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**NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry **- In the Class 12 Maths Chapter 11 NCERT solutions you will learn how to use vector algebra to study three dimensional geometry. Understanding vector algebra is a prerequisite for 3d geometry class 12. The practice of three dimensional geometry class 12 solutions will be helpful in solving three dimensional problems given in NCERT Books for Class 12.

This Story also Contains

- NCERT Three Dimensional Geometry Class 12 Questions And Answers
- NCERT Three Dimensional Geometry Class 12 Questions And Answers PDF Free Download
- NCERT Class 12 Maths Chapter 11 Question Answer - Important Formulae
- NCERT Three Dimensional Geometry Class 12 Questions And Answers (Intext Questions and Exercise)
- Three Dimensional Geometry Class 12 - Topics
- NCERT solutions for class 12 maths - Chapter wise
- NCERT solutions for class 12 subject wise
- NCERT Solutions Class Wise
- NCERT Books and NCERT Syllabus

NCERT Class 12 Maths solutions chapter 11 makes the study simple and effective. It is also very helpful to solve problems asked in the CBSE Board as well as entrance exam related to chapter three dimensional geometry class 12. NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry will build your base for many other higher-level concepts like tensors and manifolds due to that this ch 11maths class 12 becomes very important. If you want to know more about 3d geometry class 12 then you can also check NCERT solutions for other classes.

**Also read:**

- Class 12 Maths Chapter 11 Three Dimensional Geometry Notes
- Ncert Exemplar Solutions For Class 12 Maths Chapter 11 Three Dimensional Geometry

>> Distance Formula:

The distance between two points A(x_{1}, y_{1}, z_{1}) and B(x_{2}, y_{2}, z_{2}) is given by:

AB = √[(x_{2} - x_{1})² + (y_{2} - y_{1})² + (z_{2} - z_{1})²]

The distance between a point A(x, y, z) and the origin O(0, 0, 0) is given by:

OA = √(x² + y² + z²)

>> Section Formula: The coordinates of the point R, which divides the line segment joining two points P(x1, y1, z1) and Q(x2, y2, z2) internally or externally in the ratio m:n, are given by:

Internal Division: (mx_{2} + nx_{1}) / (m + n), (my_{2} + ny_{1}) / (m + n), (mz_{2} + nz_{1}) / (m + n)

External Division: (mx_{2} - nx_{1}) / (m - n), (my_{2} - ny_{1}) / (m - n), (mz_{2} - nz_{1}) / (m - n)

>> Midpoint Formula: The coordinates of the mid-point of the line segment joining (x1, y1) and (x2, y2) are:

[(x_{1} + x_{2}) / 2, (y_{1} + y_{2}) / 2]

>> Coordinates of Centroid of a Triangle: Given vertices (x_{1}, y_{1}), (x_{2}, y_{2}), and (x_{3}, y_{3}) of a triangle, the coordinates of the centroid are:

[(x_{1} + x_{2} + x_{3}) / 3, (y_{1} + y_{2} + y_{3}) / 3]

>> Incentre of a Triangle: Given vertices (x_{1}, y_{1}), (x_{2}, y_{2}), and (x_{3}, y_{3}) of a triangle, the coordinates of the incenter are:

[(ax_{1} + bx_{2} + cx_{3}) / (a + b + c), (ay_{1} + by_{2} + cy_{3}) / (a + b + c)]

>> Centroid of a Tetrahedron: Given vertices (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}), (x_{3}, y_{3}, z_{3}), and (x_{4}, y_{4}, z_{4}) of a tetrahedron, the coordinates of the centroid are:

[(x_{1} + x_{2} + x_{3} + x_{4}) / 4, (y_{1} + y_{2} + y_{3} + y_{4}) / 4, (z_{1} + z_{2} + z_{3} + z_{4}) / 4]

