# Class 12 RD Sharma Solutions- Chapter 28 The Straight Line in Space – Exercise 28.4

### 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 ADisLine AD is the perepndicular to BC so,

Hence,

coordinates of Dare:=

### 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 ADisLine 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 by: 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 PQare:As line PQ is perpendicular to the given line, so:

Therefore,

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

equation of the perpendicular PQis :

### 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 PQare: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 Qbe ::

= 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 PQis:

### 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 PQareSince, 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 Qare {}

Direction Ratios of PQare: =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 CDis

### 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