RD Sharma Class 12 Ex 28.3 Solutions Chapter 28 The Straight Line in Space

Here we provide RD Sharma Class 12 Ex 28.3 Solutions Chapter 28 The Straight Line in Space for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 28.3 Solutions Chapter 28 The Straight Line in Space book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter28
Exercise28.3
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 28.3 Solutions Chapter 28 The Straight Line in Space

Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed

Question 1. Show that the lines \frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}  and \frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}  intersect and find their point of intersection.

Solution:

Given that the coordinates of any point on the first line are 

\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}=\lambda

⇒ x = λ, y = 2λ + 2, z = 3λ – 3 

The coordinates of a general point on the second line are given by:

\frac{-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}=\mu

⇒ x = 2μ + 2, y = 3μ + 6, z = 4μ + 3

If the lines intersect, for some values of λ and μ, we must have:

λ – 2μ = 2              ……(1)

2λ – 3μ = 4          ……(2)

3λ – 4μ = 6           …..(3)

Solving this system of equations, we get

λ = 2 and μ = 0

On substituting the values in eq(3), we have

LHS = 3(2) – 4(0) 

= 6 = RHS

Thus, the given lines intersect at (2, 6, 3).

Question 2. 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.

Solution:

Given that the coordinates of any point on the first line are 

\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}=\lambda

⇒ x = 3λ + 1, y = 2λ – 1, z = 5λ + 1

The coordinates of a general point on the second line are given by:

\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}=\mu

⇒ x = 4μ – 2, y = 3μ + 1, z = -2μ – 1

If the lines intersect, for some values of λ and μ, we must have:

3λ – 4μ = -3             ……(1)

2λ – 3μ = 2              ……(2)

5λ + 2μ = -2            …..(3)

Solving this system of equations, we get

λ = -17 and μ = -12

On substituting the values in eq(3), we have

LHS = 3(-17) + 2(-12)

= -75 ≠ RHS

Thus, the given lines do not intersect with each other.

Question 3. Show that the lines \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}  and \frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}  intersect and find their point of intersection.

Solution:

Given that the coordinates of any point on the first line are 

\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}=\lambda

⇒ x = 3λ – 1, y = 5λ – 3, z = 7λ – 5

The coordinates of a general point on the second line are given by:

\frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}=\mu

⇒ x = 2μ + 2, y = 3μ + 6, z = 4μ + 3

If the lines intersect, for some values of λ and μ, we must have:

3λ – μ = 3               ……(1)

5λ – 3μ = 7            ……(2)

7λ – 5μ = 11           …..(3)

Solving this system of equations, we get

λ = 1/2 and μ = -3/2

On substituting the values in eq(3), we have

LHS = 3(2) – 4(0)

= -3/2 = RHS

Now put the value of λ in first equation and we get

x = 1/2, y = -1/2, z = -3/2

Thus, the given lines intersect at (1/2, -1/2, -3/2).

Question 4. Prove that the line through (0, -1, -1) and B(4, 5, 1) intersects the line through C(3, 9, 4) and D(-4, 4, 4). Also, find their point of intersection.

Solution:

Given that the coordinates of any point on the line AB are 

\frac{x-0}{4-0}=\frac{y+1}{5+1}=\frac{z+1}{1+1}=\lambda

⇒ x = 4λ, y = 6λ – 1, z = 2λ – 1

Also, given that the coordinates of any point on the line CD are 

\frac{x-3}{3+4}=\frac{y-9}{9-4}=\frac{z-4}{4-4}=\mu

⇒ x = 7μ + 3, y = 5μ + 9, z = 4

If the lines intersect, for some values of λ and μ, we must have:

4λ – 7μ = 3            ……(1)

6λ – 5μ = 10         ……(2)

λ = 5/2                  …..(3)

⇒ λ = 5/2 and μ = 1.

On substituting the values in eq(3), we have

LHS = 4(5/2) – 7(1)

= 3 = RHS

Now put the value of λ in line AB, we get

x = 10, y = 14, z = 4

Thus, the given lines AB and CD intersect at point (10, 14, 4).

Question 5. Prove that the line \vec{r}=(\hat{i}+\hat{j}-\hat{k})+λ(3\hat{i}-\hat{j})  and \vec{r}=(4\hat{i}-\hat{k})+μ(2\hat{i}+3\hat{k})  intersect and find their point of intersection.

Solution:

According to the question, it is given that the position vector of two points on the lines are

\vec{r}=(\hat{i}+\hat{j}-\hat{k})+λ(3\hat{i}-\hat{j})
\vec{r}=(4\hat{i}-\hat{k})+μ(2\hat{i}+3\hat{k})

If the lines intersect, then for some value of λ and μ, we must have:

(1+3\lambda)\hat{i}+(1-\lambda)\hat{j}-\hat{k}=(4+2\mu)\hat{i}+0\hat{j}+(3\mu-1)\hat{k}

Now equate the coefficient of \hat{i}, \hat{j},\hat{k} we get

1 + 3λ = 4 + 2μ    ……(1)

1 – λ = 0                …..(2)

-1 = -1 +3μ           …..(3)

On solving the equation, we get

λ = 1 and μ = 0.

Now, substituting the values in eq(1), we get

1 + 3(1) = 4 + 2(0)

4 = 4

LHS = RHS 

Thus, the coordinates of the point of intersection of the two lines are (4, 0, -1).