>> Direction Cosines of a Line: If a directed line OP makes angles α, β, and γ with the positive X-axis, Y-axis, and Z-axis, respectively, then the direction cosines l, m, and n are:

l = cos α, m = cos β, n = cos γ Also, the sum of squares of direction cosines is always 1:

l² + m² + n² = 1

>> Direction Ratios of a Line: Direction ratios of a line are denoted as a, b, and c. They are proportional to the direction cosines:

l/a = m/b = n/c

>> Angle between Two Line Segments: If a_{1}, b_{1}, c_{1}, and a_{2}, b_{2}, c_{2} are the direction ratios of two lines and θ is the acute angle between them, then:

cos θ = |(a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2}) / (√(a_{1}² + b_{1}² + c_{1}²) √(a_{2}² + b_{2}² + c_{2}²))|

>> Perpendicular and Parallel Lines: Two lines are perpendicular if: a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0

Two lines are parallel if: a_{1}/a_{2} = b_{1}/b_{2} = c_{1}/c_{2}

>> Projection of a Line Segment on a Line: Given points P(x_{1}, y_{1}, z_{1}) and Q(x_{2}, y_{2}, z_{2}) and a line with direction cosines l, m, n, the projection of PQ on the line is:

|l(x_{2} - x_{1}) + m(y_{2} - y_{1}) + n(z_{2} - z_{1})|

>> Equation of a Plane: A plane in 3-D space can be represented in various forms:

General form: ax + by + cz + d = 0 (where a, b, c are not all zero)

Normal form: lx + my + nz = p

Plane through a point (x

_{1}, y_{1}, z_{1}): a(x - x_{1}) + b(y - y_{1}) + c(z - z_{1}) = 0Intercept form: (x/a) + (y/b) + (z/c) = 1

Vector form: (r- a).n = 0 or r.n = a.n

>> Planes Parallel to Axes: Planes parallel to the X-axis, Y-axis, and Z-axis are represented as:

Plane Parallel to X-axis: by + cz + d = 0

Plane Parallel to Y-axis: ax + cz + d = 0

Plane Parallel to Z-axis: ax + by + d = 0

Free download **NCERT Class 12 Maths Chapter 11 Question Answer **for CBSE Exam.

**NCERT Class 12 Maths Chapter 11 Question Answer - Exercise: 11.1 **

** Question:1 ** If a line makes angles with the x, y and z-axes respectively, find its direction cosines.

** Answer: **

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

So, we have

Therefore the direction cosines of the lines are .

** Question:2 ** Find the direction cosines of a line which makes equal angles with the coordinate axes.

** Answer: **

If the line is making equal angle with the coordinate axes. Then,

Let the common angle made is with each coordinate axes.

Therefore, we can write;

And as we know the relation;

or

** Thus the direction cosines of the line are **

** Question:3 ** If a line has the direction ratios –18, 12, – 4, then what are its direction cosines ?

** Answer: **

GIven a line has direction ratios of -18, 12, – 4 then its direction cosines are;

Line having direction ratio ** -18 ** has direction cosine:

Line having direction ratio ** 12 ** has direction cosine:

Line having direction ratio ** -4 ** has direction cosine:

Thus, the direction cosines are .

** Question:4 ** Show that the points (2, 3, 4), (– 1, – 2, 1), (5, 8, 7) are collinear.

** Answer: **

We have the points, A (2, 3, 4),B (– 1, – 2, 1),C (5, 8, 7);

And as we can find the direction ratios of the line joining the points is given by

The direction ratios of AB are i.e.,

The direction ratios of BC are i.e., .

We can see that the direction ratios of AB and BC are proportional to each other and is -2 times.

AB is parallel to BC. and as point B is common to both AB and BC,

** Hence the points A, B and C are collinear. **

** Answer: **

Given vertices of the triangle (3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2).

Finding each side direction ratios;

Direction ratios of side AB are i.e.,

Therefore its direction cosines values are;

SImilarly for side BC;

Direction ratios of side BC are i.e.,

Therefore its direction cosines values are;

Direction ratios of side CA are i.e.,

Therefore its direction cosines values are;

** NCERT Class 12 Maths Chapter 11 Question Answer - Exercise: 11.2 **

** Question:1 ** Show that the three lines with direction cosines

** Answer: **

GIven direction cosines of the three lines;

And we know that two lines with direction cosines and are perpendicular to each other, if

Hence we will check each pair of lines:

** Lines ; **

the lines are perpendicular.

