MATH 241 Quiz 2 Solutions September 14, 2017

1. (5pts) Show that the line containing the points (2,−1, 3) and (0, 7, 9) is perpendicular to the linecontaining the points (4,−9,−3) and (7,−6,−6).

Solution. Label the pointsP1 = (2,−1, 3)

P2 = (0, 7, 9)

P3 = (4,−9,−3)

P4 = (7,−6,−6)

Then the vector~L1 = ~P1P2 = 〈−2, 8, 6〉

is parallel to the first line, and the vector

~L2 = ~P3P4 = 〈3, 3,−3〉

is parallel to the second line. It suffices to show that ~L1 ⊥ ~L2, which we can do using the dot product.

~L1 · ~L2 = (−2)(3) + (8)(3) + (6)(−3) = −6 + 24− 18 = 0

MATH 241 Quiz 2 Solutions Page 2 of 2

2. (6pts)

(a) (4pts) Find an equation of the plane that contains the point (1,−1, 2) and the line with symmetricequations

x + 2 = y + 1 =z + 5

2

Solution. Let P0 = (1,−1, 2) and let P1 = (−2,−1,−5). Since P1 is on the given line,

~P0P1 = 〈−3, 0,−7〉

is parallel to the plane. The vector~L = 〈1, 1, 2〉

is also parallel to the plane, since it is parallel to the given line contained in the plane. Therefore~L× ~P0P1 is perpendicular to the plane i.e. it is a normal vector to the plane. We compute

~n = ~L× ~P0P1 =

∣∣∣∣∣∣i j k1 1 2−3 0 −7

∣∣∣∣∣∣= [(1)(−7)− (2)(0)]i− [(1)(−7)− (2)(−3)]j + [(1)(0)− (1)(−3)]k

= −7i + j + 3k

Therefore an equation of the plane is

−7(x− 1) + (y + 1) + 3(z − 2) = 0

(b) (2pts) Sketch the plane.

Solution. We will plot the three intercepts and draw the triangle connecting them. For that itwill be easier to rewrite the plane equation as

−7x + y + 3z = −2

Then the intercepts are seen to be (by plugging in 0 for two of the variables)

(0, 0,−2/3), (0,−2, 0), (2/7, 0, 0)

The End.