vec{d_1}\| \|\vec{d_2}\|}. \]

\[ \cos \theta = \frac{20}{14 \times \sqrt{105}} = \frac{10}{\sqrt{7} \times \sqrt{105}}. \]

Thus, the angle between the lines is:

\[ \theta = \cos^{-1} \left( \frac{10}{\sqrt{7}  \sqrt{105}} \right). \]