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5. A and B are two points which have position vectors 0A = i+ 2j – 2k, OB = 2i - 3j + 6k respectively. The point P lies on AB such that AP : AB = 1 : 1 where l is a scalar. Show that OP = (1 + 1 )i + (2 – 5 A)j +(-2+ 82 ) k . (a) Given that OP is perpendicular to AB , show that A =- %3D 18 (b) Find the position vector of P for which the angle AOP and the angle POB are the same.

5. A and B are two points which have position vectors 0A = i+ 2j – 2k, OB = 2i - 3j + 6k respectively. The point P lies on AB such that AP : AB = 1 : 1 where l is a scalar. Show that OP = (1 + 1 )i + (2 – 5 A)j +(-2+ 82 ) k . (a) Given that OP is perpendicular to AB , show that A =- %3D 18 (b) Find the position vector of P for which the angle AOP and the angle POB are the same.

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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5. A and B are two points which have position vectors 0A = i+ 2j – 2k, OB = 2i - 3j + 6k respectively. The point P lies on AB such that AP : AB = 1 : 1 where l is a scalar. Show that OP = (1 + 1 )i + (2 – 5 A)j +(-2+ 82 ) k . (a) Given that OP is perpendicular to AB , show that A =- %3D 18 (b) Find the position vector of P for which the angle AOP and the angle POB are the same.
Transcribed Image Text:5. A and B are two points which have position vectors 0A = i+ 2j – 2k, OB = 2i - 3j + 6k respectively. The point P lies on AB such that AP : AB = 1 : 1 where l is a scalar. Show that OP = (1 + 1 )i + (2 – 5 A)j +(-2+ 82 ) k . (a) Given that OP is perpendicular to AB , show that A =- %3D 18 (b) Find the position vector of P for which the angle AOP and the angle POB are the same.
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