Vectors: The line l has equation r = 4i - 8k + λ(2i + j) and the plane ∏1 has equation r = ...?

Question

The line l has equation r = 4i - 8k + λ(2i + j) and the plane ∏1 has equation r = -8i + 12j + 4k + µ(2i + j) + v(2i - k), where λ, µ, v are all real numbers. The points U and V, which vary independently are on l and on ∏1, respectively. The point W which is on the line segment UV, is such that UW = 3WV.

Find an equation of the locus of W.

The points L and M have position vectors -4i + 13j + 3k and 15j + 3k, respectively.

i) Determine if the points L and M are in the plane ∏1.
ii) Find the plane ∏2, which contains the points L and M and also the line l.
iii) Find an equation of the intersection of ∏2 with the locus of W.

Um. Help, please. Seriously, I took one look at this question, saw the ∏ and the λ and promptly blanked out. They ARE just symbols, but I've been rendered pretty much clueless. Where should I start, how should I go about working out the answers? Could someone provide working, too? Thank you so very much in advance!

Answer 1

Plane ∏1 has equation r = -8i + 12j + 4k + µ(2i + j) + v(2i - k), where λ, µ, v are all real numbers.

∏1: r = <-8, 12, 4> + µ<2, 1, 0> + v<2, 0, -1>
This is a plane defined by 1 point and two directional vectors.

Given points L(-4, 13, 3) and M(0, 15, 3).

i) Determine if the points L and M are in the plane ∏1.

We have three equations in three unknowns.

Point L:
x: -8 + 2µ + 2v = -4
y: 12 + µ = 13
z: 4 - v = 3

y: µ = 1
z: v = 1

Check with x.
x: -8 + 2*1 + 2*1 = -4

This checks so point L(-4, 13, 3) is in plane ∏1.

Point M:
x: -8 + 2µ + 2v = 0
y: 12 + µ = 15
z: 4 - v = 3

y: µ = 3
z: v = 1

Check with x.
x: -8 + 2*3 + 2*1 = 0

This checks so point M(0, 15, 3) is also in plane ∏1.
_____________

Line l has equation r = 4i - 8k + λ(2i + j)
r = <4, 0, -8> + λ<2, 1, 0>
Points L(-4, 13, 3) and M(0, 15, 3)

ii) Find the plane ∏2, which contains the points L and M and also the line l.

This plane is overdefined. A plane can be defined by one line and one point not on the line. We will use point L and line l and check to see if point M is also lies in the plane.

One directional vector of the plane is v1 = <2, 1, 0>. To find a second one use the point L and the point on the line l when λ = 0.

L(-4, 13, 3) and Q(4, 0, -8).

v2 = QL = L - Q = <-4 - 4, 13 - 0, 3 - -8> = <-8, 13, 11>

The normal vector n, to the plane is orthogonal to both v1 and v2. Take the cross product.

n = v1 X v2 = <11, -22, 34>

With the normal vector and a point in the plane L, we can write the equation of the plane.

11(x + 4) - 22(y - 13) + 34(z - 3) = 0
11x + 44 - 22y + 286 + 34z - 102 = 0
11x - 22y + 34z + 228 = 0

Check to see if M lies in the plane also.

11*0 - 22*15 + 34*3 + 228 = 0

So M also lies in the plane.
__________

Don't know how to find the locus of W. Sorry.

Answer 2

i think of this mistake is rampant. that's no longer an "equation" it relatively is an expression. Equations have 2 expressions and an equivalent sign. Expressions have a simplified type to keep utilization to a uniform element. For this expression, upload the like words (those words that are further or subtracted that incorporate the vast type and form of variables). 6k + 3j - 2k + 5j must be rewirten as 6k - 2k + 3j + 5j = 4k + 8j it relatively is hypothesis that the values of ok and j are distinctive.

Answer 3

My brain feels like jelly. Or it would, if your brain had nerve endings.