P=(-3,4,5) Q=(7,-5,-4) R=(0,5,3) Determine an equation of the line containing th

Question

P=(-3,4,5) Q=(7,-5,-4) R=(0,5,3) Determine an equation of the line containing the point (1,2,3) that is parallel to the plane determined by P, Q, and R I know the equation of the plane P, Q, and R is 27x-7y+37z=76 How would I go about this? P=(-3,4,5) Q=(7,-5,-4) R=(0,5,3) Determine an equation of the line containing the point (1,2,3) that is parallel to the plane determined by P, Q, and R I know the equation of the plane P, Q, and R is 27x-7y+37z=76 How would I go about this? Q=(7,-5,-4) R=(0,5,3) Determine an equation of the line containing the point (1,2,3) that is parallel to the plane determined by P, Q, and R I know the equation of the plane P, Q, and R is 27x-7y+37z=76 How would I go about this?

Explanation / Answer

equation of plane through P=(-3,4,5), Q=(7,-5,-4), R=(0,5,3)

Let equation of plane = ax+by+cZ = d

then,

-3a+4b+5c = d

7a-5b-4c = d

5b+3c = d

solving these equations

a = 27,b = -7, c = 37, d = 76

plane = >

27x-7y+37z = 76

there are lots of line which are passing through the given point and parallel to the given plane (basically you will get a plane)

normal to the plane = <27,-7,37>

equation of plane passing though a point and pand parallel to the plane

=> <27,-7,37>.<x-1,y-2,z-3>=0

=>          27x-7y+37z=124

could you please check your question once again. Is it equation of line parallel to the plane or perpendicular to the plane?

if it is second case then,

Since we have the plane 27x-7y+37z = 76 , we know that the vector [27, -7, 37] is perpendicular to the plane.
With the point (1,2,3), we get the parametric equations for the line
x = 1 + 27t, y = 2 -7t, z = 3 +37t
The symmetric equation can be found by solving for t
t = (x - 1)/27 = (2-y)/7 = (z-3)/37

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.