VECTORS TEST. 1. Show that the vectors a and b given by a = i + j + k and b = 2i + j 3k are perpendicular. [3]

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1 VECTORS TEST Show that the vectors a and b given by a = i + j + k and b = i + j k are perpendicular [] Find a unit vector which is parallel to the vector a [] Find an equation for the line through the points A(,, ) and B(,, 0) [] Show that the line does not intersect the x-axis [] The diagram shows points A and B together with their position vectors a and b relative to an origin O A a O b B The point P lies on AB such that AP : PB = : Find in terms of a and b, an equation for the line through O and P [5] 4 The equations of two lines are r = 4 + α and r = 5 + β a Given that the lines intersect, find the value of the constant a [5] wwwmathsgurucouk Registered to : Sidcot School Page

2 5 The lines l and l are given by r = 4 + s and r = 4 respectively + t i) Find the position vector of the point of intersection of these lines [6] Find the cosine of the acute angle between l and l [5] 6 Describe, briefly, what it means for two lines to be skew [] The points A and B have position vectors i j + k and 4i + j + 4k respectively, and O is the origin i) Find an equation for the line passing through A and B [] Find the position vector of the point P on the line AB such that OP is perpendicular to AB [4] i Show that the line AB and the line with equation r = 5i + k + α(i + j + 4k) are skew lines [6] Total = 48 marks wwwmathsgurucouk Registered to : Sidcot School Page

3 Solutions Using the scalar product, a b = + + () = + = 0 () This means that a and b are indeed perpendicular as required () We have the magnitude of a is given by a = + + = () Hence the vector i + j + k is a unit vector parallel to a () The line passes through the point A with position vector a and is parallel to the vector AB = b a = = () 0 Hence, the line has equation r = a AB = () Any point along the x-axis has both the y-coordinate and the z-coordinate equal to zero () If y = 0, then λ = 0 λ = 5 If z = 0, then = 0 λ = Hence the line does not intersect the x-axis () We have that AP = = (b a) This means that the position vector of the point P is given by p = OA + AP () = a + (b a) which simplifies to give p = a + b () For the line through O and P, we can use O as a fixed point on the line and a direction vector given by OP This means that the required line has equation given by r = λop = λ( a + b) () 4 The lines meet when x = + α = β and y = 4 + α = + β () Solving simultaneously gives α =, β = () Substitute both these values into the equations for the two straight lines to get: z = + α = 5 + βa = 5 + a from which we easily obtain that a = 5 () x 5 i) We have that any point on the line l has coordinates y = 4 + s z 4 x whilst any point on the line l has coordinates y = + t () z wwwmathsgurucouk Registered to : Sidcot School Page

4 The lines therefore meet when x = s = + t and y = 4 + s = t () Solving simultaneously gives s =, t = () Substitute either s in l or t in l to get that the point of intersection has position vector i k () We require the acute angle between the direction vectors and By the scalar product, + () + = 4 4 cos θ () (θ denotes any one of the two possible angles between the lines) 5 = 4 cos θ () cos θ = 5 4 Now, a negative cosine means that the angle is obtuse and hence the cosine of the acute angle between the lines is 5 4 (which equates to cos (80 θ) ) () 6 Two lines are skew if they are i) not parallel () and they do not intersect () i) The line passes through the point A with position vector a and is parallel to the 4 vector AB = b a = = () 4 Hence, the line has equation r = a AB = () λ + First, since P lies on the line, it must be given by λ + for some value of λ () OP being perpendicular to AB means that OP AB = 0 = 0 () ( ) + ( ) + ( ) = 0 which we easily solve to give λ = 0 () This means that P has position vector (ie P = A in this case!) () wwwmathsgurucouk Registered to : Sidcot School Page 4

5 i The two lines in question are given by r = and 5 r = 0 + α 4 First, it is clear that these lines are not parallel () The lines therefore meet if, and when, x = = 5 + α and y = = α () Solving simultaneously gives λ =, α = () Substitute λ = into the equation for the line AB to get the point (7, 4, 7) Substitute α = into the equation for the second line to get the point (7, 4, 9) () This shows that the two line do not meet and hence, because they are also not parallel, the lines are skew () Total = 48 marks wwwmathsgurucouk Registered to : Sidcot School Page 5

(iii) converting between scalar product and parametric forms. (ii) vector perpendicular to two given (3D) vectors

(iii) converting between scalar product and parametric forms. (ii) vector perpendicular to two given (3D) vectors Vector Theory (15/3/2014) www.alevelmathsng.co.uk Contents (1) Equation of a line (i) parametric form (ii) relation to Cartesian form (iii) vector product form (2) Equation of a plane (i) scalar product