** Lines ; **

the lines are perpendicular.

** Lines ; **

the lines are perpendicular.

Thus, we have all lines are mutually perpendicular to each other.

** Answer: **

We have given points where the line is passing through it;

Consider the line joining the points (1, – 1, 2) and (3, 4, – 2) is AB and line joining the points (0, 3, 2) and (3, 5, 6).is CD.

So, we will find the direction ratios of the lines AB and CD;

Direction ratios of AB are

or

Direction ratios of CD are

or .

Now, lines AB and CD will be perpendicular to each other if

** Therefore, AB and CD are perpendicular to each other. **

** Answer: **

We have given points where the line is passing through it;

Consider the line joining the points (4, 7, 8) and (2, 3, 4) is AB and line joining the points (– 1, – 2, 1) and (1, 2, 5)..is CD.

So, we will find the direction ratios of the lines AB and CD;

Direction ratios of AB are

or

Direction ratios of CD are

or .

Now, lines AB and CD will be parallel to each other if

Therefore we have now;

** Hence we can say that AB is parallel to CD. **

** Answer: **

It is given that the line is passing through A (1, 2, 3) and is parallel to the vector

We can easily find the equation of the line which passes through the point A and is parallel to the vector by the known relation;

, where is a constant.

So, we have now,

** Thus the required equation of the line. **

** Answer: **

Given that the line is passing through the point with position vector and is in the direction of the line .

And we know the equation of the line which passes through the point with the position vector and parallel to the vector is given by the equation,

** So, this is the required equation of the line in the vector form. **

Eliminating , from the above equation we obtain the equation in the Cartesian form :

** Hence this is the required equation of the line in Cartesian form. **

** Answer: **

Given a line which passes through the point (– 2, 4, – 5) and is parallel to the line given by the ;

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

So, the required line is parallel to the above line.

Therefore we can take direction ratios of the required line as ** 3k ** , ** 5k ** , and ** 6k ** , where k is a non-zero constant.

And we know that the equation of line passing through the point and with direction ratios a, b, c is written by: .

Therefore we have the equation of the required line:

or

** The required line equation. **

** Question:7 ** The cartesian equation of a line is . Write its vector form .

** Answer: **

Given the Cartesian equation of the line;

Here the given line is passing through the point .

So, we can write the position vector of this point as;

And the direction ratios of the line are ** 3 ** , ** 7 ** , and ** 2. **

This implies that the given line is in the direction of the vector, .

Now, we can easily find the required equation of line:

As we know that the line passing through the position vector and in the direction of the vector is given by the relation,

So, we get the equation.

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

** Answer: **

GIven that the line is passing through the and

Thus the required line passes through the origin.

its position vector is given by,

So, the direction ratios of the line through and are,

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

Therefore the equation of the line passing through the point with position vector and parallel to is given by;

Now, the equation of the line through the point and the direction ratios a, b, c is given by;

Therefore the equation of the required line in the Cartesian form will be;

OR

** Answer: **

Let the line passing through the points and is AB;

Then as AB passes through through A so, we can write its position vector as;

Then direction ratios of PQ are given by,

Therefore the equation of the vector in the direction of AB is given by,

We have then the equation of line AB in vector form is given by,

So, the equation of AB in Cartesian form is;

or

** Question:10 ** Find the angle between the following pairs of lines:

** Answer: **

To find the angle A between the pair of lines we have the formula;

We have two lines :

and

The given lines are parallel to the vectors ;

where and respectively,

Then we have

and

Therefore we have;

or

** Question:10 ** Find the angle between the following pairs of lines:

** Answer: **

To find the angle A between the pair of lines we have the formula;

We have two lines :

and

The given lines are parallel to the vectors ;

where and respectively,

Then we have

and

Therefore we have;

or

** Question:11 ** Find the angle between the following pair of lines:

** Answer: **

Given lines are;

and

So, we two vectors which are parallel to the pair of above lines respectively.

and

To find the angle A between the pair of lines we have the formula;

Then we have

and

Therefore we have;

or

** Question:11 ** Find the angle between the following pair of lines:

** Answer: **

Given lines are;

and

So, we two vectors which are parallel to the pair of above lines respectively.

and

To find the angle A between the pair of lines we have the formula;

Then we have

and

Therefore we have;

or

** Question:12 ** Find the values of p so that the lines and are at right angles.

** Answer: **

First we have to write the given equation of lines in the standard form;

and

Then we have the direction ratios of the above lines as;

and respectively..

