Multivariable Calculus - A point perpendicular to a plane (vector)?

Question

Let P(3,2,-1), Q(7,-5,4), R(5,6,-3) be points in space.

Find scalar parametric equations for the line that passes through R perpendicular to the line determined by P and Q.

Thanks!! :)

Answer 1

P(3,2,-1), Q(7,-5,4), R(5,6,-3)

let vector PQ be u,
u = 4i-7j+5k

Line for PQ =(3,2,-1)+t(4,-7,5)

Let S(a,b,c) be a point on the line PQ and line RS is perpendicular to PQ.

Again let v represent the vector for RS,
v= (a-5)i + (b-6)j + (c+3)k

If u is perpendicular to v
then by dot product
u.v=0
(4,-7,5).(a-5,b-6,c+3)=0
4a-7b+5z = -37------------------------(1)

Next, find the point S, use scalar parametric equation from line PQ, a=3+4t , b=2-7t , c=-1+5t

Substitute into (1)
4(3+4t)-7(2-7t )+5(1+5t)=-37
t=-1/3

we have S = (5/3 , 13/3 , -8/3)

then v = (-10/3)i + (-5/3)j + (1/3)k

or simply write -10i - 5j +k for it (magnitude not important here because the term t is used)

scalar parametric equations :
x = 5 -10t
y = 6 - 5t
z = -3 +t

Answer 2

We have three points P(3,2,-1), Q(7,-5,4), R(5,6,-3). We need to find the equation of the line thru points P and Q and then find the closest point S on that line to point R.

Let u be the directional vector for the line that passes thru the points P and Q.

u = PQ = = <7-3, -5-2, 4- -1> = <4, -7, 5>

With the directional vector u and one of the points on the line (let's choose P) we can write the equation of the line.

Line PQ = OP + ku = <3, 2, -1> + t<4, -7, 5>
where k is a scalar ranging over the real numbers

The point S can be represented as a point on the line L that is a distance t from P, where t is a ratio of the distance PQ. Conceivably t is not limited to the interval [0,1].

S = P + t

Calculate the magnitude of .

|| Q - P || = || u || = √[4² + (-7)² + 5²] = √(16 + 49 + 25) = √90

Calculate .

= <5-3, 6-2, -3- -1> = <2, 4, -2>

Since RS is perpendicular to PQ the dot product of the vectors is zero.

• = 0
[R - P - t] • = 0

• = t •
• = t || ||²

t = • / || ||²
t = • u / || u ||²
t = <2, 4, -2> • <4, -7, 5> / 90 = (8 - 28 - 10)/90 = -30/90 = -1/3

Solve for S.

S = P + t = (3, 2, -1) - (1/3)<4, -7, 5>
S = (3 - 4/3, 2 + 7/3, -1 - 5/3) = (5/3, 13/3, -8/3)
_____________

Calculate the directional vector v, for the line thru R and S.

v = = <5-5/3, 6-13/3, -3- -8/3> = <10/3, 5/3, -1/3>

Any non-zero multiple of v will also be the directional vector of the line. Multiply by 3.

v = <10, 5, -1>

With the directional vector v and a point on the line R, we can write the equation of the line L thru the points R and S.

L = <5, 6, -3> + t<10, 5, -1>
where t is a scalar ranging over the real numbers

Write the equation for line L in parametric form.

L:
x = 5 + 10t
y = 6 + 5t
z = -3 - t

Answer 3

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