# RD Sharma Class 12 Ex 28.1 Solutions Chapter 28 The Straight Line in Space

Here we provide RD Sharma Class 12 Ex 28.1 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 28.1 Solutions Chapter 28 The Straight Line in Space book pdf download. Now you will get step-by-step solutions to each question.

## RD Sharma Class 12 Ex 28.1 Solutions Chapter 28 The Straight Line in Space

### Question 1. Find the vector and Cartesian equation of the line through the points (5, 2, -4) and which is parallel to the vector

Solution:

As we know that the vector equation of a line is;

Thus, the Cartesian equation of a line is;

After applying the above formulas;

The vector equation of the line is;

The Cartesian equation of a line is;

### Question 2. Find the vector equation of the line passing through the points (-1, 0, 2) and (3, 4, 6).

Solution:

Given:

Here, the direction ratios of the line are;

(3 + 1, 4 – 0, 6 – 2) = (4, 4, 4)

Thus, the given line passes through

(-1, 0, 2)

As we know that the vector equation of a line is given as;

Thus, substitute values

Hence, we get

Therefore,

Vector equation of the line is;

### Question 3. Fine the vector equation of a line which is parallel to the vectorand which passes through the point (5, -2, 4), Also, reduce it to Cartesian form.

Solution:

Consider,

The vector equation of line passing through a fixed point vector a and parallel to vector b is shown as;

Here, λ is scalar

and

The equation of the required line is;

Now substitute the value of r here

Thus, we get

Now compare the coefficients of vector

x = 5 + 2λ,y = -2 – λ,z = 4 + 3λ

After equating to λ,

We will have

Therefore,

The Cartesian form of equation of the line is;

### Question 4. A line passing through the point with position vectorand is in the direction of. Find equations of the line in vector and Cartesian form.

Solution:

Consider,

The vector equation of line passing through a fixed point vector a and parallel to vector b is shown as;

Here, λ is scalar

and

The equation of the required line is;

Now substitute the value of r here

Thus, we get

Now compare the coefficients of vector

x = 2 + 3λ,y = -3 + 4λ,z = 4 – 5λ

After equating to λ,

We will have

Therefore,

The Cartesian form of equation of the line is;

### Question 5. ABCD is a parallelogram. The position vectors of the points A, B and C are respectively, and. Find the vector equation of the line BD. Also reduce it to Cartesian form.

Solution:

Given: ABCD is a parallelogram.

Consider: AC and BD bisects each other at point O.

Thus,

Position vector of point O =

Now, Consider position vector of point O and B are represented by

and

Thus,

Equation of the line BD is the line passing through O and B is given by

[Since equation of the line passing through two pointsand]

Now, compare the coefficients of vector i, j, R

x = 2 – λ, y = -3 – 13λ, z = 4 – 17λ

After equating to λ,

We will have

Therefore,

The Cartesian form of equation of the line is;

### Question 6. Find the vector form as well as in Cartesian form, the equation of line passing through the points A(1, 2, -1) and B(2, 1, 1).

Solution:

We know that, equation of line passing though two points (x1, y1 ,z1) and (x2, y2, z2) is

Here,

(x1, y1, z1) = A(1, 2, -1)

(x2, y2 ,z2) = B(2, 1, 1)

Using equation (i), equation of line AB,

x = λ + 1, y = -λ + 2, z = 2λ – 1

Vector form of equation of line AB is,

### Question 7. Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector. Reduce the corresponding equation in Cartesian form.

Solution:

We know that vector equation of a line passing throughand parallel to the vectoris given by,

Here,

and

So, required vector equation of line is,

Now,

Equating the coefficients of

x = 1 + λ, y = 2 – 2λ, z = 3 + 3λ

x – 1 = λ,

So, required equation of line is Cartesian form,

### Question 8. Find the vector equation of a line passing through (2, −1, 1) and parallel to the line whose equations are

Solution:

We know that, equation of a line passing through a point (x1, y1, z1) and having direction ratios proportional to a, b, c is

Here,

(x1, y1, z1) = (2, -1, 1) and

Given lineis parallel to required line.

a = 2μ, b = 7μ, c = -3μ

So, equation of required line using equation (i)

x = 2λ + 2, y = 7λ – 1, z = -3λ + 1

So,

### Question 9. The Cartesian equation of a line is. Write its vector form

Solution:

The Cartesian equation of the line is

….(i)

The given line passes through the point (5, -4, 6). The position vector of this point is

Also, the direction ratios of the given line are 3, 7 and 2.