Two lines with direction ratios and are perpendicular to each other if,

Thus, the value of p is .

** Question:13 ** Show that the lines and are perpendicular to each other.

** Answer: **

First, we have to write the given equation of lines in the standard form;

and

Then we have the direction ratios of the above lines as;

and respectively..

Two lines with direction ratios and are perpendicular to each other if,

Therefore the two lines are perpendicular to each other.

** Question:14 ** Find the shortest distance between the lines

** Answer: **

So given equation of lines;

and in the vector form.

Now, we can find the shortest distance between the lines and , is given by the formula,

Now comparing the values from the equation, we obtain

Then calculating

So, substituting the values now in the formula above we get;

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

** Question:15 ** Find the shortest distance between the lines

** Answer: **

We have given two lines:

and

Calculating the shortest distance between the two lines,

and

by the formula

Now, comparing the given equations, we obtain

Then calculating determinant

Now calculating the denominator,

So, we will substitute all the values in the formula above to obtain,

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

units.

** Question:16 ** Find the shortest distance between the lines whose vector equations are and

** Answer: **

Given two equations of line

in the vector form.

So, we will apply the distance formula for knowing the distance between two lines and

After comparing the given equations, we obtain

Then calculating the determinant value numerator.

That implies,

Now, after substituting the value in the above formula we get,

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

** Question:17 ** Find the shortest distance between the lines whose vector equations are

** Answer: **

Given two equations of the line

in the vector form.

So, we will apply the distance formula for knowing the distance between two lines and

After comparing the given equations, we obtain

Then calculating the determinant value numerator.

That implies,

Now, after substituting the value in the above formula we get,

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

** NCERT class 12 three dimensional geometry ncert solutions - Exercise: 11.3 **

** Question:1(a) ** In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

** Answer: **

Equation of plane Z=2, i.e.

The direction ratio of normal is 0,0,1

Divide equation by 1 from both side

We get,

Hence, direction cosins are 0,0,1.

The distance of the plane from the origin is 2.

** Question:1(b) ** In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

** Answer: **

Given the equation of the plane is or we can write

So, the direction ratios of normal from the above equation are, .

Therefore

Then dividing both sides of the plane equation by , we get

So, this is the form of the plane, where are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

The direction cosines of the given line are and the distance of the plane from the origin is units.

** Question:1(c) ** In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

** Answer: **

Given the equation of plane is

So, the direction ratios of normal from the above equation are, .

Therefore

Then dividing both sides of the plane equation by , we get

So, this is the form of the plane, where are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

The direction cosines of the given line are and the distance of the plane from the origin is units.

** Question:1(d) ** In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

** Answer: **

Given the equation of plane is or we can write

So, the direction ratios of normal from the above equation are, .

Therefore

Then dividing both sides of the plane equation by , we get

So, this is the form of the plane, where are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

The direction cosines of the given line are and the distance of the plane from the origin is units.

** Answer: **

We have given the distance between the plane and origin equal to 7 units and normal to the vector .

So, it is known that the equation of the plane with position vector is given by, the relation,

, where d is the distance of the plane from the origin.

