NCERT Solutions for Class 12 Science Math Chapter 5 - Three Dimensional Geometry

Answer:

Let direction cosines of the line be l, m, and n.

Therefore, the direction cosines of the line are

Answer:

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

Answer:

If a line has direction ratios of −18, 12, and −4, then its direction cosines are

Thus, the direction cosines are.

Answer:

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.

Answer:

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

$\frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}$

Answer:

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

(i) For the lines with direction cosines, and , we obtain

Therefore, the lines are perpendicular.

(ii) For the lines with direction cosines, and , we obtain

Therefore, the lines are perpendicular.

(iii) For the lines with direction cosines, and , we obtain

Therefore, the lines are perpendicular.

Thus, all the lines are mutually perpendicular.

Answer:

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.

Answer:

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

Thus, AB is parallel to CD.

Answer:

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 is a constant.

This is the required equation of the line.

Answer:

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

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

Answer:

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

Answer:

The Cartesian equation of the line is

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.

Answer:

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

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

Answer:

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.,

Answer:

(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, and , respectively.

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

Answer:

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

and

The angle, Q, between the given pair of lines is given by the relation,

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

If Q is the angle between the given pair of lines, then

Answer:

The given equations can be written in the standard form as

and

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

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

Thus, the value of p is .

Answer:

The equations of the given lines areand

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.

Answer:

The equations of the given lines are

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

$d = \left|\frac{\left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right).\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)}{\left|\overrightarrow{b_1}\times \overrightarrow{b_2}\right|}\right|$

Comparing the given equations, we obtain

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

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

Answer:

The given lines are and

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

Comparing the given equations, we obtain

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

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

Answer:

The given lines are and

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

$d=\left|\frac{\left(\overrightarrow{b_1}\times \overrightarrow{b_2}\right).\left(\overrightarrow{a_2}-\overrightarrow{a_1}\right)}{\left|\overrightarrow{b_1}\times \overrightarrow{b_2}\right|}\right|$                                                ...(1)

Comparing the given equations with and , we obtain

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

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

Answer:

The given lines are

$$\begin{gathered} \overrightarrow{r}=(s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k} \\ \Rightarrow \overrightarrow{r}=(\hat{i}-\hat{j}-\hat{k})+s(\hat{i}+2\hat{j}-2\hat{k}) ...\left(2\right) \end{gathered}$$

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

For the given equations,

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

Therefore, the shortest distance between the lines isunits.

Answer:

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

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

∴

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

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.

∴

Dividing both sides of equation (1) by, 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.

Therefore, the direction cosines of the normal are and the distance of normal from the origin is units.

(c) 2x + 3y ­− z = 5 … (1)

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

Dividing both sides of equation (1) by , 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.

Therefore, the direction cosines of the normal to the plane are and the distance of normal from the origin is units.

(d) 5y + 8 = 0

⇒ 0x − 5y + 0z = 8 … (1)

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

Dividing both sides of equation (1) by 5, 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.

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

Answer:

The normal vector is,

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

This is the vector equation of the required plane.

Answer:

(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.

Answer:

(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.

Dividing both sides of equation (1) by , 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).

⇒ … (1)

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

Dividing both sides of equation (1) by 5, 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

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

… (1)

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

Dividing both sides of equation (1) by, 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).

⇒ 0x − 5y + 0z = 8 … (1)

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

Dividing both sides of equation (1) by 5, 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

Answer:

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

(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.

Answer:

(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, , and , is

This is the Cartesian equation of the required plane.

Answer:

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

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

Therefore, for the given equation,

Thus, the intercepts cut off by the plane are.

Answer:

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

Answer:

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).

Substituting in equation (1), we obtain

This is the required equation of the plane.

Answer:

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,

, where

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

Substituting in equation (3), we obtain

Substituting in equation (3), we obtain

This is the vector equation of the required plane.

Answer:

The equation of the plane through the intersection of the planes, and , is

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

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

Since the planes are perpendicular,

Substituting in equation (1), we obtain

This is the required equation of the plane.

Answer:

The equations of the given planes are and

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

Here,

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

Answer:

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

The angle between L₁ and L₂ is given by,

(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 perpendicular.

It can be seen that,

Therefore, the given planes are not parallel.

The angle between them is given by,

(b) The equations of the planes are and

Here, and

Thus, the given planes are perpendicular to each other.

(c) The equations of the given planes are and

Here, and

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

Here, and

∴

Thus, the given lines are parallel to each other.

(e) The equations of the given planes are and

Here, and

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,

Answer:

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

(b) The given point is (3, − 2, 1) and the plane is

∴

(c) The given point is (2, 3, −5) and the plane is

(d) The given point is (−6, 0, 0) and the plane is

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.

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

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°.

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

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°.

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.

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,

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

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.

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 .

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.

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).

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.

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

Answer:

The given planes are

The equation of any plane passing through the line of intersection of these planes is
$\left[\overrightarrow{r}.\left(\stackrel{⏜}{i}+\stackrel{⏜}{j}+\stackrel{⏜}{k}\right)-1\right]+\lambda \left[\overrightarrow{r}.\left(2\stackrel{⏜}{i}+3\stackrel{⏜}{j}-\stackrel{⏜}{k}\right)+4\right]=0$

$\overrightarrow{r}.\left[\left(2\lambda +1\right)\stackrel{⏜}{i}+\left(3\lambda +1\right)\stackrel{⏜}{j}+\left(1-\lambda \right)\stackrel{⏜}{k}\right]+\left(4\lambda -1\right)=0 ...\left(1\right)$

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.

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

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).

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

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.

From equations (4) and (5), we obtain

Therefore, the direction ratios of are −3, 5, and 4.

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

This is the equation of the required line.

Answer:

Let the required line be parallel to the vector given by,

The position vector of the point (1, 2, − 4) is

The equation of the line passing through (1, 2, −4) and parallel to vector is

The equations of the lines are

Line (1) and line (2) are perpendicular to each other.

Also, line (1) and line (3) are perpendicular to each other.

From equations (4) and (5), we obtain

∴Direction ratios of are 2, 3, and 6.

Substituting in equation (1), we obtain

This is the equation of the required line.

Answer:

The equation of a plane having intercepts a, b, c with x, y, and z axes respectively is given by,

The distance (p) of the plane from the origin is given by,

Answer:

The equations of the planes are

It can be seen that the given planes are parallel.

It is known that the distance between two parallel planes, ax + by + cz = d₁ and ax + by + cz = d₂, is given by,

Thus, the distance between the lines is units.

Hence, the correct answer is D.

Answer:

The equations of the planes are

2x − y + 4z = 5 … (1)

5x − 2.5y + 10z = 6 … (2)

It can be seen that,

∴

Therefore, the given planes are parallel.

Hence, the correct answer is B.