Here we provide RD Sharma Class 12 Ex 28.4 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 28.4 Solutions Chapter 28 The Straight Line in Space book pdf download. Now you will get step-by-step solutions to each question.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 28 |

Exercise | 28.4 |

Category | RD Sharma Solutions |

Table of Contents

**RD Sharma Class 12 Ex 28.4 Solutions Chapter 28 The Straight Line in Space**

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

### Question 1. Find the perpendicular distance of the point (3, -1, 11) from the line** **

**Solution:**

Let the foot of the perpendicular drawn from P (3, -1, 11) to the line is Q, so we have to find length of PQ is general point on the line

**Coordinate of Q** = , direction ratios of the given line = 2,-3,4. Since PQ is the perpendicular to the given line interface.

**So, the coordinates of Q are:**

**Distance between P and Q is given as:**

So, the required distance is units

### Question 2. Find the perpendicular distance of the point (1,0,0) from the line . Also, find the coordinates of the foot of the perpendicular and the equation of the perpendicular.

**Solution:**

Let us consider the foot of the perpendicular drawn from P (1,0,0) to the line is Q. So let us find the length of PQ i.e.

**Coordinate of Q** =

**The direction ratios of the given line: **

**So the Coordinates of Q are as follows:**

**Distance between P and Q is given by:**

PQ =

PQ =

PQ =

Hence, the foot of the perpendicular = (3,-4,-2);

Length of the perpendicular = units.

### Question 3. Find the foot of the perpendicular drawn from the point A(1,0,3) to the joint of the points B(4,7,1) and C(3,5,3).

**Solution:**

Let us consider, the foot of the perpendicular drawn from A(1,0,3) to the line joining

Points B(4,7,1) and C(3,5,3) be D. The equation of the line passing through

points B(4,7,1) and C(3,5,3) is

Let

So, the **direction ratio of AD** is

Line AD is the perepndicular to BC so,

Hence,** coordinates of D** are:

=

### Question 4. A (1,0,4), B (0,-11,3), C (2,-3,1) are three points and D is the foot of the perpendicular from A on BC. Find the coordinates of D.

**Solution:**

**Given:** D is the perpendicular from A(1,0,4) on BC. So,

**Equation of line passing through BC is:**

**Coordinates of D** = ( )

**Direction ratios of AD** is

Line AD is perpendicular to BS so,

So, coordinates of D are =

=

### Question 5. Find the foot of the perpendicular from the point (2,3,4) to the line . Also, find the perpendicular distance from the given point to the line.

**Solution:**

Let us consider that The foot of the perpendicular drawn from P(2,3,4) to the line

is .

Equation of the line is

Let

**Coordinates of Q** =

So, PQ is perpendicular to the given line,

**Coordinates of Q** =

=

=

**Distance between P and Q is given b**y: PQ =

=

=

Hence, perpendicular distance from (2,3,4) to the given line is units.

### Question 6. Find the equation of the perpendicular drawn from the point P (2,4,-1) to the line . Also, write down the coordinates of the foot of the perpendicular from P.

**Solution:**

Let be the foot of thr perpendicular drawn from P(2,4,-1) to the line

Given line is

**Coordinate of Q** (General point on the line) =

**Direction ratios of PQ** are:

As line PQ is perpendicular to the given line, so:

Therefore, **coordinates of foot of perpendicular** = {-4, 1, -3}

So **equation of the perpendicular PQ** is :

### Question 7. Find the length of the perpendicular drawn from the point (5,4,-1) to the line

**Solution:**

Let the foot of the perpendicular drawn from P(5,4,-1) to the given line is Q, so given equation of line is:

Equating the coefficients of

**Coordinate of Q** =

**Direction ratios of line PQ** are:

As line PQ is perpendicular to the given line, so:

**Coordinate of Q** = { }

=

=

**Length of perpendicular** = PQ =

=

PQ =

### Question 8. Find the foot of the perpendicular drawn from the point to the line . Also, find the length of the perpendicular.

**Solution:**

Let position vector of foot of perpendicular drawn from p on be Q . So, Q is on the line

So,** position vector of Q** =

is the position vector of Q – position vector of p =

Here, **PQ vector is perpendicular to the given line**. So,

**Position vector of Q **= {}

=

**Foot of the perpendicular **=

= Position vector of Q – Position vector of P

=

=

= units

### Question 9. Find the equation of the peprendicular drwan from the point P (-1,3,2) to the line . Also, find the coordinates of the foot of the perpendicular from P.

**Solution:**

Let Q be the perpendicular drawn from P {} on the

vector

Let the **position vector of Q** be :

:

= Position Vector of Q – Position Vector of P =

As PQ vector is perpendicular to the given line,

**Position Vector of Q** = is

**Coordinates of foot of the perpendicular:**

**Equation of PQ** is:

### Question 10. Find the foot of the perpendicular from (0,2,7) on the line

**Solution:**

Let the foot of the perpendicular drawn from (0,2,7) to the line be Q.

Given equation of the line is

**Coordinate of Q** = {}

**Direction Ratios of PQ** are

Since, PQ is perpendicular to the given line, so

**Foot of the perpendicular** = {}\

=

### Question 11. Find the foot of the perpendicular from (1,2,-3) to the line

**Solution:**

Let the foot of perpendicular from P (1,2,-3) to the line be Q.

Given the equation of line is

**Coordinates of Q** are {}

**Direction Ratios of PQ** are: =

Let PQ be the perpendicular to th egiven line, so

**Coordinate of the perpendicular:**

### Question 12. Find the equation of the line passing through the points A (0,6,-9) and B (-3, 6, 3). If D is the foot of the perpendicular drawn from a point C (7,4,-1) on the line AB, then find the coordinates of the point D and the equation of the line CD.

**Solution:**

Equation of line AB is

**Coordinate of point D** = {}

**Direction ratios of CD** =

=

As line CD is perpendicular to the line AB, so

**Coordinate of D **= {}

= {}

= (-1,2,-5)

**Equation of CD** is

### Question 13. Find the distance of the point (2,4,-1) from the line

**Solution:**

Let P = (2,4,-1)

In order to find the distance we need to find a point Q on the line. We see that line is passing through

the point Q(-5,-3,6). So, let’s take this point as the required point.

The line is also parallel to the vector

Now, =

Therefore,

### Question 14. Find the coordinates of the foot of the perpendicular drawn from point A (1,8,4) to the line joining the points B (0,-1,3) and C (2,-3,-1).

**Solution:**

Let L be the foot of the perpendicular drawn from A(1,8,4) on the line joining the points B(0,-1,3) and C(2,-3,-1).

Equation of the line passing through the points B and C is given by

Let position vector of L be,

Then, = Position vector of L – Position vector of A

Since, AL vector is perpendicular to the given line

which is parallel to

Therefore,

Putting value of lambda in Equation 1, we get:

**So, coordinates of foot of the perpendicular are**

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.