Find the image of the point ( 0 , 2 , 3 ) in the line x + 3 5 = y - 1 2 = z + 4 3

Given that Equation of line is x + 3 5 = y - 1 2 = z + 4 3 The co-ordinates of the point are ( 0 , 2 , 3 ) Let x + 3 5 = y - 1 2 = z + 4 3 = λ So the foot of the perpendicular is ( 5 λ – 3 , 2 λ + 1 , 3 λ – 4 ) The direction ratios of the perpendicular is ( 5 λ – 3 - 0 ) : ( 2 λ + 1 - 2 ) : ( 3 λ - 4 - 3 ) ⇒ ( 5 λ – 3 ) : ( 2 λ – 1 ) : ( 3 λ – 7 ) Direction ratio of the line is 5 : 2 : 3 From the direction ratio of the line and the direction ratio of its perpendicular, we have 5 ( 5 λ - 3 ) + 2 ( 2 λ – 1 ) + 3 ( 3 λ – 7 ) = 0 ⇒ 25 λ – 15 + 4 λ – 2 + 9 λ – 21 = 0 ⇒ 38 λ = 38 ⇒ λ = 1 So, the foot of the perpendicular is ( 2 , 3 , - 1 ) The foot of the perpendicular is the mid-point of the line joining ( 0 , 2 , 3 ) a n d ( α , β , γ ) So, we have α + 0 2 = 2 ⇒ α = 4 β + 2 2 = 3 ⇒ β = 4 γ + 3 2 = - 1 ⇒ γ = - 5 So, the image is ( 4 , 4 , - 5 )

Find the image of the point ( 5 , 9 , 3 ) in the line x - 1 2 = y = 2 3 = z - 3 4

Given that e quation of line is x - 1 2 = y - 2 3 = z - 3 4 The co-ordinates of the point are ( 5 , 9 , 3 ) Let x - 1 2 = y - 2 3 = z - 3 4 = λ So the foot of the perpendicular is ( 2 λ + 1 , 3 λ + 2 , 4 λ + 3 ) The direction ratios of the perpendicular is ( 2 λ + 1 - 5 ) : ( 3 λ + 2 - 9 ) : ( 4 λ + 3 - 3 ) ⇒ ( 2 λ – 4 ) : ( 3 λ – 7 ) : ( 4 λ ) Direction ratio of the line is 2 : 3 : 4 From the direction ratio of the line and the direction ratio of its perpendicular, we have 2 ( 2 λ – 4 ) + 3 ( 3 λ – 7 ) + 4 ( 4 λ ) = 0 ⇒ 4 λ – 8 + 9 λ – 21 + 16 λ = 0 ⇒ 29 λ = 29 ⇒ λ = 1 So, the foot of the perpendicular is ( 3 , 5 , 7 ) The foot of the perpendicular is the mid-point of the line joining ( 5 , 9 , 3 ) and ( α , β , γ ) So, we have α + 5 2 = 3 ⇒ α = 1 β + 9 2 = 5 ⇒ β = 1 γ + 3 2 = 7 ⇒ γ = 11 So, the image is ( 1 , 1 , 11 )

Find the image of the point ( 2 , - 1 , 5 ) in the line r → = ( 1 i ^ - 2 j ^ - 8 k ^ ) + λ ( 10 i ^ - 4 j ^ - 11 k ^ )

Given: Point ( 2 , - 1 , 5 ) Equation of line = ( 11 i ^ - 2 j ^ - 8 k ^ ) + λ ( 10 i ^ - 4 j ^ - 11 k ^ ) The equation of line can be re-arranged as x - 11 10 = x + 2 - 4 = x + 8 - 11 = r The general point on this line is ( 10 r + 11 , - 4 r – 2 , - 11 r – 8 ) Let N be the foot of the perpendicular drawn from the point P ( 2 , 1 , - 5 ) on the given line. Then, this point is N ( 10 r + 11 , - 4 r – 2 , - 11 r – 8 ) for some fixed value of r. D.r.’s of PN are ( 10 r + 9 , - 4 r - 3 , - 11 r - 3 ) D.r.’s of the given line is 10 , - 4 , - 11 . Since, PN is perpendicular to the given line, we have, 10 ( 10 r + 9 ) – 4 ( - 4 r – 3 ) – 11 ( - 11 r – 3 ) = 0 100 r + 90 + 16 r + 12 + 121 r + 33 = 0 237 r = 135 r = 135 237 Then, the image of the point is α - 11 - 11 = 0 , β + 2 7 = 1 , γ + 8 9 = 1 Therefore, the image is ( 0 , 5 , 1 ) .

E M B I B E

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>3D Geometry>Chapter 27 - Straight Line in Space>EXERCISE

R S Aggarwal Solutions for Exercise 1: EXERCISE

Author:R S Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal Solutions for Exercise 1: EXERCISE

Attempt the free practice questions from Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12 solutions are prepared by Experienced Embibe Experts.

Questions from R S Aggarwal Solutions for Exercise 1: EXERCISE with Hints & Solutions

MEDIUM

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 27 - Straight Line in Space>EXERCISE>Q 19

Find the vector and Cartesian equations of the line joining the points whose position vectors are $(\hat{i} - 2 \hat{j} + \hat{k}) and (\hat{i} + 3 \hat{j} - 2 \hat{k})$

MEDIUM

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 27 - Straight Line in Space>EXERCISE>Q 20

Find the vector equation of a line passing through the point $A (3, - 2, 1)$ and parallel to the line joining the points $B (- 2, 4, 2)$ and $C (2, 3, 3) .$ Also, find the Cartesian equations of the line.

MEDIUM

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 27 - Straight Line in Space>EXERCISE>Q 21

Find the vector equation of a line passing through the point having the position vector $(\hat{i} + 2 \hat{j} - 3 \hat{k})$ and parallel to the line joining the points with position vectors $(\hat{i} - \hat{j} + 5 \hat{k}) and (2 \hat{i} + 3 \hat{j} - 4 \hat{k})$ and Also, find the Cartesian equivalents of this equation.

HARD

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 27 - Straight Line in Space>EXERCISE>Q 22

Find the coordinates of the foot of the perpendicular drawn from the point $A (1, 2, 1)$ to the line joining the points $B (1, 4, 6) a n d C (5, 4, 4) .$

HARD

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 27 - Straight Line in Space>EXERCISE>Q 23

Find the coordinates of the foot of the perpendicular drawn from the point $A (1, 8, 4)$ to the line joining the points $B (0, - 1, 3) a n d C (2, - 3, - 1)$ .