الرياضيات البحتة (بالإنجليزية) الصف الثالث الثانوي، الأسبوع 13: A plane through three points, and a plane containing two lines

Perpendiculars to a line, distances from a line and the equation of a plane

الأسبوع 13 · مستقيم يقطع مستقيمًا معلومًا على التعامد والبعد بين نقطة ومستقيم ومعادلة المستوى بصورها المتجهة والقياسية والعامة والمستوى المار بثلاث نقط أو المحتوي على مستقيمين · 3 دروس

A plane through three points, and a plane containing two lines

Pure Mathematics Sec 3 Term 1, Unit 4, Lessons 1–2

GoalFind the equation of the plane through three points, and of the plane containing two intersecting lines.

New words

non-collinear points · three points not on one line; they fix exactly one planenormal from a cross product · n = AB × AC is perpendicular to both AB and AC, so to the planecoplanar · lying in one plane

Explanation

Three points A, B, C that are not on one line fix exactly one plane. The vectors AB and AC lie in the plane. Their vector product n = AB × AC is perpendicular to both, so it is a normal vector. If AB × AC is the zero vector, the points are on one line, and then there is no single plane. Now use the last lesson: with n and the point A, write n · r = n · A. Always check that B and C satisfy the final equation. A plane may also be given by two intersecting lines: first find their point of intersection P as in Week 12. Their direction vectors d₁ and d₂ lie in the plane, so n = d₁ × d₂ is a normal vector, and the plane is n · r = n · P. If the two lines are skew, no plane contains both. The table shows the two methods side by side.

then use any known point of the plane, and check the other points in the final equation

given	two vectors in the plane	normal vector n
three points A, B, C	AB and AC	n = AB × AC
two intersecting lines	d₁ and d₂	n = d₁ × d₂

Worked examples

Plane through A(2, 1, 0), B(0, 3, 1) and C(1, 1, 3). AB = (−2, 2, 1), AC = (−1, 0, 3). n = AB × AC = (2 × 3 − 1 × 0, −(−2 × 3 − 1 × (−1)), −2 × 0 − 2 × (−1)) = (6, 5, 2). n · A = 12 + 5 + 0 = 17, so 6x + 5y + 2z = 17. Check B: 0 + 15 + 2 = 17. Check C: 6 + 5 + 6 = 17.

Are (1, 2, 3), (3, 4, 5) and (5, 6, 7) on one plane in a unique way? AB = (2, 2, 2) and AC = (4, 4, 4) are parallel, so AB × AC is the zero vector: the three points are on one line and many planes pass through them.

Plane containing the lines r₁ = (0, 1, 2) + s(1, 2, −1) and r₂ = (3, 4, −2) + t(−1, 1, 2). They meet at s = 2, t = 1, at the point P(2, 5, 0). n = d₁ × d₂ = (2 × 2 − (−1) × 1, −(1 × 2 − (−1)(−1)), 1 × 1 − 2 × (−1)) = (5, −1, 3). n · P = 10 − 5 + 0 = 5, so 5x − y + 3z = 5. Check: the point (0, 1, 2) gives −1 + 6 = 5, and (3, 4, −2) gives 15 − 4 − 6 = 5.

Choosing the point: with the first plane, any of A, B or C gives the same right-hand side, 17. If the right-hand sides differ, something is wrong: recheck the cross product.

Class activity

Three pins and a card. Stand three pins of different heights in a lump of modelling clay and lay a stiff card on their tops: the card rests on all three, so three points fix a plane. Now add a fourth pin of any height. Does the card touch it too? (Usually not: four points are coplanar only in special cases, as in the test of Week 11.)

Practice

Find the plane through A(1, 0, 1), B(2, 2, 1) and C(0, 1, 3).

Find the plane through the origin, A(1, 2, 3) and B(2, −1, 0).

Does D(−1, −1, 3) lie in the plane of question 1?

Find the plane containing (1, 1, 0) + s(1, 0, 1) and (2, 0, 3) + t(0, 1, −2), which meet at (2, 1, 1).

Why does no plane contain both (0, 0, 0) + s(1, 0, 0) and (0, 1, 0) + t(0, 0, 1)?