my professor didn't use formula steps on all these problems but he only used answers...so please look at those all questions and include the formula steps on each of those questions.

1. Turn in: A line goes through the point A(2, 1, 3) . This line is also perpendicular to the two vectors U = it j+k . Determine the equation for this line, in both symmetric and parametric forms. V = 21 - j-3k You are missing the direction vector to construct the equation. Because the line is perpendicular to both vectors U , V , consequently, U x V will give you the direction vector: Ux V =-2i+5j -3k The equation for the line is: x =2+2t y=1-5t z=3+3t x - 5 3 Geogebra can quickly confirm your answer. The line is shown in dashed N red while the 2 vectors are shown in blue. N w 3 28. Find the equation for a plane that contains 2 points: A(0, 5, 0) , B(0, 9, 1) and is parallel to the line with x =2+3t parametric equation L =y =3-2t . Put your final answer in the form ax + by + cz +d =0 z = 4+t The equation needs a point and the normal vector n . And you do not have this normal vector. But it can be found using the cross product. From the 2 given points, you can construct: AB = 0i +4j+1k . The direction vector of the given line: r =3i-2j+1k Then: n = ABxr =6i +3j-12k Now, you can either use point A or B to write the equation. After a bit of Algebra: 2x + y-4z-5=06. Find the equation for the line that is passing through the points A(2, 4, -3) and B(3, -1, 1) . Put the answer in parametric and symmetric form. Then determine the point where this line passes through the xy- plane. You construct vector AB =i-5j +4k to obtain the direction vector. Then you can use either point A or B to construct the equation. x = 2+t Using point A, you get: y =4-5t or x - 2= 1-4 z+3 - 5 4 z =-3+4t 'ppRos ood = The xy-plane means that z =0= t = - . And from this, you get the intersection point: 11 1 4 4' 1 0 Onice apt x+ 2y-2z =-3 7. Find the distance between these 2 parallel planes: (x+ 2y-2z=3 x+2y-2z+3=080 You first need a quick rewrite to put the equations in the proper form: (x+ 2y-2z-3=0 Choose an arbitray point on one plane and then apply the distance formula D _ ax, + by. +cz. + d Vaz + 62 + c 2 From the 1st plane, (-3, 0, 0) is a good point. You should get > D =2x =t 2. Turn in: Object A travels along the line defined by L, = \\y =-t and Object B, along the line defined by z =1-t x =t-3 L2 = y =2t . Determine whether the 2 objects will cross or collide. And find the point of intersection (or z =4t-1 collision). There is a similar example in the lecture. You will find that both lines will cross (or intersect) at 14 = -2 Due to different t 12 = 1 values, there is no collision. And the intersection point is found to be: (-2, 2, 3) . 3. Turn in: The symmetric equations for 2 lines are given as: L : x =-y+2 = -z+2 L2 : x - 2 = -y+1 = z+1' . Confirm that they are P skew lines and then find the distance between them. It is left as an exercise for you to confirm that they are in fact skew lines. o Using t =0 then a point 2, on Line L, is (0, 2, 2) and a point 22 on Line Lz is (2, 1, -1) Next, do a cross product of the 2 direction vectors to find the normal vector: n = R, X R2 = (1, -1, -1) x (1, -1, 1) = (-2, -2, 0) . Using point Q2 and the above normal vector, you can now write the equation for the plane P2 a( x - x ) +b( y - yo)+c (z-z ) = 0 - 2x-2+6=0 -2(x-2)-2(y-1) +0=0 Finally, the distance from point Q, (0, 2, 2) to the plane P2 (which is also the distance between the 2 skew lines) is: D = ax +by + cz, +d -2(0 )-2(2) +0+6 2 V2 Var + b 2 + c ? V4 + 4 2Nam VOIS 4. Turn in: Find the equation for the plane that contains the 3 points: (1, 2, 1) (3, 2, 2) (4, -1, -1) . Put your final answer in the form ax + by+cz +d =0 The strategy was discussed in class. There is also example in the Lecture Note. From the 3 given points, you construct 2 vectors; then perform a cross product to determine the normal vector. With the normal vector known, you can use it with any of the 3 given points to construct the plane equation. Once you go through the Algebra to get it to the form ax + by +cz +d =0, everyone will get the same equation regardless of which point you choose. 3x+7y-62-11=0 A quick graph using Geogebra will further visually confirm this plane with 3 points on it. x =1-t 5. Turn in: Determine the point where the line _ y = -3+ 2t passes through (z= t the plane x+ y+z-4=0 . And find the angle that this line makes with the plane using the dot product. You may express the angle relative to the vertical or horizontal, but be sure to be specific. Take the line and plug it into the plane equation: (1-t)+(-3+2t)+t-4=0 and then find t. You will get =3 1 -> Plugging this t = 3 back into the line and you will find the line passes through the plane at: (-2, 3, 3) The direction for the line is: r = -i +2j + k The normal vector of the plane is: n=itjthe store- ron (-1, 2, 1) . (1, 1, 1) cos 0 = 2 16 V3 /18 The angle is: 0 = cos- V18 ~ 61.90 Note that this is the angle with the normal vector. Relative to the plane, the angle is: 90 -61.9 = 28. 10