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![Find the scalar equation of the plane = [-2,2,7] + s[3, −2,1] + t[−2,1,5]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2ef31c9f-1e44-437f-a71e-09574dc73129%2F1957ce7b-4985-432e-a0b6-9f09749cc4ab%2F4vyy6e_processed.png&w=3840&q=75)
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- Find a scalar equation of the plane that contains the points (−1, −3, 1), (1, 0, 3), and (2, 7, 0).Draw v = [ 4/1 ] and w = [ -2/2 ] and v+w and v-w in a single xy plane.Calculate the triple scalar products w. (v × u) and u ⋅ (w × v), where u = (7, 2, −1), v = w. (vx u) = 330 = (2, 8, -3), and w = u. (w x v) = 350 = (9, 8, -10).
- Find the vector equation for the line of 5x + 3y - 2z = −1 and 5x + 5z = -4 intersection of the planes r = = 0,0) + t(15,00)Determine the scalar equation of the plane that contains A(-1, 3, 1) and is perpendicular to both 5x - 2y+z-6=0 and 4x + y 2z+ 9 = 0.Find the scalar equation of the plane that contain the line * = [2, 1,7] + t[0, 1, 0], t e R and is parallel to the line 7 = [3,0, 4] + s[2,–1,0], s e R. 1.
- Consider the plane with a scalar equation 4x-2y+z-19=0. Find the point of intersection of this plane with the line passing through the points A(1, -6, -5) and B(3, -3, 3).Suppose the straight lines L₁ and L₂ have respective vector equations 5 r1 = -3 + t -0-0--0- and 4 3 where s and t are scalar parameters. 9 = 8 +s -2 -5 -5 2 a) Given that L₁ and L2 intersect at some point P, find its coordinates. P =( )Find the equation of the plane that passes through the point (−2, 8, 4) and contains vectors ⃗a (2,4,6) and ⃗b(4, -2,6)
- 3. If a = 3x - 5ỹ and b = -2x +9y, write the vector 5a-86 in terms of x and yThe velocity vector of a particle in the xy-plane is (√7, 9e3-3) for all real t. If the position of the particle at t = 1 is (9, 8), then the position vector of the particle is Select one: O.a. ( 21, 3 + 3/3, 303³1-3 + 5) (4) b. 3 4 ○ c. d. 9³-3-1) (+37³½³, 30³-3) 4 Od (-1.3-5) 4 1, 3e
