Topic 4: Vectors Markscheme 4.6 Intersection of Lines and Planes Paper 2

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1 Topic : Vectors Markscheme. Intersection of Lines and Planes Paper. Using an elimination method, x y + z x y z x y x + y 8 y Therefore x, y, z Using matrices, x y z x y z 5 (using a graphic display calculator) Therefore x, y, z. For the line of intersection: x + y + z x y + z x + z 8x + y + z x y + z x y The equation of the line of intersection is x y + z Let x y + z z, z, y (,, ) x + y Let z x y x, x, y (,, ) y + z + (or equivalent) (C) The equation of the line of intersection is r + λ (or equivalent) (C) Higher Level [] []

2 . METHOD R : R : 5 R : R R : (M) Let z t, then y t + and x t. Therefore the line of intersection is x t, y t +, z t (or equivalent). (C) METHOD Let z > x + y x + y >x, y The direction vector of the line of intersection is i j k 5 i j k Therefore the line of intersection is r + t (or equivalent) (A) (C). (a) u v i j k i j + k. (M)(AG) [] i + j 5k (b) w λ + µ λ µ λ + µ (C) The line of intersection of the planes is parallel to u v. Now, w. (u v) λ + µ + 8λ µ 5λ µ for all λ, µ. (C) Therefore, w is perpendicular to the line of intersection of the given planes. (AG) The line of intersection of the planes is perpendicular to u and to v, (M) so it will be perpendicular to the plane containing u and v, that is, (R) to all vectors of the form λu + µv w. (C) []

3 5. (a) The system is x y z x y. z Therefore, the solution is x, y, z. (G) The system of equations is: R R R R R R 5 9 R R R 5 9 (M) Back substitution gives x, y, z. (C) x, y, z. (G) (b) i j k v i j + k i j 5k. (C) (c) u m(i + j k) + n(i + j + k) (m + n)i + (m + n)j + ( m + n)k (C) Therefore, v. u (m + n) (m + n) 5( m + n) m + n m n + m 5n, for all m and n. (C) That is, v is perpendicular to u for all values of m and n. (AG) (d) v is perpendicular to both a and b [from part (b)]. Therefore, v. a v. b, so v. u m(v. a) + n(v. b), and hence v is perpendicular to u for all values of m and n. (R)(AG) The normal to the plane, i j + k, and v are both perpendicular to the required line, l. Therefore, the direction of l is given by i j k 5 5 v (i j + k) 5 i j + k i j + k (C) Thus, an equation for l is r i j + k + λ(i + j 5k), where λ is a scalar. [Any form of the correct answer is quite acceptable.] (C) []

4 . (a) (i) Using row reduction, x + y + z k y + z k y z k 9 x + y + z k y + z k z k Not a unique solution because the coefficient of z in the third equation is zero. (ii) In order for the system to have a solution, (R) k, consistent for k. (i) (ii) Consider The zero value confirms that the equations do not have a unique solution. (R) Consider k 5 9 9k Consistent when this determinant is zero, ie k. (R) ( λ ) ( ) (b) The general solution is z λ, y, x λ. [9]

5 5. (a) x + y + z I x + y z II 5x + y + bz III Solving for z III II x + bz + z IV also II I x z V V + IV bz 5z z 5 b (b) If b 5, z is undefined. (R) Hence equation has unique solution if b 5. [] 8. (a) z y x x y + z (b) (i) + P lies in π (AG) (ii) r + λ 5 (c) cos θ cos θ (.59) θ.5 radians (or θ. ) The angle between the planes is π.5.98 radians (or ) (N) []

6 9. (a) (i) For points which lie in π and π () λ + µ + s+ t () + λ µ s+ t () + 8λ 9µ + s+ t subtracting () from () λ + µ λ µ (AG) N (ii) On the line of intersection λ µ an equation of the line is r (b) The plane π contains, eg the point (,, ). + λ + λ 8 9 x The equation of the plane is y 5. z + λ N 5 The cartesian equation of the plane is x y+ z 5. N (c) Intersection between line r + λ and π. x y+ z 5 ( λ) ( λ) + λ 5 This equation is satisfied by any real value of λ the planes intersect at the line r + λ. (R) N []. (a) Substituting for a, b and c into c ma + nb Forming any of the following equations AA m + n Eq() m + n 5 Eq() m + n Eq() Note: Accept equations in vector form. Solving for m and n m and n A N

7 (b) METHOD a b i j k i j + k A Attempting to find a b ( 5 ) u METHOD 5 i j + k (i j + k) ( ) Stating equations derived from a u and b u where u xi + yj + zk. x y + z x + y + z Eq() Eq() Attempting to solve the above system of equations Solution sets include x z and y z A A y x and z x A z y and x y Note: Accept any correct numerical solution such as x, y, z. Using x + y + z (ie u ) to find values for x, y and z. Either u ( i + j k) or u ( i j + k) (c) (i) METHOD A A Equation of π is of the form x + y + z d Substituting (,, ) ( d ) M ( x + y + z A N METHOD r (i + j + k) d Evaluate the scalar product a (i + j + k) ( ) M x + y + z A N (ii) L(,, ), M,, and N,, A A N N

8 (d) (i) P has coordinates (x, y, z) ( λ, λ, λ) Substituting the coordinates of P into the equation of π λ + λ + λ A λ P,, A N (ii) Distance + + M (e) or equivalent (.55) A N (Given θ is the angle between π and a line and α is the angle π between the normal and a line) cos α cos θ sin θ Using the scalar product eg ( sin ) ( i j + k) ( i + j + k) a b sin θ or a b (R) a b cosα M a b θ i j + k i + j + k θ.88 (. ) (or equivalent) arcsin A N []. (from GDC) x + λ A y λ A r i j + λ i + j + k AAA N [] 8