Distances to planes and lines In this note we will look at distances to planes and lines. Our approach is geometric. Very broadly, we will draw a sketch and use vector techniques. Please note is that our sketches are not oriented, drawn to scale or drawn in perspective. Rather they are a simple ’cartoon’ which shows the important features of the problem. 1. Distance: point to plane: → − Ingredients: i) A point P , ii) A plane with normal N and containing a point Q. � → �� −−→ − −−→ N� � The distance from P to the plane is d = |PQ| cos θ = �PQ · �. � |N| � We will explain this formula by way of the following example. Example 1: Let P = (1, 3, 2). Find the distance from P to the plane x + 2y = 3. Answer: First we gather our ingredients. Q = (3, 0, 0) is a point on the plane (it is easy to ﬁnd such a point). N = normal to plane = i + 2j. R = point on plane closest to P (this is point unknown and we do not need to ﬁnd it to ﬁnd the distance). The ﬁgure shows that � � �−−→� �−−→ N � � � � �. distance = |P R| = �PQ� cos θ = �PQ · |N| � −−→ Computing PQ = 2i − 3j − 2k gives � � � � �−−→ N � � �1, 2, 0� �� 4 � � � √ distance = �PQ · = �2, −3, −2� · =√ . � |N| � � 5 5 P θ N Q R 2. Distance: point to line: Ingredients: i) A point P , ii) A line with direction vector v and containing a point Q. � � �−−→ v �� The distance from P to the line is d = |QP| sin θ = ��QP × . |v| � We will explain this formula by way of the following example. Example 2: Let P = (1, 3, 2), ﬁnd the distance from the point P to the line through (1, 0, 0) and (1, 2, 0). Answer: First we gather our ingredients. Q = (1, 0, 0) (this is easy to ﬁnd). v = �1, 2, 0� − �1, 0, 0� = 2j is parallel to the line. R = point on line closest to P (this is point is unknown). Using the relation |A × B| � = |A||B| � sin θ, the ﬁgure shows that �−−→� �−−→ � θ v � � �. distance = |P R| = �PQ� sin θ = ��QP × v Q |v| � �−−→ � −−→ � v � Computing: PQ = 3j + 2k, which implies �QP × |v| � = |(3j + 2k) × j| = | − 2i| = 2. P R 3. Distance between parallel planes: The trick here is to reduce it to the distance from a point to a plane. Example 3: Find the distance between the planes x + 2y − z = 4 and x + 2y − z = 3. Both planes have normal N = i + 2j − k so they are parallel. Take any point on the ﬁrst plane, say, P = (4, 0, 0). Distance between planes = distance from P to second plane. Choose Q = (1, 0, 0) = point on second plane √ √ −−→ N ⇒ d = |QP · |N| | = |3i · (i + 2j − k)|/ 6 = 6/2. 4. Distance between skew lines: We place the lines in parallel planes and ﬁnd the distance between the planes as in the previous example As usual it’s easy to ﬁnd a point on each line. Thus, to ﬁnd the parallel planes we only need to ﬁnd the normal. N = v1 × v2 , where v1 and v2 are the direction vectors of the lines.