Question Solved1 Answer Find a vector parametric equation π‘Ÿβƒ— (𝑑) for the line through the points 𝑃=(βˆ’4,1,0) and 𝑄=(βˆ’9,5,5) for each of the given conditions on the parameter 𝑑. (a) If π‘Ÿβƒ— (0)=βŒ©βˆ’4,1,0βŒͺ and π‘Ÿβƒ— (8)=βŒ©βˆ’9,5,5βŒͺ, then π‘Ÿβƒ— (𝑑)= (b) If π‘Ÿβƒ— (3)=𝑃 and π‘Ÿβƒ— (5)=𝑄, then π‘Ÿβƒ— (𝑑)= (c) If the points 𝑃 and 𝑄 correspond to the parameter values 𝑑=0 and 𝑑=βˆ’4, respectively, then π‘Ÿβƒ— (𝑑)=

M5U5P1 The Asker Β· Calculus

Find a vector parametric equation 𝑟⃗ (𝑡) for the line through the points 𝑃=(−4,1,0) and 𝑄=(−9,5,5) for each of the given conditions on the parameter 𝑡.

(a) If 𝑟⃗ (0)=〈−4,1,0〉 and 𝑟⃗ (8)=〈−9,5,5〉, then
𝑟⃗ (𝑡)=

(b) If 𝑟⃗ (3)=𝑃 and 𝑟⃗ (5)=𝑄, then
𝑟⃗ (𝑡)=

(c) If the points 𝑃 and 𝑄 correspond to the parameter values 𝑡=0 and 𝑡=−4, respectively, then
𝑟⃗ (𝑡)=

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Ques Given Points are P(-4,1,0) andQ=(-9,5,5)a) If vec(r)(0)=(:-4,1,0:) and vec(r)(8)=(:-9,5,5:)We know that genral form of 3D-vector equation of line is a point plus vector multiplid by a scalar, t :r(z)=(x_(p),y_(p),z_(p))+t(x_(v),y_(v),z_(v))-**the parametric equations are -{:[x=tx_(v)+x_(p)],[y=ty_(v)+yp],[z=tz_(v)+z_(p)]:}at t=0 and at t=8 from(1), (2) f (3) we havefrom(1){:[-4=0*x_(v)+x_(p)-(4)],[-9=8*x_(v)+x_(p)-(5]:}b){:[" If " vec(r)(3)=P=(:-4","1","0:)],[" and " vec(r)(5)=Q=(:-9","5","5:)]:}then from (1) we have.{:[-4=3x_(v)+x_(p)quad" (10) "],[-9=5x_(v)+x_(p)],[+-11],[5=-2x_(v)=>x_(v)=-(5)/(2)],[" and "x_(p)=(7)/(2)]:}from (2) we have -{:[1=3y_(v)+y_(p)-12],[5=5y_(v)+y_(p)-(13],[-4=-2y_(v)=>y_(v)=2],[" and "y_(p)=-5],[" from (3) we have- "],[0=3.zv+z_(p)-(14)],[5=5z_(v)+z_(p)- ... See the full answer