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Three Dimensional Geometry - Angle between two lines, two planes, a line and a plane

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Direction Vectors and Ratios: In 3D geometry, every line has a direction vector b⃗=ai^+bj^+ck^\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}, where a,b,ca, b, c are the direction ratios. Visually, imagine a vector pointing along the line's path; these ratios define its orientation relative to the xx, yy, and zz axes.

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Angle between Two Lines: The angle θ\theta between two lines depends solely on their direction vectors b1⃗\vec{b_1} and b2⃗\vec{b_2}. If the lines are represented as r⃗=a1⃗+λb1⃗\vec{r} = \vec{a_1} + \lambda \vec{b_1} and r⃗=a2⃗+μb2⃗\vec{r} = \vec{a_2} + \mu \vec{b_2}, we treat them as if they both pass through the origin to find the angle between them using the dot product formula.

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Normal Vectors of Planes: A plane is uniquely defined by its normal vector n⃗=Ai^+Bj^+Ck^\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}, which is a vector sticking straight out (perpendicular) from the surface of the plane. Visually, if you place a flat sheet of paper on a table, the normal vector would be a pencil standing vertically on it.

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Angle between Two Planes: The angle between two intersecting planes is defined as the angle between their respective normal vectors. Imagine two flat surfaces (like an open book); the angle between the surfaces is the same as the angle between two lines drawn perpendicular to each surface from the same point on the spine.

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Angle between a Line and a Plane: This is the angle θ\theta between the line and its projection on the plane. Visually, think of a rod piercing a surface; the angle is measured between the rod and its shadow on the surface. Because the formula uses the normal vector of the plane (which is 90∘90^\circ to the surface), we use sin⁡θ\sin \theta instead of cos⁡θ\cos \theta.

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Condition for Perpendicularity: Two lines (or two planes) are perpendicular if the dot product of their direction vectors (or normal vectors) is zero. In Cartesian form, this means a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0. Visually, this occurs when the vectors meet at a 90∘90^\circ angle.

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Condition for Parallelism: Two lines (or two planes) are parallel if their direction ratios (or normal ratios) are proportional, meaning a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}. Visually, parallel lines or planes never meet and maintain a constant distance between each other.

📐Formulae

Angle between two lines (Vector): cos⁡θ=∣b1⃗⋅b2⃗∣∣b1⃗∣∣b2⃗∣\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}

Angle between two lines (Cartesian): cos⁡θ=∣a1a2+b1b2+c1c2∣a12+b12+c12a22+b22+c22\cos \theta = \frac{|a_1 a_2 + b_1 b_2 + c_1 c_2|}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}}

Angle between two planes (Vector): cos⁡θ=∣n1⃗⋅n2⃗∣∣n1⃗∣∣n2⃗∣\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}

Angle between two planes (Cartesian): cos⁡θ=∣A1A2+B1B2+C1C2∣A12+B12+C12A22+B22+C22\cos \theta = \frac{|A_1 A_2 + B_1 B_2 + C_1 C_2|}{\sqrt{A_1^2 + B_1^2 + C_1^2} \sqrt{A_2^2 + B_2^2 + C_2^2}}

Angle between a line and a plane (Vector): sin⁡θ=∣b⃗⋅n⃗∣∣b⃗∣∣n⃗∣\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}

Angle between a line and a plane (Cartesian): sin⁡θ=∣aA+bB+cC∣a2+b2+c2A2+B2+C2\sin \theta = \frac{|aA + bB + cC|}{\sqrt{a^2 + b^2 + c^2} \sqrt{A^2 + B^2 + C^2}}

💡Examples

Problem 1:

Find the angle between the pair of lines given by: x−22=y−15=z+3−3\frac{x-2}{2} = \frac{y-1}{5} = \frac{z+3}{-3} and x+1−1=y−48=z−54\frac{x+1}{-1} = \frac{y-4}{8} = \frac{z-5}{4}.

Solution:

  1. Identify the direction ratios of the two lines: b1=(2,5,−3)b_1 = (2, 5, -3) and b2=(−1,8,4)b_2 = (-1, 8, 4).
  2. Calculate the dot product: a1a2+b1b2+c1c2=(2)(−1)+(5)(8)+(−3)(4)=−2+40−12=26a_1 a_2 + b_1 b_2 + c_1 c_2 = (2)(-1) + (5)(8) + (-3)(4) = -2 + 40 - 12 = 26.
  3. Calculate the magnitude of b1b_1: 22+52+(−3)2=4+25+9=38\sqrt{2^2 + 5^2 + (-3)^2} = \sqrt{4 + 25 + 9} = \sqrt{38}.
  4. Calculate the magnitude of b2b_2: (−1)2+82+42=1+64+16=81=9\sqrt{(-1)^2 + 8^2 + 4^2} = \sqrt{1 + 64 + 16} = \sqrt{81} = 9.
  5. Use the formula: cos⁡θ=26938\cos \theta = \frac{26}{9\sqrt{38}}.
  6. Therefore, θ=cos⁡−1(26938)\theta = \cos^{-1} \left(\frac{26}{9\sqrt{38}}\right).

Explanation:

To find the angle between two lines in Cartesian form, we extract the denominators as direction ratios, treat them as vectors, and apply the cosine dot product formula.

Problem 2:

Find the angle between the line r⃗=(i^+2j^−k^)+λ(i^−j^+k^)\vec{r} = (\hat{i} + 2\hat{j} - \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k}) and the plane r⃗⋅(2i^−j^+k^)=4\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 4.

Solution:

  1. Identify the line's direction vector: b⃗=i^−j^+k^\vec{b} = \hat{i} - \hat{j} + \hat{k}.
  2. Identify the plane's normal vector: n⃗=2i^−j^+k^\vec{n} = 2\hat{i} - \hat{j} + \hat{k}.
  3. Calculate b⃗⋅n⃗=(1)(2)+(−1)(−1)+(1)(1)=2+1+1=4\vec{b} \cdot \vec{n} = (1)(2) + (-1)(-1) + (1)(1) = 2 + 1 + 1 = 4.
  4. Calculate ∣b⃗∣=12+(−1)2+12=3|\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}.
  5. Calculate ∣n⃗∣=22+(−1)2+12=6|\vec{n}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}.
  6. Use the sine formula: sin⁡θ=43⋅6=418=432=223\sin \theta = \frac{4}{\sqrt{3} \cdot \sqrt{6}} = \frac{4}{\sqrt{18}} = \frac{4}{3\sqrt{2}} = \frac{2\sqrt{2}}{3}.
  7. Therefore, θ=sin⁡−1(223)\theta = \sin^{-1} \left(\frac{2\sqrt{2}}{3}\right).

Explanation:

When finding the angle between a line and a plane, we use the sine function because the angle between the line and the plane is the complement of the angle between the line and the plane's normal vector.