Processing math: 100%

User interface language: English | Español

Date May 2011 Marks available 2 Reference code 11M.2.hl.TZ2.11
Level HL only Paper 2 Time zone TZ2
Command term Find Question number 11 Adapted from N/A

Question

The points P(−1, 2, − 3), Q(−2, 1, 0), R(0, 5, 1) and S form a parallelogram, where S is diagonally opposite Q.

Find the coordinates of S.

[2]
a.

The vector product PQ×PS=(137m). Find the value of m .

[2]
b.

Hence calculate the area of parallelogram PQRS.

[2]
c.

Find the Cartesian equation of the plane, 1 , containing the parallelogram PQRS.

[3]
d.

Write down the vector equation of the line through the origin (0, 0, 0) that is perpendicular to the plane 1 .

[1]
e.

Hence find the point on the plane that is closest to the origin.

[3]
f.

A second plane, 2 , has equation x − 2y + z = 3. Calculate the angle between the two planes.

[4]
g.

Markscheme

PQ=(113) , SR=(0x5y1z)     (M1)

point S = (1, 6, −2)     A1

[2 marks]

a.

PQ=(113)PS=(241)     A1

PQ×PS=(1372)

m = −2     A1

[2 marks]

b.

area of parallelogram PQRS =|PQ×PS|=(13)2+72+(2)2     M1

=222=14.9     A1

[2 marks]

c.

equation of plane is −13x + 7y − 2z = d     M1A1

substituting any of the points given gives d = 33

−13x + 7y − 2z = 33     A1

[3 marks]

d.

equation of line is r=(000)+λ(1372)     A1

Note: To get the A1 must have r= or equivalent.

 

[1 mark]

e.

169λ+49λ+4λ=33     M1

λ=33222 (=0.149)     A1

closest point is (14374,7774,1137) (=(1.93, 1.04,  - 0.297))     A1

[3 marks]

f.

angle between planes is the same as the angle between the normals     (R1)

cosθ=13×1+7×22×1222×6     M1A1

θ=143 (accept θ=37.4 or 2.49 radians or 0.652 radians)     A1

[4 marks]

g.

Examiners report

This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = ...

a.

This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = ...

b.

This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = ...

c.

This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = ...

d.

This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r= ...

e.

This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = ...

f.

This was a multi-part question that was well answered by many candidates. Wrong answers to part (a) were mainly the result of failing to draw a diagram. Follow through benefitted many candidates. A high proportion of candidates lost the mark in (e) by not writing their answer as an equation in the form r = ...

g.

Syllabus sections

Topic 4 - Core: Vectors » 4.1 » Concept of a vector.

View options