Three-Dimensional Geometry - Angle between Lines, Planes, and a Line and a Plane

Find the angle between the two lines whose direction ratios are $(1, 1, 2)$ and $(\sqrt{3}-1, -\sqrt{3}-1, 4)$.

1. Let the direction ratios be $(a_1, b_1, c_1) = (1, 1, 2)$ and $(a_2, b_2, c_2) = (\sqrt{3}-1, -\sqrt{3}-1, 4)$.
2. Use the formula $\cos \theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$.
3. Calculate the numerator: $1(\sqrt{3}-1) + 1(-\sqrt{3}-1) + 2(4) = \sqrt{3} - 1 - \sqrt{3} - 1 + 8 = 6$.
4. Calculate the denominators: $\sqrt{1^2+1^2+2^2} = \sqrt{6}$ and $\sqrt{(\sqrt{3}-1)^2 + (-\sqrt{3}-1)^2 + 4^2} = \sqrt{(3+1-2\sqrt{3}) + (3+1+2\sqrt{3}) + 16} = \sqrt{4+4+16} = \sqrt{24} = 2\sqrt{6}$.
5. $\cos \theta = \frac{6}{\sqrt{6} \cdot 2\sqrt{6}} = \frac{6}{2 \cdot 6} = \frac{1}{2}$.
6. Therefore, $\theta = \cos^{-1}(\frac{1}{2}) = 60^{\circ}$ or $\frac{\pi}{3}$.

To find the angle between two lines, we apply the cosine formula using their direction ratios. The calculation involves finding the dot product of the direction vectors and dividing by the product of their magnitudes.

Find the angle between the line $\frac{x-2}{3} = \frac{y+1}{-1} = \frac{z-3}{2}$ and the plane $3x + 4y + z + 5 = 0$.

1. Identify the direction ratios of the line: $(a, b, c) = (3, -1, 2)$.
2. Identify the direction ratios of the normal to the plane: $(A, B, C) = (3, 4, 1)$.
3. Use the formula for the angle between a line and a plane: $\sin \theta = \frac{|Aa + Bb + Cc|}{\sqrt{a^2+b^2+c^2}\sqrt{A^2+B^2+C^2}}$.
4. Calculate the numerator: $|(3)(3) + (-1)(4) + (2)(1)| = |9 - 4 + 2| = 7$.
5. Calculate the denominators: $\sqrt{3^2+(-1)^2+2^2} = \sqrt{9+1+4} = \sqrt{14}$ and $\sqrt{3^2+4^2+1^2} = \sqrt{9+16+1} = \sqrt{26}$.
6. $\sin \theta = \frac{7}{\sqrt{14}\sqrt{26}} = \frac{7}{\sqrt{364}}$.
7. $\theta = \sin^{-1}(\frac{7}{\sqrt{364}})$.

The angle between a line and a plane is calculated using the sine of the angle, relating the line's direction vector and the plane's normal vector. Note that we take the absolute value to ensure we find the acute angle.

1. Identify the normal vectors of the planes: $\vec{n_1} = 2\hat{i} - \hat{j} + \hat{k}$ $\vec{n_2} = \hat{i} + \hat{j} + 2\hat{k}$
2. Use the formula $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$
3. Calculate dot product: $\vec{n_1} \cdot \vec{n_2} = (2)(1) + (-1)(1) + (1)(2) = 2 - 1 + 2 = 3$
4. Calculate magnitudes: $|\vec{n_1}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$ $|\vec{n_2}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}$
5. Substitute values: $\cos \theta = \frac{3}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$
6. Therefore, $\theta = \cos^{-1}(\frac{1}{2}) = 60^\circ$.

The angle between two planes is the angle between their normal vectors. We extract the normal vectors from the coefficients of $x, y, z$ and apply the cosine dot product formula.

1. Identify the direction vector of the line: $\vec{b} = \hat{i} - \hat{j} + \hat{k}$
2. Identify the normal vector of the plane: $\vec{n} = 2\hat{i} - \hat{j} + \hat{k}$
3. Use the formula $\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$
4. Calculate dot product: $\vec{b} \cdot \vec{n} = (1)(2) + (-1)(-1) + (1)(1) = 2 + 1 + 1 = 4$
5. Calculate magnitudes: $|\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$ $|\vec{n}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$
6. Substitute: $\sin \theta = \frac{4}{\sqrt{3} \cdot \sqrt{6}} = \frac{4}{\sqrt{18}} = \frac{4}{3\sqrt{2}} = \frac{2\sqrt{2}}{3}$
7. Thus, $\theta = \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)$.

For the angle between a line and a plane, we use the sine of the angle because the geometric angle is the complement of the angle between the line and the plane's normal.