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MCV4U Equations of Lines & Planes
Grade 12 Unit 7 — practice questions with full solutions

Vector, parametric, symmetric and scalar equations, intersections, distances and angles — where vectors turn back into geometry.

MCV4U · Grade 12 Geometry and Algebra of Vectors 5 core 1 challenge
1Core
Write the vector equation and the parametric equations of the line through (1, −2, 3) with direction vector (2, 1, −4).
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  1. A vector equation has the form r = r0 + tm, where r0 is a point on the line and m its direction.
  2. So r = (1, −2, 3) + t(2, 1, −4).
  3. Reading off each coordinate separately gives the parametric form.
  4. x = 1 + 2t, y = −2 + t, z = 3 − 4t.
Final answerr = (1, −2, 3) + t(2, 1, −4); x = 1 + 2t, y = −2 + t, z = 3 − 4t
2Core
Write the symmetric equation of the line through (1, −2, 3) with direction vector (2, 1, −4).
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  1. Start from the parametric equations and solve each for t.
  2. From x = 1 + 2t: t = (x − 1) ÷ 2.
  3. From y = −2 + t: t = y + 2.
  4. From z = 3 − 4t: t = (z − 3) ÷ (−4).
  5. Setting the three expressions equal gives the symmetric form.
Final answer(x − 1)⁄2 = (y + 2)⁄1 = (z − 3)⁄(−4)
3Core
Find the scalar equation of the plane with normal vector (3, −1, 2) passing through the point (2, 0, −1).
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  1. A plane with normal (A, B, C) through (x0, y0, z0) satisfies A(x − x0) + B(y − y0) + C(z − z0) = 0.
  2. Substitute: 3(x − 2) − 1(y − 0) + 2(z + 1) = 0.
  3. Expand: 3x − 6 − y + 2z + 2 = 0.
  4. Collect constants: 3x − y + 2z − 4 = 0.
Final answer3x − y + 2z − 4 = 0
4Core
Find the point where the line x = 1 + 2t, y = −2 + t, z = 3 − 4t meets the plane 3x − y + 2z − 4 = 0.
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  1. Substitute the parametric expressions into the plane equation.
  2. 3(1 + 2t) − (−2 + t) + 2(3 − 4t) − 4 = 0.
  3. Expand: 3 + 6t + 2 − t + 6 − 8t − 4 = 0.
  4. Collect: 7 − 3t = 0, so t = 7⁄3.
  5. Substitute back: x = 1 + 14⁄3 = 17⁄3, y = −2 + 7⁄3 = 1⁄3, z = 3 − 28⁄3 = −19⁄3.
Final answer(17⁄3, 1⁄3, −19⁄3)
5Core
Find the distance from the point (4, 1, −2) to the plane 2x − y + 2z − 6 = 0.
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  1. Distance from (x1, y1, z1) to Ax + By + Cz + D = 0 is |Ax1 + By1 + Cz1 + D| ÷ √(A2 + B2 + C2).
  2. Numerator: |2(4) − 1 + 2(−2) − 6| = |8 − 1 − 4 − 6| = |−3| = 3.
  3. Denominator: √(4 + 1 + 4) = √9 = 3.
  4. Distance = 3 ÷ 3 = 1.
Final answer1 unit
6Challenge
Find the acute angle between the planes with normal vectors (1, 2, −1) and (2, −1, 3), to one decimal place.
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  1. The angle between two planes equals the angle between their normals.
  2. Dot product: (1)(2) + (2)(−1) + (−1)(3) = 2 − 2 − 3 = −3.
  3. |n1| = √6 ≈ 2.4495 and |n2| = √14 ≈ 3.7417.
  4. cos θ = −3 ÷ (2.4495 × 3.7417) ≈ −0.3273, giving θ ≈ 109.1°.
  5. Because the angle is obtuse, the acute angle between the planes is 180° − 109.1° ≈ 70.9°.
Final answer≈ 70.9°

MCV4U Equations of Lines & Planes — common questions

Short answers to the things students ask most about this unit.

What is the difference between vector, parametric and symmetric equations of a line?

They carry the same information in different formats: the vector form uses a point plus a multiple of the direction, the parametric form lists each coordinate separately, and the symmetric form eliminates the parameter.

How do I find where a line meets a plane?

Substitute the line's parametric expressions into the scalar equation of the plane, solve the resulting equation for the parameter, then substitute that value back into the line to get the point.

More MCV4U practice — grade 12 calculus and vectors

MCV4U Rates of Change & LimitsMCV4U DerivativesMCV4U Applications of Derivatives & Curve SketchingMCV4U Derivatives of Exponential & Trigonometric FunctionsAll MCV4U units
← Unit 6: Cartesian Vectors, Dot & Cross ProductsBack to all assignments →
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