Chapter 11 Three Dimensional Geometry

7/31/2019 Chapter 11 Three Dimensional Geometry

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Q. 1.

Find the equation of the plane, which is at a distance of 5 unit from the origin and has

as a normal vector.

Ans.

Here, p = 5 and ,

Q. 2.

Find the equation of the plane whose distance from the origin is 8 units and the direction

ratios of the normal are 6, – 3, – 2.

Ans.

Here, p = 8, a = 6, b = – 3, c = – 2.

Therefore, the required equation of the plane is:

Q. 3.

Find the equation of the plane is . Find the perpendicular distance of the

plane from the origin.

Ans.



The given equation of the plane is .

Here,

Perpendicular distance of the plane from the origin is

units.

Q. 4.

Write the normal and Cartesian form of the plane .

Ans.

The given equation of the plane is .

Here,

Now, Cartesian form of the equation of this plane is:

Q. 5.



Find the equation of the plane passing through the point (3, – 3, 1) and perpendicular to the

line joining the points (3, 4, – 1) and (2, – 1, 5).

Ans.

The equation of the plane passing through the point (3, – 3, 1) is:

a(x – 3) + b(y + 3) + c(z – 1) = 0 and the direction ratios of the line joining the points(3, 4, – 1) and (2, – 1, 5) is 2 – 3, – 1 – 4, 5 + 1, i.e., – 1, – 5, 6.Since the plane is perpendicular to the line whose direction ratios are – 1, – 5, 6, therefore, direction ratiosof the normal to the plane is – 1, – 5, 6.So, required equation of plane is: – 1(x – 3) – 5(y + 3) + 6(z – 1) = 0i.e., x + 5y – 6z + 18 = 0.

Q. 6.

If is the normal from the origin to the plane, and is the unit vector along . P(x, y,

z) be any point on the plane and is perpendicular to . Find the equation of plane in

the normal form.

Ans.

Here, and is perpendicular to . Therefore, .



Q. 7.

Let be the vector normal to the plane and be the position vector of the point throughwhich the plane passes. Then find the equation of plane.

Ans.

Let P be a arbitrary point on the plane with position vector .



Thus, equation of the plane with as a normal vector and as a position vector of the point throughwhich the plane passes is

Q. 8.

Find the equation of the plane passing through the points (–1, 4, – 3), (3, 2, – 5) and (– 3, 8, –

5).

Ans.

Let .

The equation of the plane is:

Q. 9.

Find the co-ordinates of the foot of the perpendicular drawn from the origin to the plane 2x –

3y + 6z + 7 = 0.

Ans.



Let the co-ordinates of the foot of the perpendicular drawn from the origin to the plane 2x – 3y +6z + 7 = 0 is (x1, y1, z1).

Now, the equation in the normal form is:

Substituting these in the equation of the plane, we get

Hence, the foot of the perpendicular is

Q. 10.

Find the equation of the plane passing through the points (2, 5, – 8) and perpendicular to

each one of the planes 2x – 3y + 4z + 1 = 0 and 4x + y – 2z + 6 = 0.

Ans.

Equation of the plane passing through the point (2, 5, – 8) is:

a(x – 2) + b(y – 5) + c(z + 8) = 0 …(1)If the plane perpendicular to the plane 2x – 3y + 4z + 1 = 0, then2a – 3b + 4c = 0 …(2) andIf the plane perpendicular to the plane 4x + y – 2z + 6 = 0, then4a + b – 2c = 0 …(3)On solving equations (2) and (3), we get

On substituting the proportional values of a, b and c in (1), we get(x – 2) + 10(y – 5) + 7(z + 8) = 0



Q. 1.

The equation of the plane which makes the intercepts 2, 3, 4 with X-axis, Y-axis and Z-axis

respectively.

Ans.

Equation of the plane making the intercepts 2, 3, 4 with co-ordinate axes is:

Q. 2.

Reduce the equation of the plane 3x + 4y – 6z = 6 to intercept form and find the intercepts

made by the plane with the co-ordinate axes.

Ans.

The given equation of the plane is 3x + 4y – 6z = 6.

It is the required equation in the intercept form and it makes the intercepts 2, 3/2 and – 1 with the X-axis, Y-axis and Z-axis respectively.

Q. 3.

Write the equation of plane passing through the intersection of two given planes and

.

Ans.

If and are two planes.



Then equation of plane passing through the intersection of two given planes is:

Q. 4.

Show that the lines and are coplanar.

