number 10 and 12 plz Find the parametric equation of the line of interaction of

Question

number 10 and 12 plz

Find the parametric equation of the line of interaction of the place r_1: 2r + 3y - 7a + 7 - 0 and r_2: - 3a + p + 5c + 6 - 0 Determine whether the line l_1: r - 1 + l, y = - 3 + 21, t = - 2 - 1 and t_2: x - 17 + 3a, M = 4 + p, s = - 8 - s have a point of interaction? Find the distance from the point p (3 - 8, 1) to the line (x - 3)/2 - (y + 7)/(- 1) - (z + 2)/5. Find the distance from the given point p to the given place p - (2, - 1, 4) and pi: 3x - v + T^z - 2. Show that line l; (1, 3, - 1) + t (0, 3, 5) line entirely in the plane which passes through the points (- 5, 0, 0), (0, - 1, - 7) and (- 2, 0, - 4).

Explanation / Answer

10. The two planes are 2x+3y-7z = -7 and -3x+y+5 z = -6

These two planes have normal vectors a = (2, 3, -7) and b = (-3, 1, 5), respectively. Let L denote the line of intersection. Then the vector v = a × b = (22,11,11) is parallel to L. Now, we need to find a point P on L. To find P, we will solve the system of equations of the planes 2x+3y-7z = -7 and -3x + y +5 z = -6

Let us consider P to be the point of L where z = 0. Thus, on substituting z = 0 in the above system we get 2x +3y =-7 and -3x + y = -6. On solving these equations, we get x = 1 and y = -3. Thus P: (1,-3,0) is a point on L. Then the parametric equations of the line L are x = 1+22t, y = -3+11t, z = 11t.

12. The direction vector of the given line is v = (2,-1,5). Also, the co-ordinates f a point q on the given line are (3,-7,-2). Then the vector u connecting q and p is ( 3-3,-7+8, -2-1)= ( 0,1,-3). Further uxv = (-2,6,2). Then the distance from the point p to the given line is |uxv|/|v| = ([(-2)2+62 +22]/ [22 +(-1)2 +52] = 44/30 = (44/30) = (22/15).

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