NCERT Solutions for Class 12 Maths Chapter 11: Three Dimensional Geometry

A-line makes 90°, 135°, 45°with x, y and z  axes respectively.

Therefore, Direction cosines of the line are cos 90°, cos 135°, and cos 45°

⇒ Direction cosines of the line are 0,-1/√2 and 1/√2

 

Let the direction cosines of the line make an angle α with each of the coordinate axes.

∴ l = cos α, m = cos α, n = cos α

Thus, the direction cosines of the line, which is equally inclined to the coordinate axes, are 

The direction ratios of the lines are -18, 12, -4

Direction cosines of the lines are

Hence, the direction cosine of the line are -9/11, 6/11 -2/11.

 

The given points are A (2, 3, 4), B(−1, −2, 1) and C (5, 8, 7)

It is known that the direction ratios of line joining the points, (x1, y1, z1) and (x2, y2, z2), are given by, x2 − x1, y2 − y1, and z2 − z1.

The direction ratios of AB are (−1 − 2), (−2 − 3), and (1 − 4) i.e., −3, −5, and −3.

The direction ratios of BC are (5 − (− 1)), (8 − (− 2)), and (7 − 1) i.e., 6, 10, and 6.

It can be seen that the direction ratios of BC are −2 times that of AB i.e., they are proportional.

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

 

The vertices of ΔABC are A (3, 5, −4), B (−1, 1, 2), and C (−5, −5, −2).

The direction ratios of side AB are (−1 − 3), (1 − 5), and (2 − (−4)) i.e., −4, −4, and 6.

Therefore, the direction cosines of AB are

The direction ratios of BC are (−5 − (−1)), (−5 − 1), and (−2 − 2) i.e., −4, −6, and −4.

Therefore, the direction cosines of BC are

The direction ratios of CA are 3−(−5), 5−(−5) and −4−(−2) i.e. 8, 10 and -2.

Therefore the direction cosines of CA are

 

 

The required line passes through the origin. Therefore, its position vector is given by,

= 0            …..(1)

The direction ratios of the line through origin and (5, −2, 3) are

(5 − 0) = 5, (−2 − 0) = −2, (3 − 0) = 3

The line is parallel to the vector given by the equation,

The equation of the line in vector form through a point with position vector and parallel to is,

The equation of the line through the point (x1, y1, z1) and direction ratios a, b, c is given by,

Therefore, the equation of the required line in the Cartesian form is

 

Two lines with direction cosines, l1, m1, n1 and l2, m2, n2, are perpendicular to each other, if l1l2 + m1m2 + n1n2 = 0

Therefore, the lines are perpendicular.

Thus, all the lines are mutually perpendicular.

 

Let AB be the line joining the points, (1, −1, 2) and (3, 4, − 2), and CD be the line joining the points, (0, 3, 2) and (3, 5, 6).

The direction ratios, a1, b1, c1, of AB are (3 − 1), (4 − (−1)), and (−2 − 2) i.e., 2, 5, and −4.

The direction ratios, a2, b2, c2, of CD are (3 − 0), (5 − 3), and (6 −2) i.e., 3, 2, and 4.

AB and CD will be perpendicular to each other, if a1a2 + b1b2+ c1c2 = 0

a1a2 + b1b2+ c1c2 = 2 × 3 + 5 × 2 + (− 4) × 4

= 6 + 10 − 16

= 0

Therefore, AB and CD are perpendicular to each other.

 

Let AB be the line through the points, (4, 7, 8) and (2, 3, 4), and CD be the line through the points, (−1, −2, 1) and (1, 2, 5).

The directions ratios, a1, b1, c1, of AB are (2 − 4), (3 − 7), and (4 − 8) i.e., −2, −4, and −4.

The direction ratios, a2, b2, c2, of CD are (1 − (−1)), (2 − (−2)), and (5 − 1) i.e., 2, 4, and 4.

AB will be parallel to CD, if a1/a2 = b1/b2 = c1/c2

a1/a2 = -2/2 = -1

b1/b2 = -4/4 = -1

c1/c2 = -4/4 = -1

∴a1/a2 = b1/b2 = c1/c2

Thus, AB is parallel to CD.

 

It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector through A is

It is known that the line which passes through point A and parallel to is given by λ is a constant.

This is the required equation of the line.

 

It is given that the line passes through the point with position vector

It is known that a line through a point with position vector and parallel to is given by the equation,

This is the required equation of the line in vector form.

Eliminating λ, we obtain the Cartesian form equation as

 

It is given that the line passes through the point (−2, 4, −5) and is parallel to

The direction ratios of the line, , are 3, 5, and 6.

The required line is parallel to

Therefore, its direction ratios are 3k, 5k, and 6k, where k ≠ 0

It is known that the equation of the line through the point (x1, y1, z1) and with direction ratios, a, b, c, is given by

Therefore the equation of the required line is

 

The Cartesian equation of the line is

        ….(1)

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 vector and in the direction of the vector is given by the equation,

This is the required equation of the given line in vector form.

