MATH1131 Assignment Questions.docx - MATH1131 Assignment Questions Question 1 Ans a A possible n1 is 1 2 \u22121 \u00d7 \u22121 0 = 3 2 1 2 A possible n2 is 1 2 1

# MATH1131 Assignment Questions.docx - MATH1131 Assignment...

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MATH1131 Assignment Questions Question 1 Ans: a. A possible n 1 is ( 1 1 2 ) × ( 2 0 1 ) = ( 1 3 2 ) A possible n 2 is ( 1 2 1 ) × ( 2 1 0 ) = ( 1 2 3 ) As the n 1 is not a multiple of n 2 and n 1 ≠λ n 2 , which implies that Π 1 , Π 2 are not parallel planes. Therefore, they must intersect in a line. b. Using the point-normal form n∙ ( x a ) = 0 Cartesian equation for Π 1: ( 1 3 2 ) ( x ( 0 1 1 ) ) = 0 ( 1 3 2 ) x = ( 1 3 2 ) ( 0 1 1 ) ( 1 3 2 ) ( x 1 x 2 x 3 ) = ( 1 3 2 ) ( 0 1 1 ) Rearranging and simply x 1 + 3 x 2 + 2 x 3 = 5 Cartesian equation for Π 2: ( 1 2 3 ) ( x ( 0 0 0 ) ) = 0 ( 1 2 3 ) x = ( 1 2 3 ) ( 0 0 0 ) ( 1 2 3 ) ( x 1 x 2 x 3 ) = ( 1 2 3 ) ( 0 0 0 ) Rearranging and simplify
x 1 + 2 x 2 + 3 x 3 = 0 c. First method Let x 1 be the parameter ω R For Π 1 ω + 5 = 3 x 2 + 2 x 3 For Π 2 ω =− 2 x 2 3 x 3 Now expressions of x 2 , x 3 in terms of ω are x 2 = ω + 3 x 3 =− ω 2 Therefore, the parametric vector form of the line of intersection L can be found: x = ( x 1 x 2 x 3 ) = ( ω ω + 3 ω 2 ) = ( 0 3 2 ) + ω ( 1 1 1 ) for ω R d. Second method From Π 2 x 1 = μ 1 + 2 μ 2 x 2 =− 2 μ 1 μ 2 x 3 = μ 1 Substitute expressions into Cartesian equation for Π 1 ( μ 1 + 2 μ 2 ) + 3 ( 2 μ 1 μ 2 ) +
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