Find a point on the line x + 2 3 = y + 1 2 = z - 3 2 at a distance of 3 2 units from the point ( 1 , 2 , 3 ) .

Given that, x + 2 3 = y + 1 2 = z - 3 2 = λ (say) ∴ ( 3 λ - 2 , 2 λ - 1 , 2 λ + 3 ) is any general point on the line. Now, the distance of the point ( 3 λ - 2 , 2 λ - 1 , 2 λ + 3 ) from the point ( 1 , 2 , 3 ) is 3 2 units, then ( 3 λ - 2 - 1 ) 2 + ( 2 λ - 1 - 2 ) 2 + ( 2 λ + 3 - 3 ) 2 = ( 3 2 ) ⇒ ( 3 λ - 3 ) 2 + ( 2 λ - 3 ) 2 + 4 λ 2 = 18 ⇒ 9 λ 2 - 18 λ + 9 + 4 λ 2 - 12 λ + 9 + 4 λ 2 = 18 ⇒ 17 λ 2 - 30 λ = 0 ⇒ λ ( 17 λ - 30 ) = 0 ⇒ λ = 0 or λ = 30 17 Hence, the required point on the line is ( - 2 , - 1 , 3 ) or 56 17 , 43 17 , 77 17

Show that the lines x - 4 1 = y + 3 - 4 = z + 1 7 and x - 1 2 = y + 1 - 3 = z + 10 8 intersect. Also find the coordinates of their point of intersection.

The given lines are x - 4 1 = y + 3 - 4 = z + 1 7 = λ (say) . . . 1 and x - 1 2 = y + 1 - 3 = z + 10 8 = μ (say) . . . 2 . Then, any point on the line 1 is P λ + 4 , - 4 λ - 3 , 7 λ - 1 . . . 3 . And any point on the line 2 is Q 2 μ + 1 , - 3 μ - 1 , 8 μ - 10 . . . ( 4 ) . Clearly lines 1 and 2 will intersect if P = Q . i.e., λ + 4 , - 4 λ - 3 , 7 λ - 1 = 2 μ + 1 , - 3 μ - 1 , 8 μ - 10 For some particular value of λ and μ . ⇒ λ + 4 = 2 μ + 1 , - 4 λ - 3 = - 3 μ - 1 and 7 λ - 1 = 8 μ - 10 ⇒ λ - 2 μ = - 3 . . . 5 ⇒ - 4 λ + 3 μ = 2 . . . 6 ⇒ 7 λ - 8 μ = - 9 . . . 7 On multiplying the equation 5 by 4 and then adding equation 6 , we get ⇒ 4 λ - 2 μ + - 4 λ + 3 μ = 4 - 3 + 2 4 λ - 8 μ - 4 λ + 3 μ = - 12 + 2 - 5 μ = - 10 μ = 2 On putting the value of μ in equation 5 , we get λ - 2 2 = - 3 ⇒ λ - 4 = - 3 ⇒ λ = 1 On putting the value of λ and μ in equation 7 , we get 7 1 - 8 2 = - 9 ⇒ 7 - 16 = - 9 ⇒ - 9 = - 9 , which is true. Hence, lines 1 and 2 intersect and their point of intersection is P 1 + 4 , - 4 1 - 3 , 7 1 - 1 i.e., P 5 , - 7 , 6 .

Show that the lines r → = 3 i ^ + 2 j ^ - 4 k ^ + λ ( i ^ + 2 j ^ + 2 k ^ ) ; r → = 5 i ^ - 2 j ^ + μ ( 3 i ^ + 2 j ^ + 6 k ^ ) are intersecting. Hence, find their point of intersection.