 

 

The equations of the given lines are

The direction ratios of the given lines are 7, −5, 1 and 1, 2, 3 respectively.

Two lines with direction ratios, a1, b1, c1 and a2, b2, c2, are perpendicular to each other, if a1a2 + b1 b2 + c1c2 = 0

∴ 7 × 1 + (−5) × 2 + 1 × 3

= 7 − 10 + 3

= 0

Therefore, the given lines are perpendicular to each other.

 

Let the line passing through the points, P (3, −2, −5) and Q (3, −2, 6), be PQ.

Since PQ passes through P (3, −2, −5), its position vector is given by,

The direction ratios of PQ are given by,

(3 − 3) = 0, (−2 + 2) = 0, (6 + 5) = 11

The equation of the vector in the direction of PQ is

The equation of PQ in vector form is given by,

The equation of PQ in Cartesian form is

i.e.,

 

 

(i) Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by,

The given lines are parallel to the vectors, , respectively.

(ii) The given lines are parallel to the vectors, respectively.

 

 

Let 1 and 2 be the vectors parallel to the pair of lines, , respectively.

(ii) Let 12 be the vectors parallel to the given pair of lines, , respectively.

 

The given equations can be written in the standard form as

The direction ratios of the lines are respectively

Two lines with direction ratios, a1, b1, c1 and a2, b2, c2, are perpendicular to each other, if a1a2 + b1 b2 + c1c2 = 0

 

 

The equations of the given lines are

Since, the shortest distance between the two skew lines is given by

Therefore, the shortest distance between the two lines is 3√2/2 units.

 

 

The given lines are

It is known that the shortest distance between the two lines, , is given by

Comparing this equation we have

Since distance is always non-negative, the distance between the given lines is 2√29 units.

 

The given lines are

It is known that the shortest distance between the lines,   is given by,

…..(1)

Comparing the given equations with , we obtain

Substituting all the values in equation (1), we obtain

Therefore, the shortest distance between the two given lines is 3/√19 units.

 

 

The given lines are

It is known that the shortest distance between the lines, is given by,

For the given equations,

Substituting all the values in equation (3), we obtain

Therefore, the shortest distance between the lines is 8/√29 units.

 

(a) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1, y1, z1).

2x + 3y + 4z − 12 = 0

⇒ 2x + 3y + 4z = 12 … (1)

The direction ratios of normal are 2, 3, and 4.

∴ √ 2² + 3² + 4² = √29

Dividing both sides of equation (1) by √29, we obtain

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by

(ld, md, nd).

Therefore, the coordinates of the foot of the perpendicular are

(b) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1, y1, z1).

3y + 4z – 6 = 0

⇒ 0x + 3y + 4y = 6 … (1)

The direction ratios of the normal are 0, 3, and 4.

∴ √0 + 3² + 4² = 5

Dividing both sides of equation (1) by 5, we obtain

0x + 3/5y + 4/5z = 6/5

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by

(ld, md, nd).

Therefore, the coordinates of the foot of the perpendicular are

(c) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1, y1, z1).

x + y + z = 1… (1)

The direction ratios of the normal are 1, 1, and 1.

√1² + 1² + 1² = √3

Dividing both sides of equation (1) by √3, we obtain

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by

(ld, md, nd).

Therefore, the coordinates of the foot of the perpendicular are

(d) Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1, y1, z1).

5y + 8 = 0

⇒ 0x − 5y + 0z = 8 … (1)

The direction ratios of the normal are 0, −5, and 0.

√0 + (-5)2 + 0 = 5

Dividing both sides of equation (1) by 5, we obtain

-y = 8/5

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by

(ld, md, nd).

Therefore, the coordinates of the foot of the perpendicular are

 

(The equation of the plane is z = 2 or 0x + 0y + z = 2 … (1)

The direction ratios of normal are 0, 0, and 1.

∴ √02 + 02 + 12 = 1

Dividing both sides of equation (1) by 1, we obtain

0.x + 0.y + 1.z = 2

This is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.

(b) x + y + z = 1 … (1)

The direction ratios of normal are 1, 1, and 1.

∴ √(1)² + (1)² + (1)² = √3

Dividing both sides of equation (1) by √3, we obtain

1/√3x + 1/√3y + 1/√3z = 1/√3           ….(2)

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal are 1/√3,1/√3 and 1/√3the distance of normal from the origin is 1/√3 units.

(c) 2x + 3y ­− z = 5 … (1)

The direction ratios of normal are 2, 3, and −1.

∴ √(2)² + (3)² + (-1)² = √14

Dividing both sides of equation (1) by √14 , we obtain

2/√14x + 3/√14y – 1/√14z = 5/√14

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane are  2/√14x + 3/√14y – 1/√14z and the distance of normal from the origin is 5/√14  units.

(d) 5y + 8 = 0

⇒ 0x − 5y + 0z = 8 … (1)

The direction ratios of normal are 0, −5, and 0.

∴√0 +(-5)² + 0 = 5

Dividing both sides of equation (1) by 5, we obtain

-y = 8/5

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane are 0, −1, and 0 and the distance of normal from the origin is 8/5 units.

