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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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The angle θ\theta between two lines is defined as the angle between their direction vectors b1⃗\vec{b_1} and b2⃗\vec{b_2}. If the lines are given by 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}, then cos⁡θ=∣b1⃗⋅b2⃗∣∣b1⃗∣∣b2⃗∣\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}. For Cartesian form, use the direction ratios (a1,b1,c1)(a_1, b_1, c_1) and (a2,b2,c2)(a_2, b_2, c_2). Note that lines are perpendicular if a1a2+b1b2+c1c2=0a_1 a_2 + b_1 b_2 + c_1 c_2 = 0.

Diagram showing two intersecting lines L1 and L2 with an angle theta between them.
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The angle between two planes r⃗⋅n1⃗=d1\vec{r} \cdot \vec{n_1} = d_1 and r⃗⋅n2⃗=d2\vec{r} \cdot \vec{n_2} = d_2 is equal to the angle between their normal vectors n1⃗\vec{n_1} and n2⃗\vec{n_2}. The planes are parallel if their normal vectors are proportional, and perpendicular if the dot product of their normal vectors is zero (A1A2+B1B2+C1C2=0A_1 A_2 + B_1 B_2 + C_1 C_2 = 0).

A plane with its normal vector shown perpendicular to the surface.
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The angle between a line r⃗=a⃗+λb⃗\vec{r} = \vec{a} + \lambda \vec{b} and a plane r⃗⋅n⃗=d\vec{r} \cdot \vec{n} = d is the complement of the angle between the line and the normal to the plane. Thus, we use the sine function: sin⁡θ=∣b⃗⋅n⃗∣∣b⃗∣∣n⃗∣\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}. If the line is parallel to the plane, b⃗⋅n⃗=0\vec{b} \cdot \vec{n} = 0. If the line is perpendicular to the plane, b⃗\vec{b} is parallel to n⃗\vec{n}.

Illustration showing a line intersecting a plane and its normal, highlighting the angle theta.
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Direction Cosines (DC) and Direction Ratios (DR) provide a way to express orientation in 3D. If (l,m,n)(l, m, n) are DCs, then l2+m2+n2=1l^2 + m^2 + n^2 = 1. If (a,b,c)(a, b, c) are DRs, they are proportional to DCs. The cosine of the angle between two lines can be simply written as cos⁡θ=∣l1l2+m1m2+n1n2∣\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2| using direction cosines.

📐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.

Problem 3:

Find the angle between the two planes 3x−6y+2z=73x - 6y + 2z = 7 and 2x+2y−2z=52x + 2y - 2z = 5.

Two intersecting planes representing the calculated angle.

Solution:

  1. Identify normal vectors: n1⃗=3i^−6j^+2k^\vec{n_1} = 3\hat{i} - 6\hat{j} + 2\hat{k} and n2⃗=2i^+2j^−2k^\vec{n_2} = 2\hat{i} + 2\hat{j} - 2\hat{k}.
  2. Calculate dot product: n1⃗⋅n2⃗=(3)(2)+(−6)(2)+(2)(−2)=6−12−4=−10\vec{n_1} \cdot \vec{n_2} = (3)(2) + (-6)(2) + (2)(-2) = 6 - 12 - 4 = -10.
  3. Calculate magnitudes: ∣n1⃗∣=32+(−6)2+22=9+36+4=7|\vec{n_1}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9+36+4} = 7. ∣n2⃗∣=22+22+(−2)2=4+4+4=12=23|\vec{n_2}| = \sqrt{2^2 + 2^2 + (-2)^2} = \sqrt{4+4+4} = \sqrt{12} = 2\sqrt{3}.
  4. Use the formula: cos⁡θ=∣−10∣7×23=10143=573\cos \theta = \frac{|-10|}{7 \times 2\sqrt{3}} = \frac{10}{14\sqrt{3}} = \frac{5}{7\sqrt{3}}.
  5. θ=cos⁡−1(5321)\theta = \cos^{-1}\left(\frac{5\sqrt{3}}{21}\right).

Explanation:

The angle between two planes is the acute angle between their normal vectors. We apply the cosine dot product formula using the coefficients of x,y,zx, y, z as the components of the normal vectors.

Problem 4:

Find the angle between the line x−23=y+14=z−212\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12} and the plane x+y+z=10x + y + z = 10.

Line passing through a plane with normal vector shown.

Solution:

  1. Direction vector of the line: b⃗=3i^+4j^+12k^\vec{b} = 3\hat{i} + 4\hat{j} + 12\hat{k}.
  2. Normal vector of the plane: n⃗=1i^+1j^+1k^\vec{n} = 1\hat{i} + 1\hat{j} + 1\hat{k}.
  3. Dot product: b⃗⋅n⃗=(3)(1)+(4)(1)+(12)(1)=3+4+12=19\vec{b} \cdot \vec{n} = (3)(1) + (4)(1) + (12)(1) = 3 + 4 + 12 = 19.
  4. Magnitudes: ∣b⃗∣=32+42+122=9+16+144=169=13|\vec{b}| = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9+16+144} = \sqrt{169} = 13. ∣n⃗∣=12+12+12=3|\vec{n}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}.
  5. Use formula: sin⁡θ=∣b⃗⋅n⃗∣∣b⃗∣∣n⃗∣=19133\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|} = \frac{19}{13\sqrt{3}}.
  6. θ=sin⁡−1(19133)\theta = \sin^{-1}\left(\frac{19}{13\sqrt{3}}\right).

Explanation:

To find the angle between a line and a plane, we use the sine formula because we are calculating the angle between the line and its projection on the plane, which is the complement of the angle between the line and the plane's normal.