Question

Find a unit vector normal to the plane containing the points P=(5,−4,0),Q=(2,0,−5) and R=(−1,−2,−3). A unit vector normal to plane containing the points P=(5,−4,0),Q=(2,0,−5) and R=(−1,−2,−3) is (Type a vector in terms of i,j and k.)

Answer 1

To find a unit vector normal to the plane containing points P, Q, and R, calculate the vectors between these points, find the cross product of these vectors to get a normal vector, and then normalize it to have magnitude 1.

Step-by-step explanation:

To find a unit vector normal to the plane containing points P=(5,−4,0), Q=(2,0,−5), and R=(−1,−2,−3), you need to perform several steps.

Firstly, determine two vectors that are in the plane. This can be done by subtracting the coordinates of point Q from point P to get vector PQ, and subtracting the coordinates of point R from point Q to get vector QR.

PQ = P - Q = (5 - 2, -4 - 0, 0 - (-5)) = (3, -4, 5)

QR = Q - R = (2 - (-1), 0 - (-2), -5 - (-3)) = (3, 2, -2)

Next, calculate the cross product of vectors PQ and QR to find a vector normal to the plane :

N = PQ × QR = (i, j, k)
(3, -4, 5)
(3, 2, -2)

N = i((-4)×(-2) - 5×2) - j(3×(-2) - 5×3) + k(3×2 - (-4)×3) = i(8 - 10) - j(-6 - 15) + k(6 + 12)

N = -2i + 21j + 18k

The magnitude of N is calculated by √((-2)^2 + 21^2 + 18^2) = √(4 + 441 + 324) = √769 ≈ 27.73

Finally, divide the normal vector by its magnitude to get the unit vector normal to the plane:

Unit Normal Vector = ±(1/27.73)(-2i + 21j + 18k) = ±(-0.072i + 0.757j + 0.649k)

Therefore, the unit vectors normal to the plane are -0.072i + 0.757j + 0.649k and its opposite.