Question 6. Determine whether the following pairs of lines intersect or not:

(i) \vec{r}=(\hat{i}-\hat{j})+\lambda( 2\hat{i}+\hat{k})  and \vec{r}=(2\hat{i}-\hat{j})+\mu(\hat{i}-\hat{j}-\hat{k}).

Solution:

Given that:

\vec{r}=(\hat{i}-\hat{j})+\lambda( 2\hat{i}+\hat{k})
\vec{r}=(2\hat{i}-\hat{j})+\mu(\hat{i}-\hat{j}-\hat{k}).

If the lines intersect, then for some value of λ and μ, we must have:

(1+2\lambda)\hat{i}-\hat{j}+\lambda\hat{k}=(2+\mu)\hat{i}+(-1+\mu)\hat{j}-\mu\hat{k}

Now equate the coefficient of \hat{i}, \hat{j},\hat{k} we get

1 + 2λ = 2 + μ      …..(1)

-1 = -1 + μ           …..(2)

λ = -μ                 …..(3)

On solving the equations, we get

λ = 0 and μ = 0.

Now, substitute the values in eq(1), we get

1 + 2λ = 2 + μ

1 + 2(0) = 2 + 0

1 ≠ 2

LHS ≠ RHS 

Thus, the given lines do not intersect.

(ii) \frac{x-1}{2}=\frac{y+1}{3}=z  and \frac{x+1}{5}=\frac{y-2}{1};z=2

Solution:

Given that the coordinates of any point on the line AB are 

\frac{x-1}{2}=\frac{y+1}{3}=z=\lambda

⇒ x = 2λ + 1, y = 3λ – 1, z = λ 

The coordinates of a general point on the second line are given by

\frac{x+1}{5}=\frac{y-2}{1}=\mu;z=2

⇒ x = 5μ – 1, y = μ + 2, z = 2

If the lines intersect, for some values of λ and μ, we must have:

2λ – 5μ = -2             ……(1)

3λ – μ = 3                ……(2)

λ = 2                        …..(3)

Solving this system of equations, we get

λ = 2 and μ = 3

On substituting the values in eq(3), we have

LHS = 2(2) – 5(3)

= -2 ≠ RHS

Thus, the given lines do not intersect each other.

(iii) \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}  and \frac{x-4}{2}=\frac{y-0}{0}=\frac{z+1}{3}

Solution:

Given that the coordinates of any point on the line AB are 

\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}=\lambda

⇒ x = λ, y = 2λ + 2, z = 3λ – 3

The coordinates of a general point on the second line are given by

\frac{x-4}{2}=\frac{y-0}{0}=\frac{z+1}{3}=\mu

⇒ x = 2μ + 4, y = 0, z = 3μ – 1

If the lines intersect, for some values of λ and μ, we must have:

λ – 2μ = 2            …….(1)

2λ – 3μ = 4         ……(2)

3λ – 4μ = 6         ……(3)

On solving this system of equations, we get

λ = 1 and μ = 0

On substituting the values in eq(3), we have

LHS = 3(1) – 2(0)

= 3 = RHS

Thus, the given lines intersect at (4, 0, -1).

(iv) \frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}  and \frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}

Solution:

Given that the coordinates of any point on the line AB are 

\frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}=\lambda

⇒ x = 4λ + 5, y = 4λ + 7, z = -5λ – 3

The coordinates of a general point on the second line are given by:

\frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}=\mu

⇒ x = 7μ + 8, y = μ + 4, z = 3μ + 5

If the lines intersect, for some values of λ and μ, we must have:

4λ – 7μ = 3            …….(1)

4λ – μ = -3            ……(2)

5λ + 3μ = -8        ……(3)

On solving this system of equations, we get

λ = -1 and μ = -1

On substituting the values in eq(3), we have

LHS = 5(-1) – 3(-1)

= -8 = RHS

Thus, the given lines intersect at (1, 3, 2).

Question 7. Show that the lines \vec{r}=(3\hat{i}+2\hat{j}-4\hat{k})+\lambda( \hat{i}+2\hat{j}+2\hat{k})  and \vec{r}=(5\hat{i}-2\hat{j})+\mu(3\hat{i}+2\hat{j}+6\hat{k})  are intersecting. Hence, find their point of intersection.

Solution:

Given that,

\vec{r}=(3\hat{i}+2\hat{j}-4\hat{k})+\lambda( \hat{i}+2\hat{j}+2\hat{k})
\vec{r}=(5\hat{i}-2\hat{j})+\mu(3\hat{i}+2\hat{j}+6\hat{k})

If the lines intersect, then for some value of λ and μ, we must have:

(3+\lambda)\hat{i}-\hat{j}+(2+2\lambda)\hat{j}+(2\lambda-4)\hat{k}=(2+\mu)\hat{i}+(-1+\mu)\hat{j}-\mu\hat{k}

Now equate the coefficient of \hat{i}, \hat{j},\hat{k} we get

3 + λ = 5 + 3μ       ……..(1)

2 + 2λ = -2 + 2μ   ……..(2)

2λ – 4 = 6μ           ……..(3)

Solving the equation, we have:

λ = -4 and μ = -2.

On substituting the values, we get

LHS = 2(-4) – 4 

= -12

RHS = 6(-2)

= -12

Thus, the given lines intersect at point(-1, -6, -12).

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