Ans.

Here, x1 = 1, y1 = – 2, z1 = – 5, a1 = 1, b1 = – 2, c1 = – 5 and

x2 = 3, y2 = – 1, z2 = – 5, a2 = 3, b2 = – 1, c2 = – 5.

Therefore, the lines are coplanar.

Q. 5.

Find the equation of the plane passing through the line of intersection of the planes 3x – 5y +

4z + 11 = 0, 2x – 7y + 4z – 3 = 0 and the point (– 2, 1, 3).

Ans.

The equation of the plane passing through the line of intersection of the planes 3x – 5y + 4z +

11 = 0, 2x – 7y + 4z – 3 = 0 is:

. Also, the plane passing through the point (– 2, 1, 3),

Now, the required equation of the plane is:



Q. 8.

Reduce the equation 2x + 3y - 4z = 12 to intercept form and find its intercepts on the

coordinate axes.

Ans.

Q. 9.

A plane which remains at a constant distance 3p from the origin cuts the co-ordinate axes at

A, B and C. Show that the locus of the centroid of triangle ABC is x-2 + y-2 + z-2 = p-2.

Ans.

Let a, b, c are the intercepts made by the plane with the co-ordinate axes, then the co-ordinatesof A, B and C are (a, 0, 0), (0, b, 0) and (0, 0, c) respectively.



Then the equation of the plane is . …(1) Also, the perpendicular distance of this plane from the origin is 3p.

Now, let (x, y, z) be the co-ordinates of the centroid of the triangle ABC, then

On putting the values of a, b and c in (2), we get

Q. 10.

Find the equation of the plane passing through the line of intersection of the planes

and the point .

Ans.

The equation of the plane passing through the planes and is:

.

Also, the plane passing through the point .



Now, the required equation of the plane is:

Q. 1.

Write the condition for the plane and are perpendicular.

Ans.

Let be the angle between the planes and . Also, are the normal

to the planes and . Then the angle between the planes is given by

.

Also, both planes are perpendicular, if

Q. 2.

Find the angle between the planes 2x + y – z = 4 and x – 2y – z = 7 using vector method.

Ans.

The angle between the two planes is the angle between their normal and

.



Q. 3.

Find the distance of the point (3, – 2, 5) from the plane .

Ans.

Here, , and d = 5.

Now, the distance of the point (3, – 2, 5) from the plane is:

Q. 4.

What is the length of the perpendicular from the origin to the plane ?

Ans.

The length of the perpendicular from the origin to the plane is:

.

Q. 5.



Show that the planes 3x + 4y – 5z + 7 = 0 and x + 3y + 3z + 7 = 0 are perpendicular.

Ans.

The direction ratios of the normals to the planes are 3, 4, – 5 and 1, 3, 3.

Now, (3)(1) + (4)(3) + (– 5)(3) = 3 + 12 – 15 = 15 – 15 = 0.Therefore, the planes are perpendicular to each other.

Q. 6.

A variable plane is at a constant distance p from the origin and meets the axes in points

A,B,C. Through A,B, C planes are drawn parallel to the coordinate planes. Prove that the

locus of their point of intersection is x-2 + y-2+z-2 = p-2.

Ans.

Q. 7.

Find the equation of the plane which passes through the point (3,4,-1) and is parallel to the

plane 2x - 3y + 5z + 7 = 0. Also find the distance between the two planes.

Ans.



Q. 8.

Find the angle between the line and the plane 5x – 4y + 7z + 10 = 0.

Ans.

Let be the angle between the given line and the plane. The vector equations of the line and

plane are: and .

Here, and .Now, angle between the line and the plane is given by



Q. 9.

Find the angle between the planes 3x – 2y + 6z = 8 and 4x – 8y + z = 13.

Ans.

Let the angle between the planes be .

Also, here, A1 = 3, B1 = – 2, C1 = 6 and A2 = 4, B2 = – 8, C2 = 1.

Q. 10.

The foot of the perpendicular drawn from the origin to a plane is (2, 1, 5). Find the equation

of the plane.

Ans.

Since, the foot of the perpendicular to the plane is A(2, 1, 5). Therefore, (2, 1, 5) is the point onthe plane.

So, equation of the plane passing through the point (2,.1, 5) is:a(x – 2) + b(y – 1) + c(z – 5) = 0.Now, the direction ratios of the perpendicular line OA = 2 – 0, 1 – 0, 5 – 0, i.e., 2, 1, 5.