This means that the line is in the direction of vector,

It is known that the line through position vectorand in the direction of the vectoris given by the equation,

### Question 10. Find the Cartesian equation of a line passing through (1, -1, 2) and parallel to the line whose equations are . Also, reduce the equation obtained in vector form.

Solution:

We know that, equation of a line passing through a point (x1, y1, z1) and having direction ratios proportional to a, b, c is

Here,

(x1, y1, z1) = (1, -1, 2) and

Given lineis parallel to required line,

So,

a = μ, b = 2μ, c = -2μ

So, equation of required line using equation (i) is,

x = λ + 1, y = 2λ – 1, z = -2λ +2

So,

### Question 11. Find the direction cosines of the line. Also, reduce it to vector form

Solution:

Given:

x = -2λ + 4, y = 6λ, z = -3λ + 1

So,

Direction ratios of the line are = -2, 6, -3

Direction cosines of the lines are,

### Question 12. The Cartesian equations of a line are x = ay + b, z = cy + d. Find its direction ratios and reduce it to vector form.

Solution:

x = ay + b

z = cy + d

So, DR’s of line are (a, 1, c)

From above equation, we can write

x = aλ + b

y = λ

z = cλ + d

So vector equation of line is

### Question 13. Find the vector equation of a line passing through the point with position vectorand parallel to the line joining the points with the position vectorand. Also, find the Cartesian equivalent of this equation.

Solution:

We know that, equation of a line passing throughand parallel to vectoris

……. (i)

Here,

and, \vec{b} = line joiningand

Equation of the line is

For Cartesian form of equation put

Equating coefficients of

x = 1 + λ, y = -2 + 2λ, z = -3 – 2λ

### Question 14. Find the points on the lineat a distance of 5 units from the points P(1, 3, 3).

Solution:

Given, line is

General points Q on line is (3λ – 2, 2λ -1), 2λ + 3)

Distance of points P from Q =

PQ =

(5)2 = (3λ -3)2 + (2λ – 4) + (2λ)2

25 = 9λ2 + 9 – 18λ + 4λ2 + 16 – 16λ + 4λ2

17λ2 – 34λ = 0

17λ (λ – 2) = 0

λ = 0 or 2

So, points on the line are (3(0) – 2, 2(0) – 1, 2(0) + 3)

(3(2) – 2, 2(2) – 1, 2(2) + 3)

= (-2, -1, 3), (4, 3, 7)

### Question 15. Show that the points whose position vectors areandare collinear.

Solution:

Let the given points are A,B.C with position vectorsrespectively.

We know that, equation of a line passing throughandare

If A, B, C are collinear thenmust satisfy equation (i)

Equation the coefficients of

-2 + 3 = 7 , λ = 3

3 – λ = 0 , λ = 3

3λ = -1 , λ =

Since, value of λ are not equal, so,

Given points are collinear.

### Question 16. Find the Cartesian and vector equations of a line which passes through the points (1, 2, 3) and is parallel to the line

Solution:

We know that, equation of a line passing through a point (x1, y1, z1) and having direction ratios proportional to a, b, c is

Here,

(x1, y1, z1) = (1, 2, 3) and

Given line

Its parallel to the required line, so

a = μ , b = 7μ, c =μ

So, equation of required line using equation (i) is,

Multiplying the denominators by 2

x = -2λ + 1, y = 14λ + 2, z = 3λ + 3

So, vector form of the equation of required line,

### Question 17. The Cartesian equations of a line are 3x + 1 = 6y – 2 = 1 – z. Find the fixed point through which it passes, its direction ratios, and also its vector equation.

Solution:

Given equation of line is,

3x + 1 = 6y -2 = 1 – z

Dividing all by 6

Comparing it with equation of line equation of line passing through (x,1 y1, z1) and the direction ratios a, b, c,

a = 2, b = 1, -6

So, direction ratios of the line are -2, 1, -6

From equation (i)

So, vector equation of the given line is,

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