Given that, r → = 3 i ^ + 2 j ^ - 4 k ^ + λ ( i ^ + 2 j ^ + 2 k ^ ) r → = ( 3 + λ ) i ^ + 2 + 2 λ j ^ + ( - 4 + 2 λ ) k ^ r → = 5 i ^ - 2 j ^ + μ ( 3 i ^ 2 j ^ + 6 k ^ ) r → = 5 + 3 μ i ^ + - 2 + 2 μ j ^ + 6 μ k ^ . . . 1 Any point on line (1) is 3 + λ i ^ + ( 2 + 2 λ ) j ^ + ( - 4 + 2 λ ) k ^ . . . 2 It will lie on line (2) iff 3 + λ i ^ + ( 2 + 2 λ ) j ^ + ( - 4 + 2 λ ) k ^ = 5 + 3 μ i ^ + ( - 2 + 2 μ ) j ^ + 6 μ k ^ is consistent Comparing coefficients of i ^ , j ^ , k ^ 3 + λ = 5 + 3 μ , 2 + 2 λ = + 2 + 2 μ , - 4 + 2 λ = 6 μ are consistent Or λ - 3 μ = 2 , 2 λ - 2 μ = - 4 2 λ - 6 μ = 4 are consistent λ - 3 μ = 2 λ - μ = - 2 → λ - ( - 2 ) = - 2 - + + μ = - 2 λ + 2 = - 2 λ = - 4 - 2 μ = 4 Put λ & μ i n λ - 3 μ = 2 - 4 - 3 ( - 2 ) = 2 - 4 + 6 = 3 2 = 2 Which is True Hence the lines in Intersects & point of Intersection is for it Put λ = - 4 3 - 4 i + 2 + 2 × ( - 4 ) j + - 4 + 2 × ( - 4 ) k Or - i - 6 j - 12 k

Show that the lines x - 1 3 = y + 1 2 = z - 1 5 and x + 2 4 = y - 1 3 = z + 1 - 2 do not intersect.

The coordinates of any point on the first line are given by x - 1 3 = y + 1 2 = z - 1 5 = λ ⇒ x = 3 λ + 1 y = 2 λ - 1 z = 5 λ + 1 The coordinates of a general point on the first line are 3 λ + 1 , 2 λ - 1 , 5 λ + 1 The coordinates of any point on the second line are given by x + 2 4 = y - 1 3 = z + 1 - 2 = μ ⇒ x = 4 μ - 2 y = 3 μ + 1 z = - 2 μ - 1 The coordinates of a general point on the second line are 4 μ - 2 , 3 μ + 1 , - 2 μ - 1 If the lines intersect, then they have a common point. So, for some values of λ and μ we must have 3 λ + 1 = 4 μ - 2 , 2 λ - 1 = 3 μ + 1 , 5 λ + 1 = - 2 μ - 1 ⇒ 3 λ - 4 μ = - 3 . . . ( 1 ) 2 λ - 3 μ = 2 . . . ( 2 ) 5 λ + 2 μ = - 2 . . . ( 3 ) Solving 1 and 2 , we get λ = - 17 μ = - 12 Substituting λ = - 17 and μ = - 12 in ( 3 ) , we get L H S = 5 λ + 2 μ = 5 - 17 + 2 - 12 = - 109 ≠ - 2 ⇒ L H S ≠ R H S Hence, the given lines do not intersect.

A line with direction 2 , 7 , - 5 is drawn to intersect the lines x - 5 3 = y - 7 - 1 = z + 2 1 and x + 3 - 3 = y - 3 - 3 = z - 6 4 at P and Q respectively. Then find length of P Q .

The given equation are, x - 5 3 = y - 7 - 1 = z + 2 1 = t → 1 x + 3 - 3 = y - 3 - 3 = z - 6 4 = s → 2 The general points on the given lines are P ( 5 + 3 t , 7 - t , - 2 + t ) and Q ( - 3 - 3 s , 3 + 2 s , 6 + 4 s ) The direction ratios of P Q are < 3 - 3 s - 5 - 3 t , 3 + 2 s - 7 + t , 6 + 4 s + 2 - t > or < - 8 - 3 s - 3 t , - 4 + 2 s + t , + 8 + 4 s - t > If P Q is the desired line then the direction ratios of P Q Should be proportional to < 2 , 7 , - 5 > Therefore, - 8 - 3 s - 3 t 2 = - 4 + 2 s + t 7 = 8 + 4 s - t - 5 Taking first and second numbers, then - 56 - 21 s - 21 t = - 8 + 4 s + 2 t ⇒ 25 s + 23 t = - 48 … ( i ) Taking second and third numbers, then 20 - 10 s - 5 t = 56 + 28 s - 7 t ⇒ 38 s - 2 t = - 36 … i i Solving ( i ) and ( i i ) , then we get: s = - 1 and t = - 1 So, the coordinates of P and Q are ( 5 + 3 ( - 1 ) , 7 - ( - 1 ) , - 2 - 1 ) = ( 2 , 8 , - 3 ) and ( - 3 - 3 ( - 1 ) , 3 + 2 ( - 1 ) , 6 + 4 ( - 1 ) ) = ( 0 , 1 , 2 ) Thus, the length of the line intercepted between the lines is | P Q | = ( 0 - 2 ) 2 + ( 1 - 8 ) 2 + ( 2 + 3 ) 2 = 78 units.

