"Parameterize the plane through the point (−1,−3,1) with the normal vector ⟨−4,4,3⟩

r⃗ (s,t)=

(Use s and t for the parameters in your parameterization, and enter your vector as a single vector, with angle brackets: e.g., as < 1 + s + t, s - t, 3 - t >.)"

Question

28-12-2016
Mathematics

Answered

"Parameterize the plane through the point (−1,−3,1) with the normal vector ⟨−4,4,3⟩

r⃗ (s,t)=

(Use s and t for the parameters in your parameterization, and enter your vector as a single vector, with angle brackets: e.g., as < 1 + s + t, s - t, 3 - t >.)"

Verified Answer :

Related Questions

The volume of a square prism is 144x3 216 x2 81x What is an expression that could describe the perimeter of one of the prism s square faces Plz include steps of

During which of the following chemical changes does a precipitate form Select one a vinegar is added to baking powder b lemon juice is added to water c lemon ju

What was the first financially successful English colony in the new world A Canada B Jamestown C Maryland D Plymouth

What mass in grams of Ag is produced from reaction of 3175 g of Cu

Organize for whether it is equivalent to 98 89 or neither 18 x19 8 divided by 9 18 divided by 16 118 x 16 18 x 9 1618 Thank chu

Why topics are called Cancer Copics Capricorn

from the mantle rises and flows out through a vent onto Earths surface as This molten material sometimes along with ash cinders and rock builds up a mountain a

The majority of renewable power in South America comes from which source nuclear power plants coal hydroelectric dams solar panels

Compare the health and environmental risks associated with groundlevel ozone to the health and environmental risks associated with ozone layer depletion

lily has a 100 gift card to her favorite pastry shopshe spends 4 a day at the shop for the next 12 days

nobillionaireNobley nobillionaireNobley · Answer 1 · 2017-01-07T07:19:44-05:00

The equation of the required plane can be obtained thus:
-4(x + 1) + 4(y + 3) + 3(z - 1) = 0
-4x - 4 + 4y + 12 + 3z - 3 = 0
4x - 4y - 3z = 5

Let x = 1, y = 2, then 4(1) - 4(2) - 3z = 5
z = (4 - 8 - 5)/3 = -9/3 = -3
Thus, point (1, 2, -3) is a point on the plane.

Let a = (a1, a2, a3) and b = (b1, b2, b3) be vectors parallel to the plane.
Then, -4a1 + 4a2 + 3a3 = 0 and -4b1 + 4b2 + 3b3 = 0
Let a1 = 2, a2 = -1, then a3 = (4(2) - 4(-1))/3 = (8 + 4)/3 = 12/3 = 4 and let b1 = -1 and b2 = 2, then b3 = (4(-1) - 4(2))/3 = (-4 - 8)/3 = -12/3 = -4
Thus a = (2, -1, 4) and b = (-1, 2, -4)

Therefore, the required parametric equation is r(s, t) = s(2, -1, 4) + t(-1, 2, -4) + (1, 2, -3) = (2s, -s, 4s) + (-t, 2t, -4t) + (1, 2, -3) = (2s - t + 1, -s + 2t + 2, 4s - 4t - 3)

"Parameterize the plane through the point (−1,−3,1) with the normal vector ⟨−4,4,3⟩r⃗ (s,t)= (Use s and t for the parameters in your parameterization, and enter your vector as a single vector, with angle brackets: e.g., as < 1 + s + t, s - t, 3 - t >.)"

Verified Answer :

Related Questions

"Parameterize the plane through the point (−1,−3,1) with the normal vector ⟨−4,4,3⟩

r⃗ (s,t)=

(Use s and t for the parameters in your parameterization, and enter your vector as a single vector, with angle brackets: e.g., as < 1 + s + t, s - t, 3 - t >.)"