 

The normal vector is,

It is known that the equation of the plane with position vector is given by, . = d

This is the vector equation of the required plane.

 

(a) It is given that equation of the plane is

For any arbitrary point P (x, y, z) on the plane, position vector is given by,

Substituting the value of in equation (1), we obtain

This is the Cartesian equation of the plane.

(b)

For any arbitrary point P (x, y, z) on the plane, position vector is given by,

Substituting the value of in equation (1), we obtain

This is the Cartesian equation of the plane.

(c)

For any arbitrary point P (x, y, z) on the plane, position vector is given by,

Substituting the value of in equation (1), we obtain

This is the Cartesian equation of the given plane.

 

(a) The position vector of point (1, 0, −2) is

The normal vector N perpendicular to the plane is

The vector equation of the plane is given by,

is the position vector of any point P (x, y, z) in the plane.

Therefore, equation (1) becomes

This is the Cartesian equation of the required plane.

(b) The position vector of the point (1, 4, 6) is

The normal vector perpendicular to the plane is

The vector equation of the plane is given by,

is the position vector of any point P (x, y, z) in the plane.

Therefore, equation (1) becomes

This is the Cartesian equation of the required plane.

 

We know that through three collinear points A, B, C i.e., through a straight line, we can pass an infinite number of planes.

(a) The given points are A (1, 1, −1), B (6, 4, −5), and C (−4, −2, 3).

Since A, B, C are collinear points, there will be infinite number of planes passing through the given points.

(b) The given points are A (1, 1, 0), B (1, 2, 1), and C (−2, 2, −1).

Therefore, a plane will pass through the points A, B, and C.

It is known that the equation of the plane through the points, is

This is the Cartesian equation of the required plane.

 

 

2x + y – z = 5                  ………(1)

Dividing both sides of equation (1) by 5, we obtain

It is known that the equation of a plane in intercept form is x/a + y/b + z/c = 1, where a, b, c are the intercepts cut off by the plane at x, y, and z axes respectively.

Therefore, for the given equation,

a = 5/2, b = 5 and c = -5

Thus, the intercepts cut off by the plane are 5/2, 5 and -5.

 

The equation of the plane ZOX is

y = 0

Any plane parallel to it is of the form, y = a

Since the y-intercept of the plane is 3,

∴ a = 3

Thus, the equation of the required plane is y = 3

The equation of any plane through the intersection of the planes,

3x − y + 2z ­− 4 = 0 and x + y + z − 2 = 0, is

The plane passes through the point (2, 2, 1). Therefore, this point will satisfy equation (1).

 

 

The equations of the planes are

The equation of any plane through the intersection of the planes given in equations (1) and (2) is given by,

The plane passes through the point (2, 1, 3). Therefore, its position vector is given by,

Substituting in equation (3), we obtain

This is the vector equation of the required plane.

 

The equation of the plane through the intersection of the planes, x + y + z = 1 and 2x + 3y + 4z = 5, is

(x + y + z – 1)  + λ(2x + 3y + 4z – 5)

⇒(2λ + 1)x + (3λ + 1)y + (4λ + 1)z – (5λ + 1) = 0             …..(1)

The direction ratios, a1, b1, c1, of this plane are (2λ + 1), (3λ + 1), and (4λ + 1).

The plane in equation (1) is perpendicular to x – y + z = 0

Its direction ratios, a2, b2, c2, are 1, −1, and 1.

Since the planes are perpendicular,

Substituting λ = -1/3 in equation (1), we obtain

This is the required equation of the plane.

 

 

The equations of the given planes are

It is known that if n1 and n2 are normal to the planes, then the angle between them, Q, is given by,

 

The direction ratios of normal to the plane,, are a1, b1, c1 and .

(a) The equations of the planes are 7x + 5y + 6z + 30 = 0 and

3x − y − 10z + 4 = 0

Here, a1 = 7, b1 =5, c1 = 6

Therefore, the given planes are not parallel.

The angle between them is given by,

(b) The equations of the planes are 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0

Thus, the given planes are perpendicular to each other.

(c) The equations of the given planes are 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1

Here,

Thus, the given planes are not perpendicular to each other.

Thus, the given planes are parallel to each other.

(d) The equations of the planes are and 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0

Thus, the given lines are parallel to each other.

(e) The equations of the given planes are 4x + 8y + z – 8 = 0 and y + z – 4 = 0

Therefore, the given lines are not perpendicular to each other.

Therefore, the given lines are not parallel to each other.

The angle between the planes is given by,

 

It is known that the distance between a point, p(x1, y1, z1), and a plane, Ax + By + Cz = D, is given by,

(a) The given point is (0, 0, 0) and the plane is 3x – 4y + 12z = 3

(b) The given point is (3, − 2, 1) and the plane is 2x – y + 2z + 3 = 0

(c) The given point is (2, 3, −5) and the plane is x + 2y – 2z = 9

(d) The given point is (−6, 0, 0) and the plane is 2x – 3y + 6z – 2 = 0