help Symmetric equations for the line that passes through P (3, 7, 2) and is par

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Symmetric equations for the line that passes through P (3, 7, 2) and is parallel to the line: r (t) = (2,-l,4)+t(3,4,-6) are: x - 2/3=y +1=z - 4/-6 b. x = 2 + 3t,y = -1+ 4t,z = 4-6t c. x - 3/3=y - 7/4=z-2/-6 d. x = 3 + 3t, y = 7 + 4t, z = 2 - 6t e.None of these Find an equation for the plane which passes through the point P(-1, 2, -3) and is parallel to the plane x - 2y + 4z -3 = 0 a. (x+1) - 2(y - 2) + 4(z + 3) = 0 b. (x-l)-2(y + 2) + 4(z-3) = 0 c. (x+l) + 2(y-2)-4(z+3) = 0 d. (x-l) + 2(y + 2)-4(z-3) = 0 e. None of these

Explanation / Answer

6)The directional vector of the given line is:

v = <3,4,-6>

A point P, on the line can be found by letting parameter t = 0.
P(2, -1, 4)

A second directional vector v, of the plane can be found from the two identified points in the plane P(2,-1,4) and Q(3,7,2).

PQ = <Q - P> = <3-2, 7+1, 2-4> = <1,8,-2>

The normal vector n, of the plane is orthogonal to both directional vectors. Take the cross product.

n = v X PQ = <3,4,-6> X <1,8,-2>=<40,0,20>

With the normal vector of the plane and a point in the plane we can write the equation of the plane. Let's choose point P(2,-1,4). Remember, the normal vector of a plane is orthogonal to any vector that lies in the plane. And the dot product of orthogonal vectors is zero. Define R(x,y,z) to be an arbitrary point in the plane. Then vector PR lies in the plane.

n

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