Therefore, the required plane is:2(x – 2) + 1(y – 1) + 5(z – 5) = 0i.e, 2x + y + 5z = 30.



Q. 1.

Find the direction cosines of the line whose direction ratios are 4, – 3, 2 .

Ans.

Let l, m, n be the direction cosines, then

Q. 2.

Find the direction cosines of the vector .&

Ans.

Direction ratios of the vector = 6, – 2, – 3.

Q. 3.

Find the direction cosines of the line segment joining the points (2, 0, 1) and (–1, 3, –2).

Ans.

The direction ratios of the line segment joining the points (2, 0, 1) and (–1, 3, –2) are (– 1 – 2), (3 – 0), (– 2 – 1) i.e.,

–3, 3, –3.



Q. 4.

Find the angle inclined to Z-axis, where line is inclined to X-axis at 45o and to Y-axis at 60o.

Ans.

Q. 5.

If a line in the XY-plane makes an angle 60o with X-axis. Find the direction cosines of this

line.

Ans.

A line makes an angle 60o with X-axis and 30o with Y-axis, then it will make an angle 90o with Z-axis.

Thus, the direction cosines of this line are .

Q. 6.



If a line makes angles with OX, OY and OZ respectively. Prove that

.

Ans.

Let be the direction cosines of the given line. Then

.

Q. 7.

If a line makes angles with OX, OY and OZ respectively. Prove that

.

Ans.

Let be the direction cosines of the given line. Then

.

Q. 8.

The co-ordinates of the vertices of the triangle are A(–1 0, 3), B(2, – 3, 6) and C(–1, 3, 2). Find

the direction cosines of the medians of the triangle



Ans.

The co-ordinates of the vertices of the triangle are A (5, – 1, – 2), B (1, – 3, 6) and C (–1, 3, 2). Let mid-points of

BC, CA and AB be D, E and F respectively.

Co-ordinates of point D = ,

Co-ordinates of point E = and

Co-ordinates of point F = .

Q. 9.

Show that the points A (2, 3, – 4), B (1, – 2, 3) and C (3, 8, – 11) are collinear.

Ans.

Direction ratios of line joining A and B are 1 – 2, – 2 – 3, 3 + 4 i.e., – 1, – 5, 7.

The direction ratios of line joining B and C are 3 –1, 8 + 2, – 11 – 3, i.e., 2, 10, – 14.

It is clear that direction ratios of AB and BC are proportional, hence, AB is parallel

to BC.

But point B is common to both AB and BC. Therefore, A, B, C are collinear points.

Q. 10.

The vertices of a triangle ABC are A (–1, 2, –3), B (5, 0, –6) and



C (0, 4, –1). Find the direction cosines of the bisector of the angle BAC.

Ans.

Here, AB = .

AC = .

By geometry, the bisector of will divide the side BC in the ratio AB : AC i.e., in

the ratio 7 : 3 internally. Let the bisector of meets the side BC at point D.

Therefore, D divides BC in the ratio 7 : 3.

Coordinates of D are i.e., .

Therefore, direction ratios of the bisector AD are .

Hence, direction cosines of the bisector AD are

, , i.e., .

Q. 1.

Find the Cartesian equation of the straight line passing through the point (1, 3, – 2) and

parallel to the vector .

Ans.

The equation passing through the point (1, 3, – 2) and the direction ratios of the vector to which

line is parallel are 2, 3, – 1.

Cartesian equation of line is:

Q. 2.



Find the Cartesian equation of line which passes through the points (2, 0, 5) and (4, – 3, – 2).

Ans.

Cartesian equation of line passing through the points (2, 0, 5) and (4, – 3, – 2) is:

Q. 3.

Find the Cartesian equation of the line which passes through the origin and parallel to the

line .

Ans.

The direction ratios of the line, which parallel to the line is (– 1, – 2, 5).

The Cartesian equation of the line which passes through the origin and parallel to the line

is .

Q. 4.

Find the Cartesian equation of line which passes through the origin and the point (2, 1, 5).

Ans.

Cartesian equation of line passing through the points (0, 0, 0) and (2, 1, 5) is:



Q. 5.

Prove that the points (7, – 2, 5), (–2, 4, 2) and (4, 0, 4) are collinear. Also find the equation of

the line containing these points.

Ans.

Let the points A, B and C are (7, – 2, 5), (–2, 4, 2) and (4, 0, 4) respectively.

