Check here the Latest detailed Answers and Solutions as per the latest guidelines of Miscellaneous Exercise - Chapter 11 Three Dimensional Geometry class 12 ncert solutions Maths - [2024-2025]

Miscellaneous Exercise - Chapter 11 Three Dimensional Geometry class 12 ncert solutions Maths - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 1:

Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, −1), (4, 3, −1).

Answer:

Let OA be the line joining the origin, O (0, 0, 0), and the point, A (2, 1, 1).

Also, let BC be the line joining the points, B (3, 5, −1) and C (4, 3, −1).

The direction ratios of OA are 2, 1, and 1 and of BC are (4 − 3) = 1, (3 − 5) = −2, and (−1 + 1) = 0

OA is perpendicular to BC, if a₁a₂ + b₁b₂ + c₁c₂ = 0

∴ a₁a₂ + b₁b₂ + c₁c₂ = 2 × 1 + 1 (−2) + 1 ×0 = 2 − 2 = 0

Thus, OA is perpendicular to BC.

Question 2:

If l₁, m₁, n₁ and l₂, m₂, n₂ are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m₁n₂ − m₂n₁, n₁l₂ − n₂l₁, l₁m₂ − l₂m₁.

Answer:

It is given that l₁, m₁, n₁ and l₂, m₂, n₂ are the direction cosines of two mutually perpendicular lines. Therefore,

Let l, m, n be the direction cosines of the line which is perpendicular to the line with direction cosines l₁, m₁, n₁ and l₂, m₂, n₂.

l, m, n are the direction cosines of the line.

∴l²+ m² + n² = 1 … (5)

It is known that,

∴

Substituting the values from equations (5) and (6) in equation (4), we obtain

Thus, the direction cosines of the required line are

Question 3:

Find the angle between the lines whose direction ratios are a, b, c and b − c,

c − a, a − b.

Answer:

The angle Q between the lines with direction cosines, a, b, c and b − c, c − a,

a − b, is given by,

Thus, the angle between the lines is 90°.

Question 4:

Find the equation of a line parallel to x-axis and passing through the origin.

Answer:

The line parallel to x-axis and passing through the origin is x-axis itself.

Let A be a point on x-axis. Therefore, the coordinates of A are given by (a, 0, 0), where a ∈ R.

Direction ratios of OA are (a − 0) = a, 0, 0

The equation of OA is given by,

Thus, the equation of line parallel to x-axis and passing through origin is

Question 5:

If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

Answer:

The coordinates of A, B, C, and D are (1, 2, 3), (4, 5, 7), (−4, 3, −6), and

(2, 9, 2) respectively.

The direction ratios of AB are (4 − 1) = 3, (5 − 2) = 3, and (7 − 3) = 4

The direction ratios of CD are (2 −(− 4)) = 6, (9 − 3) = 6, and (2 −(−6)) = 8

It can be seen that,

Therefore, AB is parallel to CD.

Thus, the angle between AB and CD is either 0° or 180°.

Question 6:

If the lines and are perpendicular, find the value of k.

Answer:

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

It is known that two lines with direction ratios, a₁, b₁, c₁ and a₂, b₂, c₂, are perpendicular, if a₁a₂ + b₁b₂ + c₁c₂ = 0

Therefore, for, the given lines are perpendicular to each other.

Question 7:

Find the vector equation of the plane passing through (1, 2, 3) and perpendicular to the plane

Answer:

The position vector of the point (1, 2, 3) is

The direction ratios of the normal to the plane, , are 1, 2, and −5 and the normal vector is

The equation of a line passing through a point and perpendicular to the given plane is given by,

Question 8:

Find the equation of the plane passing through (a, b, c) and parallel to the plane

Answer:

Any plane parallel to the plane, , is of the form

The plane passes through the point (a, b, c). Therefore, the position vector of this point is

Therefore, equation (1) becomes

Substituting in equation (1), we obtain

This is the vector equation of the required plane.

Substituting in equation (2), we obtain

Question 9:

Find the shortest distance between lines

and.

Answer:

The given lines are

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

Comparing to equations (1) and (2), we obtain

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

Therefore, the shortest distance between the two given lines is 9 units.

Question 10:

Find the coordinates of the point where the line through (5, 1, 6) and

(3, 4, 1) crosses the YZ-plane

Answer:

It is known that the equation of the line passing through the points, (x₁, y₁, z₁) and (x₂, y₂, z₂), is

The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by,

Any point on the line is of the form (5 − 2k, 3k + 1, 6 −5k).

The equation of YZ-plane is x = 0

Since the line passes through YZ-plane,

5 − 2k = 0

Therefore, the required point is .

Question 11:

Find the coordinates of the point where the line through (5, 1, 6) and

(3, 4, 1) crosses the ZX − plane.

