MAH 050 4.4 Additional Examples Page 4.4 4.4 Additional Examples for Chapter 4 Example 4.4. Prove that the line joining the midpoints of two sides of a triangle is parallel to and exactly half as long as the third side of that triangle. We need to prove that DE BC. BC BO+ OC OC OB OD OA + OB and OE ( OA + OC) DE DO + OE OE OD OA + OC OA OB OC OB BC Example 4.4. Find the coordinates of the point P that is one-fifth of the way from A(,, 3) to B(7, 4, 9). OA [ 3 ], OB [ 7 4 9 ] P splits the line segment AB in the ratio r:s :4. he general formula for the location of such a point is s r OP OA + OB r+ s r+ s [Page 4.04 of these lecture notes] herefore 7 4 OP 4 + 5 5 5 4 3 9 3 OR 6 5 OP OA + AP OA + 5 AB + 6 4 5 5 5 3 3 herefore the point P is located at,,. 4 3 5 5 5
MAH 050 4.4 Additional Examples Page 4.5 Example 4.4.3 he points P(, 3, ), Q(4, 7, ), R(, 5, 3) and S are the four vertices of a parallelogram PQSR, with sides PQ and PR meeting at vertex P. Find the coordinates of point S. Following the path OQS: QS PR OR OP [ ] OS OQ + QS + [ ] 4 7 [ ] [ 3 9 4 ] herefore the point S is at (3, 9, 4). [One could follow the path ORS instead.] Example 4.4.4 Find the parametric and symmetric equations of the line L that passes through the points Q(, 5, 3) and R(4, 7, ). Find the distance r of the point P(, 7, 0) from the line and find the coordinates of the nearest point N on the line to the point P. he line direction vector is d QR [ 3 4 ] Either Q or R may serve as the known point on the line. Choosing Q, the vector equation of the line is x 3 p OQ + t d y 5 + t, t z 3 4 he Cartesian parametric equations are x + 3 t, y 5+ t, z 34 t, t from which the symmetric form follows: ( 5) x y z3 3 4
MAH 050 4.4 Additional Examples Page 4.6 Example 4.4.4 (continued) he vector from Q (a known point on the line) to P is v QP [ 7 ] he shadow of this vector on the line is the projection 3 3 3 3 44 8 proj vd i u v d d 9 + 44 + 6 i 69 d 7 4 4 4 3 3 69 u 69 4 4 3 4 NP NQ + QP u+ v + 0 4 7 3 r NP 6 + 0 + 9 5 5 OR riangle PNQ is right-angled at N r v u + 44 + 49 9 + 44 + 6 5 r 5 he location of N can be found from 3 ON OQ + QN 5 + 7 3 4 7 [or one may use ON OP + PN instead] herefore the point N is at (, 7, 7).
MAH 050 4.4 Additional Examples Page 4.7 Example 4.4.5 x y z Show that the lines L : 3 skew and find the distance between them. x y0 z and L : are d 3 and passes through point P (,, ). Line L has line direction vector [ ] Line L has line direction vector [ ] d and passes through point P (, 0, ). Clearly d is not a multiple of d. herefore the two lines are not parallel. OR he angle between the lines,θ, is also the acute angle between the direction vectors of the lines. d i d + ( ) + 3 4 + 3 6 d 4+ + 9 4, d 4+ + 6 6 3 cos θ d d i ± 0 and 4 6 7 θ θ π d d 3 he lines are therefore at an angle of θ cos cos 0.65465 49. 7 Upon finding a non-zero distance between the lines, we will complete the proof that these lines are skew.
MAH 050 4.4 Additional Examples Page 4.8 Example 4.4.5 (continued) he vector n d d is orthogonal to both lines. he length of the projection of PP onto n is the distance r between the lines. i n j ( 3) i ( 6) j+ ( + ) k 4 k 3 PP PO + OP [ ] + [ 0 ] [ ] 4 i i PP i n 4 r proj PP n PP n n 4 + + + + 3 herefore the distance between the two non-parallel lines is and the two lines are skew. 4 3 r.309 3 Example 4.4.6 Find two non-zero vectors that are orthogonal to each other and to [ ] u 3 0. It is easy to construct a non-zero vector v whose dot product with u is zero: v 0 0 v i u 0+ 0 + 0 0 [ ] For the third vector, just take the cross product of the first two vectors: i 3 0 0 3 0 3 0 w u v j 0 i 3 0 0 j 0 + k 0 i j + k k 0 herefore v [ 0 0 ] and w [ 3 0 ].