E M B I B E

Understanding ISC Mathematics Class 12 Volume 2>Chapter 2 - Three Dimensional Geometry>EXERCISE

M L Aggarwal Solutions for Chapter: Three Dimensional Geometry, Exercise 2: EXERCISE

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Three Dimensional Geometry, Exercise 2: EXERCISE

Attempt the practice questions on Chapter 2: Three Dimensional Geometry, Exercise 2: EXERCISE with hints and solutions to strengthen your understanding. Understanding ISC Mathematics Class 12 Volume 2 solutions are prepared by Experienced Embibe Experts.

Questions from M L Aggarwal Solutions for Chapter: Three Dimensional Geometry, Exercise 2: EXERCISE with Hints & Solutions

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 2 - Three Dimensional Geometry>EXERCISE>Q 15

If the points $(- 1, 3, 2), (- 4, 2, - 2) and (5, 5, λ)$ are collinear, then find the value of $λ$ .

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 2 - Three Dimensional Geometry>EXERCISE>Q 16

Find a point on the line $\frac{x + 2}{3} = \frac{y + 1}{2} = \frac{z - 3}{2}$ at a distance of $3 \sqrt{2}$ units from the point $(1, 2, 3) .$

HARD

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 2 - Three Dimensional Geometry>EXERCISE>Q 17

Find the points on the line through the points $A (3, 4, - 1) and B (5, 1, 5)$ at a distance of $7$ units from the midpoint of line segment $A B .$

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 2 - Three Dimensional Geometry>EXERCISE>Q 18

Show that the lines $\frac{x - 4}{1} = \frac{y + 3}{- 4} = \frac{z + 1}{7}$ and $\frac{x - 1}{2} = \frac{y + 1}{- 3} = \frac{z + 10}{8}$ intersect. Also find the coordinates of their point of intersection.

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 2 - Three Dimensional Geometry>EXERCISE>Q 18

Show that the lines $\vec{r} = 3 \hat{i} + 2 \hat{j} - 4 \hat{k} + λ (\hat{i} + 2 \hat{j} + 2 \hat{k})$ ; $\vec{r} = 5 \hat{i} - 2 \hat{j} + μ (3 \hat{i} + 2 \hat{j} + 6 \hat{k})$ are intersecting. Hence, find their point of intersection.

HARD

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 2 - Three Dimensional Geometry>EXERCISE>Q 19

Prove that the line through $A (0, - 1, - 1) and B (4, 5, 1)$ intersects the line through $C (3, 9, 4) and D (- 4, 4, 4)$ .

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 2 - Three Dimensional Geometry>EXERCISE>Q 19

Show that the lines $\frac{x - 1}{3} = \frac{y + 1}{2} = \frac{z - 1}{5}$ and $\frac{x + 2}{4} = \frac{y - 1}{3} = \frac{z + 1}{- 2}$ do not intersect.

EASY

12th ICSE

IMPORTANT

Understanding ISC Mathematics Class 12 Volume 2>Chapter 2 - Three Dimensional Geometry>EXERCISE>Q 20

A line with direction $2, 7, - 5$ is drawn to intersect the lines $\frac{x - 5}{3} = \frac{y - 7}{- 1} = \frac{z + 2}{1}$ and $\frac{x + 3}{- 3} = \frac{y - 3}{- 3} = \frac{z - 6}{4}$ at $P and Q$ respectively. Then find length of $P Q$ .

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