Answer:

It is known that the equation of the line passing through the points, (x₁, y₁, z₁) and (x₂, y₂, z₂), is

The line passing through the points, (5, 1, 6) and (3, 4, 1), is given by,

Any point on the line is of the form (5 − 2k, 3k + 1, 6 −5k).

Since the line passes through ZX-plane,

Therefore, the required point is.

Question 12:

Find the coordinates of the point where the line through (3, −4, −5) and (2, − 3, 1) crosses the plane 2x + y + z = 7).

Answer:

It is known that the equation of the line through the points, (x₁, y₁, z₁) and (x₂, y₂, z₂), is

Since the line passes through the points, (3, −4, −5) and (2, −3, 1), its equation is given by,

Therefore, any point on the line is of the form (3 − k, k − 4, 6k − 5).

This point lies on the plane, 2x + y + z = 7

∴ 2 (3 − k) + (k − 4) + (6k − 5) = 7

Hence, the coordinates of the required point are (3 − 2, 2 − 4, 6 × 2 − 5) i.e.,

(1, −2, 7).

Question 13:

Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

Answer:

The equation of the plane passing through the point (−1, 3, 2) is

a (x + 1) + b (y − 3) + c (z − 2) = 0 … (1)

where, a, b, c are the direction ratios of normal to the plane.

It is known that two planes, and , are perpendicular, if

Plane (1) is perpendicular to the plane, x + 2y + 3z = 5

Also, plane (1) is perpendicular to the plane, 3x + 3y + z = 0

From equations (2) and (3), we obtain

Substituting the values of a, b, and c in equation (1), we obtain

This is the required equation of the plane.

Question 14:

If the points (1, 1, p) and (−3, 0, 1) be equidistant from the plane , then find the value of p.

Answer:

The position vector through the point (1, 1, p) is

Similarly, the position vector through the point (−3, 0, 1) is

The equation of the given plane is

It is known that the perpendicular distance between a point whose position vector is and the plane, is given by,

Here,and d

Therefore, the distance between the point (1, 1, p) and the given plane is

Similarly, the distance between the point (−3, 0, 1) and the given plane is

It is given that the distance between the required plane and the points, (1, 1, p) and (−3, 0, 1), is equal.

∴ D₁ = D₂

Question 15:

Find the equation of the plane passing through the line of intersection of the planes and and parallel to x-axis.

Answer:

The given planes are

The equation of any plane passing through the line of intersection of these planes is

r→.i⏜+j⏜+k⏜-1+λr→.2i⏜+3j⏜-k⏜+4=0r→.2λ+1i⏜+3λ+1j⏜+1-λk⏜+4λ-1=0 …(1)

Its direction ratios are (2λ + 1), (3λ + 1), and (1 − λ).

The required plane is parallel to x-axis. Therefore, its normal is perpendicular to x-axis.

The direction ratios of x-axis are 1, 0, and 0.

Substituting in equation (1), we obtain

Therefore, its Cartesian equation is y − 3z + 6 = 0

This is the equation of the required plane.

Question 16:

If O be the origin and the coordinates of P be (1, 2, −3), then find the equation of the plane passing through P and perpendicular to OP.

Answer:

The coordinates of the points, O and P, are (0, 0, 0) and (1, 2, −3) respectively.

Therefore, the direction ratios of OP are (1 − 0) = 1, (2 − 0) = 2, and (−3 − 0) = −3

It is known that the equation of the plane passing through the point (x₁, y₁ z₁) is

where, a, b, and c are the direction ratios of normal.

Here, the direction ratios of normal are 1, 2, and −3 and the point P is (1, 2, −3).

Thus, the equation of the required plane is

Question 17:

Find the equation of the plane which contains the line of intersection of the planes , and which is perpendicular to the plane .

Answer:

The equations of the given planes are

The equation of the plane passing through the line intersection of the plane given in equation (1) and equation (2) is

The plane in equation (3) is perpendicular to the plane,

Substituting in equation (3), we obtain

This is the vector equation of the required plane.

The Cartesian equation of this plane can be obtained by substituting in equation (3).

Question 18:

Find the distance of the point (−1, −5, −10) from the point of intersection of the line and the plane.

Answer:

The equation of the given line is

The equation of the given plane is

Substituting the value of from equation (1) in equation (2), we obtain

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

This means that the position vector of the point of intersection of the line and the plane is

This shows that the point of intersection of the given line and plane is given by the coordinates, (2, −1, 2). The point is (−1, −5, −10).

The distance d between the points, (2, −1, 2) and (−1, −5, −10), is

Question 19:

Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes and .

Answer:

Let the required line be parallel to vector given by,

The position vector of the point (1, 2, 3) is

The equation of line passing through (1, 2, 3) and parallel to is given by,

The equations of the given planes are

The line in equation (1) and plane in equation (2) are parallel. Therefore, the normal to the plane of equation (2) and the given line are perpendicular.