MAH 050 4.4 Additional Examples Page 4.9 Example 4.4.7 Find the Cartesian equation of the plane that passes through the points A(3, 0, 0), B(, 4, 0) and C(, 5, 3) and find the coordinates of the nearest point N on the plane to the origin and find the distance of the plane from the origin. u AB 4 0 wo vectors in the plane are [ ] v AC [ 5 3 ] A normal to the plane is. and i u v j 4 5 k 0 3 4 4 5 i j + k 3 3 0 3 0 3 4 5 3 n 4 [ 3 0 0 ] a OA n i a 3 4+ 0+ 0 he Cartesian equation of the plane is 4x+ y + z herefore a normal vector to the plane is [ ] he displacement vector for N is the projection of the displacement vector of any point on the plane onto the plane s normal vector. OAin ON proj OA ( OA ) n inn n n 3 4 4 4 4 0 4 i 8 + + 3 0 herefore N is the point 8 ( 3, 3, 3) and r ON 4 + + 8 3 3
MAH 050 4.4 Additional Examples Page 4.30 Example 4.4.8 Find all vectors w that are orthogonal to both [ ] u 3 and v [ 4 3 ]. One vector that is orthogonal to both u and v is i 4 3 4 4 u v j 3 i 5 0 5 3 j 3 + k 3 i + j k k 3 Any multiple of this vector is also orthogonal to both u and v. herefore the set of vectors w is t, t. { [ ] } Check: wu i t i 3 4+ 3 t 0 t [ ] [ ] and wv i t i 4 3 4 6+ t 0 t [ ] [ ] Example 4.4.9 he vertices of a triangle ABC are at A(, 0, ), B(,, ) and C(3,, ). Find the angle at vertex A (correct to the nearest degree). u AB 3 0 9+ + 0 0 v AC 4+ 4+ 9 3 uv i 3 0 i 6 + 0 8 [ ] u [ ] v [ ] [ ] Let θ be the angle at vertex A, then uv i 8 cosθ 0.8437 u v 3 0 θ 47.4875 47
MAH 050 4.4 Additional Examples Page 4.3 Example 4.4.0 (extbook, page 79, exercises 4., question 8) Show that every plane containing the points P(,, ) and Q(, 0, ) must also contain the point R(, 6, 5). PQ [ ] 3 6 6 3 3PQ and QR [ ] [ ] Points P and Q are in a plane all points on the line through P and Q are in any plane containing P and Q. But QR 3PQ point R is on the line through P and Q herefore every plane containing the points P and Q must also contain the point R. Example 4.4. (extbook, page 80, exercises 4., question 44(a)) Prove the Cauchy-Schwarz inequality uv i u v. Let θ be the angle between vectors u and v. uv i u v cosθ, but cosθ uv i u v
MAH 050 4.4 Additional Examples Page 4.3 Example 4.4. Find the point of intersection of the lines x y 6 z 7 3. x 3 y3 z0 and Line : x 3+ s, y 3 + s, z 0 s Line : x t, y 6 + t, z 7 + 3t At the point of intersection x 3+ st s+ t 4 y 3+ s 6+ t s t 3 z s 7+ 3t s + 3t 7 Solving the over-determined linear system for s and t, 4 3 3 R R R R 3 4 0 5 0 R3 R 3 7 3 7 0 5 0 3 3 R 5 0 5 0 0 R3 R 0 0 0 0 0 0 0 R+ R 0 s and t 0 0 0 Unique solution a single point of intersection does exist. x 3+ ( ) y 3+ ( ) z ( ) herefore the two lines intersect at the point (,, ).