I am uploading this question for PART A. I know you guys cannot just go and answer a question all the way through so I am uploading individual parts of these questions so it doesn't get rejected and y'all get more money!! My tutor flaked on me today (not associated with bartleby). This is not a test and I am not copying answers but I need a reference sheet which I usually get from my tutor who isn't here. Thanks! The diagram below shows a cuboid (rectangular solid) OJKLMNPQ. The vertex O is (0, 0, 0), Jis (6, 0, 0), K is (6, 0, 10), M is (0, 7,0) and Q is (0, 7, 10).Q (0, 7, 10)|K(6, 0, 10)M(0.7.0)- NO(0, 0, 0)J(6, 0, 0)(a) (i)Show that JQ10(ii)Find MK.(0)-6)(b)An equation for the line (MK) is r 7 +s 710(i)Write down an equation for the line (JQ) in the form r= a+tb.(ii) Find the acute angle between (JQ) and (MK).(c)The lines (JQ) and (MK) intersect at D. Find the position vector of D.

Name: I am uploading this question for PART A. I know you guys cannot just g
Uploaded: 2024-03-20T07:07:37.000Z
Channel: Bartleby
Description: I am uploading this question for PART A. I know you guys cannot just go and answer a question all the way through so I am uploading individual parts of these questions so it doesn't get rejected and y'all get more money!! My tutor flaked on me today

We have given vertices,O≡(0, 0, 0), J≡(6, 0, 0), K ≡(6, 0, 10), M ≡ (0, 7, 0) and Q ≡ (0, 7, 10). Formulae: 1. If A≡(a1, a2, a3) and B = (b1, b2, b3) then AB→ =b1 - a1b2 - a2b3 - a3. 2. Vector equation of line passing through position vector a→ and b→ is r = a→+λ(b⇀-a⇀). 3. If m1, m2, m3 and n1, n2, n3 are direction vectors of two lines then angle between them is given by cos(θ) = m1n1 +m2n2 +m3n3 m12 + m22+m32n12 + n22+n32. a) To show: JQ→=-6710 We have given J = (6, 0, 0) and Q = (0, 7, 10). Consider, JQ→=0 - 67 - 010 - 0 = -6710. To find MK→: We have given M=(0, 7, 0) and K = (6, 0, 10). Therefore, MK→ = 6 - 00 - 710 - 0=6-710.b) i) To write an equation for the line JQ in the form r = a + tb. We have given J = (6, 0, 0) and Q = (0, 7, 10). Therefore, r = 600+t0 - 67 - 010 - 0 = 600+t- 67 10 . ⇒r = 600+t- 67 10 . ii) To find angle between JQ and MK. We gave direction vector of line JQ is (-6, 7, 10) and direction vectors of line MK is (6, -7, 10). Therefore, angle between JQ and MK is cos(θ) = (-6)(6) + 7(-7) + 10(10)(-6)2 + 72 +10262 + (-7)2 +102 =-36 -49 +10036 + 49 + 10036 + 49+ 100 = 15185185 = 15185 = 0.83 ⇒θ = cos-1(0.83) ⇒θ = 33.56° Therefore, angle between JQ and MK is 33.56 degree.We have equations of line JQ and MK are r = 600+t- 67 10 and r = 070+s 6-7 10 respectively. Equating the values of r we get, 600+t- 67 10 = 070+s 6-7 10 ⇒t- 67 10 -s6-7 10 = 070- 600 ⇒t-6t - 6s7t + 7s 10 t - 10s = -670 ⇒ -6t - 6s = -6, 7t + 7s = 7, 10t - 10s = 0 ⇒t +s = 1, t - s = 0 Solving t + s = 1 and t - s = 0 we get t = s = 12. Substituting values of t and s in either equation of lines we get, r = 600+12- 67 10 = 6 - 6/20 + 7/20 + 10/2 = 37/25 Therefore, D = 37/25. Answer: a) ii) MK→ =6-710 b) i) Equation for the line JQ in the form r = a + tb is r = 600+t- 67 10 ii) Angle between JQ and MK is 33.56 degree. c) The point of intersection of JQ and MK is D = 37/25.