Let $P$, $Q$ and $R$ be the points $$P = \left (-3, -1, 2\right ), Q = \left (-1, 2, -2\right ), R = \left (2, 3, 0\right ).$$

Find a unit vector orthogonal to the plane through $P$, $Q$ and $R$ that has a positive 1st component.

What is its 3rd component?

Question

Let $P$, $Q$ and $R$ be the points $$P = \left (-3, -1, 2\right ), Q = \left (-1, 2, -2\right ), R = \left (2, 3, 0\right ).$$

Find a unit vector orthogonal to the plane through $P$, $Q$ and $R$ that has a positive 1st component.

What is its 3rd component?

BlackTom AI · Accepted Answer

The problem asks for a unit vector orthogonal to the plane through P, Q, and R with a positive first component, and then to report its third component.

First, compute the vectors along the plane from P to Q and from P to R:
- PQ = Q - P = (-1 - (-3), 2 - (-1), -2 - 2) = (2, 3, -4)
- PR = R - P = (2 - (-3), 3 - (-1), 0 - 2) = (5, 4, -2)

Next, find a normal vector to the plane by taking the cross product n = PQ × PR:
- n_x = 3*(-2) - (-4)*4 = -6 - (-16) = 10
- n_y = -(2*(-2) - (-4)*5) = -(-4 - (-20)) = -(16) = -16
- n_z = 2*4 - 3*5 = 8 - 15 = -7
So a normal vector is n = (10, -16, -7).

To obtain a unit vector, divide by its magnitude. The magnitude is ||n|| = sqrt(10^2 + (-16)^2 + (-7)^2) = sqrt(100 + 256 + 49) = sqrt(405) = 9*sqrt(5).

Thus, a unit normal with a positive first component is u = n / ||n|| = (10, -16, -7) / (9*sqrt(5)).

Reading off the third component of this unit vector gives u_z = -7 / (9*sqrt(5)).

Therefore, the third component is -7/(9√5).

MATH1061/1002/1021 (ND) MATH1061 Canvas Quiz 5

Let \(P\), \(Q\) and \(R\) be the points \[P = \left (-3, -1, 2\right ), Q = \left (-1, 2, -2\right ), R = \left (2, 3, 0\right ).\] Find a unit vector orthogonal to the plane through \(P\), \(Q\) and \(R\) that has a positive 1st component. What is its 3rd component?

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