krit.club logo

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

Grade 12ICSE

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

🔑Concepts

•

The direction of a line in 3D space is defined by its Direction Ratios (a,b,c)(a, b, c) or Direction Cosines (l,m,n)(l, m, n). Visually, a line can be thought of as a vector b⃗=ai^+bj^+ck^\vec{b} = a\hat{i} + b\hat{j} + c\hat{k} extending infinitely in both directions. The angle between two lines is the angle between their respective direction vectors.

•

When two lines intersect, they form two pairs of vertically opposite angles. By convention, the 'angle between two lines' refers to the acute angle θ\theta. If the lines are skew (do not intersect and are not parallel), the angle is defined as the angle between two intersecting lines that are respectively parallel to the given skew lines.

•

Two lines with direction ratios (a1,b1,c1)(a_1, b_1, c_1) and (a2,b2,c2)(a_2, b_2, c_2) are perpendicular if the dot product of their direction vectors is zero, satisfying a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0. Visually, they form a 90∘90^{\circ} angle at their point of intersection.

•

The orientation of 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 perpendicular to every line lying on the plane. The angle between two planes is defined as the acute angle between their normal vectors. If you imagine two intersecting sheets of paper, the angle between them is the same as the angle between two pencils held perpendicular to each sheet.

•

The angle θ\theta between a line and a plane is the complement of the angle ϕ\phi between the line and the normal to the plane. Visually, if you project the line onto the plane, θ\theta is the angle the line makes with its projection. Because we use the normal vector for calculation, we use the sine function instead of cosine: sin⁡θ=cos⁡(90∘−θ)\sin \theta = \cos(90^{\circ} - \theta).

•

Condition for Parallelism: Two planes are parallel if their normal vectors are proportional, meaning A1A2=B1B2=C1C2\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}. A line is parallel to a plane if its direction vector is perpendicular to the plane's normal vector, i.e., aA+bB+cC=0aA + bB + cC = 0.

📐Formulae

Angle θ\theta between two lines with direction ratios (a1,b1,c1)(a_1, b_1, c_1) and (a2,b2,c2)(a_2, b_2, c_2): cos⁡θ=∣a1a2+b1b2+c1c2∣a12+b12+c12a22+b22+c22\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}}

Angle θ\theta between two lines using vector form 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}: cos⁡θ=∣b1⃗⋅b2⃗∣∣b1⃗∣∣b2⃗∣\cos \theta = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}

Angle θ\theta between two planes A1x+B1y+C1z+D1=0A_1x + B_1y + C_1z + D_1 = 0 and A2x+B2y+C2z+D2=0A_2x + B_2y + C_2z + D_2 = 0: cos⁡θ=∣A1A2+B1B2+C1C2∣A12+B12+C12A22+B22+C22\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}}

Angle θ\theta between a line with direction ratios (a,b,c)(a, b, c) and a plane Ax+By+Cz+D=0Ax + By + Cz + D = 0: 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}}

Angle θ\theta 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: sin⁡θ=∣b⃗⋅n⃗∣∣b⃗∣∣n⃗∣\sin \theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}

💡Examples

Problem 1:

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

Solution:

  1. Let the direction ratios be (a1,b1,c1)=(1,1,2)(a_1, b_1, c_1) = (1, 1, 2) and (a2,b2,c2)=(3−1,−3−1,4)(a_2, b_2, c_2) = (\sqrt{3}-1, -\sqrt{3}-1, 4).
  2. Use the formula cos⁡θ=∣a1a2+b1b2+c1c2∣a12+b12+c12a22+b22+c22\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(3−1)+1(−3−1)+2(4)=3−1−3−1+8=61(\sqrt{3}-1) + 1(-\sqrt{3}-1) + 2(4) = \sqrt{3} - 1 - \sqrt{3} - 1 + 8 = 6.
  4. Calculate the denominators: 12+12+22=6\sqrt{1^2+1^2+2^2} = \sqrt{6} and (3−1)2+(−3−1)2+42=(3+1−23)+(3+1+23)+16=4+4+16=24=26\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⁡θ=66⋅26=62⋅6=12\cos \theta = \frac{6}{\sqrt{6} \cdot 2\sqrt{6}} = \frac{6}{2 \cdot 6} = \frac{1}{2}.
  6. Therefore, θ=cos⁡−1(12)=60∘\theta = \cos^{-1}(\frac{1}{2}) = 60^{\circ} or π3\frac{\pi}{3}.

Explanation:

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.

Problem 2:

Find the angle between the line x−23=y+1−1=z−32\frac{x-2}{3} = \frac{y+1}{-1} = \frac{z-3}{2} and the plane 3x+4y+z+5=03x + 4y + z + 5 = 0.

Solution:

  1. Identify the direction ratios of the line: (a,b,c)=(3,−1,2)(a, b, c) = (3, -1, 2).
  2. Identify the direction ratios of the normal to the plane: (A,B,C)=(3,4,1)(A, B, C) = (3, 4, 1).
  3. Use the formula for the angle between a line and a plane: 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}}.
  4. Calculate the numerator: ∣(3)(3)+(−1)(4)+(2)(1)∣=∣9−4+2∣=7|(3)(3) + (-1)(4) + (2)(1)| = |9 - 4 + 2| = 7.
  5. Calculate the denominators: 32+(−1)2+22=9+1+4=14\sqrt{3^2+(-1)^2+2^2} = \sqrt{9+1+4} = \sqrt{14} and 32+42+12=9+16+1=26\sqrt{3^2+4^2+1^2} = \sqrt{9+16+1} = \sqrt{26}.
  6. sin⁡θ=71426=7364\sin \theta = \frac{7}{\sqrt{14}\sqrt{26}} = \frac{7}{\sqrt{364}}.
  7. θ=sin⁡−1(7364)\theta = \sin^{-1}(\frac{7}{\sqrt{364}}).

Explanation:

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.