Direction ratios of the line AB = – 2 – 7, 4 + 2, 2 – 5 = – 9, 6, – 3 andDirection ratios of the line BC = 4 + 2, 0 – 4, 4 – 2 = 6, – 4, 2.It is clear that direction ratios of AB and BC are proportional.Hence, AB is parallel to BC. But point B is common to both AB and BC. Therefore, A, B, C are collinear points.

Cartesian equation of the line ABC: Equation of the line passing through C(4, 0, 4) andparallel to the line whose direction ratios are – 9, 6, – 3 is

.

Q. 6.

Find the Cartesian form of the line and find its direction cosines.

Ans.

The given equation is: . This line passes through (2, – 1, 0) andparallel to the line whose ratios are (1, 2, 3).

Thus, the Cartesian equation of the line is:

Now, direction ratios of the line are 1, 2, 3.



So, the direction cosines are

.

Q. 7.

Find the Cartesian equation of the line which passes through the point (3, 2, – 1) and equally

inclined to the axes.

Ans.

Let the direction ratios of the line which is equally inclined to the axes.

Now, the equation of the line passing through (3, 2, – 1) and parallel to the line whose direction ratios are 1,1, 1. is:

.

Q. 8.

Find the equation of the line passes through the origin and in the direction of the line which

passes through the points (0, 2, – 5) and (7, 5, 2).

Ans.

Direction ratios of the line which passes through the point (0, 2, – 5) and (7, 5, 2) = 7 – 0, 5 – 2,2 + 5 = 7, 3, 7.

Now, the equation of the line passes through the origin and parallel to the line whose direction ratios are 7,

3, 7 is:

Q. 9.

Find the equation of the line in vector and in Cartesian form that passes through the point

with position vector and is in the direction .



Ans.

The vector form of the equation passes through and parallel to the vector

is: .

The Cartesian form of this equation is: .

Q. 1.

Let be the direction ratios of two lines and be the angle between

them. Then find the angle between them in the form of direction ratios. Also find the

condition that the lines are perpendicular.

Ans.

Let be the direction ratios of two lines, then the angle between the

vectors is .

If the lines are perpendicular, i.e.,

Q. 2.

Find the angle between the pair of lines given by and

.



Ans.

Here, and . Let the angle between the lines be .

Q. 3.

Find the angle between the lines and

Ans.

The direction ratios of the first line are 2, – 3, 6 and the direction ratios of the second line are –

2, 1, 2. If the angle between the lines is , then

Q. 4.

Show that the lines and are perpendicular.



Ans.

The direction ratios of the first line are 3, – 2, 5 and the direction ratios of the second line are 1,

– 1, 1. If the angle between the lines is , then

Thus, the lines are perpendicular.

Q. 5.

Show that the line segment joining the points (– 1, 2, – 3) and (1, 4, 1) is parallel to the line

segment joining the points (1, 2, 3) and (2, 3, 5).

Ans.

The direction ratios of the first line segment is 1 + 1, 4 – 2, 1 + 3 i.e., 2, 2, 4 and the directionratios of the second line segment is 2 – 1, 3 – 2, 5 – 3 i.e., 1, 1, 2.

Let the angle between the both line segments be , then

Thus, the lines segments are parallel.

Q. 6.

Find the distance between the lines and

.



Ans.

The both lines are parallel, because are proportional.

Therefore, the distance between the parallel lines is given by

Q. 7.

Find the shortest distance between the lines: and .

Ans.

The vector equations of these equations are:



and

Q. 8.

Find the shortest distance between the lines: and

.

Ans.

The given equations can be written as

and



Q. 9.

Show that the line 6x = – y = – 4z is perpendicular to the 2x = 3y = – z.

Ans.

The given lines are 6x = – y = – 4z and 2x = 3y = – z.

Therefore, the direction ratios of these lines are

Thus, the lines are perpendicular.

Q. 10.



Find the angle between any two diagonals of a cube.

Ans.

Let O, one vertex of a cube, be the origin and three edges through O be the

Co-ordinate axes. The four diagonals are OP, AA', BB' and CC'. Let ‘a’ be the length of each edge. Then the co-ordinates of P, A, A' are (a, a, a), (a, 0, 0), (0, a, a).

The direction ratios of OP are a, a, a.

The direction cosines of OP are i.e., .

Similarly, direction cosines of AA' are .

Let be the angle between the diagonals OP and AA'. Then

.

Thus the angle between any two diagonals of a cube is .

Chapter 11 Three Dimensional Geometry

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