The points A, B, and C have position vectors i + 2j + 3k, -4 + 5j - k and 4i - k respectively...?

Question

The points A, B, and C have position vectors i + 2j + 3k, -4 + 5j - k and 4i - k respectively. The point D is on the line segment AB such that 2AD = DB.

i) Find the position vector of D.
ii) Find the equation of the plane which passes through D and is perpendicular to AB.
iii) Find the position vector of the point E in which the line through C parallel to the vector i - 2j - k meets the plane.
iv) Find the perpendicular distance of E from the line AB.

I apologise for the sheer number of sub-questions, but I'm really confused. My lecturer's workings are elegant, but lack detail, and I really don't get them. Any help'd be very much appreciated. If you were to answer this question, could you include your working? I'd like to study others' workings when it comes to solving vector questions instead of just my own, my lecturer's, or my classmates'. Thanks!

Answer 1

Given
A<1, 2, 3>
B<-4, 5, -1>
C<4, 0, -1>

i) Find the position vector of D.
D is 1/3 the distance from A to B.

D = A + (1/3)AB = A + (1/3) = (2A + B)/3
D = <2-4, 4+5, 6-1>/3 = <-2, 9, 5>/3 = <-2/3, 3, 5/3>
____________

ii) Find the equation of the plane which passes through D and is perpendicular to AB.

AB is the normal vector n, to the plane.

n = AB = B - A = <-4-1, 5-2, -1-3> = <-5, 3, -4>

Any non-zero multiple of n is also a normal vector to the plane. Multiply by -1.

n = <5, -3, 4>

With the normal vector n and the point D we can now write the equation of the plane.

5(x + 2/3) - 3(y - 3) + 4(z - 5/3) = 0

Multiply by 3 to eliminate fractions.

5(3x + 2) - 3(3y - 9) + 4(3z - 5) = 0
15x + 10 - 9y + 27 + 12z - 20 = 0
15x - 9y + 12z + 17 = 0
_________

iii) Find the position vector of the point E in which the line through C parallel to the vector i - 2j - k meets the plane.

L = C + t<1, -2, -1>
L = <4, 0, -1> + t<1, -2, -1>
L = <4 + t, -2t, -1 - t>
where t is a scalar ranging over the real numbers

I assume L intersects the plane found in ii), namely
15x - 9y + 12z + 17 = 0

Express the equation of the plane in terms of t.

15x - 9y + 12z + 17 = 0
15(4 + t) - 9(-2t) + 12(-1 - t) + 17 = 0
60 + 15t + 18t - 12 - 12t + 17 = 0
21t = -65
t = -65/21

E:
x = 4 + t = 4 - 65/21 = 19/21
y = -2t = -2(-65/21) = 130/21
z = -1 - t = -1 - (-65/21) = (65 - 21)/21 = 44/21

E(19/21, 130/21, 44/21)
___________

iv) Find the perpendicular distance of E from the line AB.

Line AB = A + tn = <1, 2, 3> + t<5, -3, 4>
where t is a scalar ranging over the real numbers

E(19/21, 130/21, 44/21)

Define vector v = EA.
v = EA = A - E = <1 - 19/21, 2 - 130/21, 3 - 44/21>
v = <2/21, -88/21, 19/21>

Distance = | v X n | / || n ||
Distance = | <5, -3, 4> X <2/21, -88/21, 19/21> / || n ||
Distance = | 10/21 + 264/21 + 76/21 | / √(5² + (-3)² + 4²)
Distance = (350/21) / √50 = 50/3 / √50 = √50 / 3
Distance = 5√2 / 3
________

Answer 2

Let us put in vector notation the points:
A=(1;2;3) B=(-4;5;-1) and C=(4;0;-1)
So, according to the entitlement,
2(D-A) = B-D
3D = B+2A
D= (B+2A)/3
So, D's coordinates are: (1/3)*(-4+2;5+4;-1+6)
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We have the normal vector (B-A) and a point the plane passes through. As B-A = (-5;3;-4),
the plane has an equation like: -5x+3y-4z+K=0
As we know the point D satisfies the equation, just subs there the point to find the K constant.
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Two lines are parallel when their direction lines are multiple of some other vector. To get easier, we take THAT one as direction line.
So, we know that the line which passes has the equation (parametric)
(x;y;z) = C+t(1;-2;-1) , t a real number.
2 things intercept when for the same variables, they satisfy both equations. so we have to solve, simultanely, the 3 equations of the straight line, and the equation of the plane. The point (x,y,z) found as solution is the Point E.
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For the last one, use the triangle AEB, calculate its area (with cross product) , and remember the Base times the the height is also the area.

Answer 3

the circulate product would be perpendicular to the two vector a and vector b so c is vector [ -18, - 33 , -6 ] yet c is a unit vector so we ought to continuously divide via the value of vector c c = a million/ [3 sqrt(161) ] [ -18, -33 , - 6 ] considering that ok = [ 0, 0 ,a million ] then c.ok = - 2 sqrt (161) / 161 and the respond is absolutely the of that value or 2 sqrt (161) / 161