Calculating ;

** is the vector equation of the required plane. **

** Question:3(a) ** Find the Cartesian equation of the following planes:

** Answer: **

Given the equation of the plane

So we have to find the Cartesian equation,

Any point on this plane will satisfy the equation and its position vector given by,

Hence we have,

Or,

** Therefore this is the required Cartesian equation of the plane. **

** Question:3(b) ** Find the Cartesian equation of the following planes:

** Answer: **

Given the equation of plane

So we have to find the Cartesian equation,

Any point on this plane will satisfy the equation and its position vector given by,

Hence we have,

Or,

** Therefore this is the required Cartesian equation of the plane. **

** Question:3(c) ** Find the Cartesian equation of the following planes:

** Answer: **

Given the equation of plane

So we have to find the Cartesian equation,

Any point on this plane will satisfy the equation and its position vector given by,

Hence we have,

Or,

** Therefore this is the required Cartesian equation of the plane. **

** Question:4(a) ** In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

** Answer: **

Let the coordinates of the foot of perpendicular P from the origin to the plane be

Given a plane equation ,

Or,

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

Therefore

So, now dividing both sides of the equation by we will obtain,

This equation is similar to where, are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by .

The coordinates of the foot of the perpendicular are;

or

** Question:4(b) ** In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

** Answer: **

Let the coordinates of the foot of perpendicular P from the origin to the plane be

Given a plane equation ,

Or,

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

Therefore

So, now dividing both sides of the equation by we will obtain,

This equation is similar to where, are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by .

The coordinates of the foot of the perpendicular are;

or

** Question:4(c) ** In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

** Answer: **

Let the coordinates of the foot of perpendicular P from the origin to the plane be

Given plane equation .

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

Therefore

So, now dividing both sides of the equation by we will obtain,

This equation is similar to where, are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by .

The coordinates of the foot of the perpendicular are;

or ..

** Question: ** ** 4(d) ** In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

** Answer: **

Let the coordinates of the foot of perpendicular P from the origin to the plane be

Given plane equation .

or written as

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

Therefore

So, now dividing both sides of the equation by we will obtain,

Then finding the coordinates of the foot of the perpendicular are given by .

The coordinates of the foot of the perpendicular are;

or .

** Answer: **

Given the point and the normal vector which is perpendicular to the plane is

The position vector of point A is

So, the vector equation of the plane would be given by,

Or

where is the position vector of any arbitrary point in the plane.

Therefore, the equation we get,

or

** So, this is the required Cartesian equation of the plane. **

** Question:5(b) ** Find the vector and cartesian equations of the planes

that passes through the point (1,4, 6) and the normal vector to the plane is .

** Answer: **

Given the point and the normal vector which is perpendicular to the plane is

The position vector of point A is

So, the vector equation of the plane would be given by,

Or

where is the position vector of any arbitrary point in the plane.

Therefore, the equation we get,

** So, this is the required Cartesian equation of the plane. **

** Question:6(a) ** Find the equations of the planes that passes through three points.

(1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3)

** Answer: **

The equation of the plane which passes through the three points is given by;

Determinant method,

Or,

Here, these three points A, B, C are collinear points.

** Hence there will be an infinite number of planes possible which passing through the given points. **

** Question:6(b) ** Find the equations of the planes that passes through three points.

(1, 1, 0), (1, 2, 1), (– 2, 2, – 1)

** Answer: **

The equation of the plane which passes through the three points is given by;

Determinant method,

As determinant value is not equal to zero hence there must be a plane that passes through the points A, B, and C.

Finding the equation of the plane through the points,

After substituting the values in the determinant we get,

** So, this is the required Cartesian equation of the plane. **

** Question:7 ** Find the intercepts cut off by the plane 2x + y – z = 5.

** Answer: **

Given plane

We have to find the intercepts that this plane would make so,

Making it look like intercept form first:

By dividing both sides of the equation by 5 (as we have to make the R.H.S =1) , we get then,

So, as we know that from the equation of a plane in intercept form, where a,b,c are the intercepts cut off by the plane at x,y, and z-axes respectively.

Therefore after comparison, we get the values of a,b, and c.

.

** Hence the intercepts are . **

** Question:8 ** Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

** Answer: **

Given that the plane is parallel to the ZOX plane.

So, we have the equation of plane ZOX as .

And an intercept of 3 on the y-axis

Intercept form of a plane given by;

So, here the plane would be parallel to the x and z-axes both.

we have any plane parallel to it is of the form, .

Equation of the plane required is .

** Answer: **

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

Can be written in the form of; , where

So, the plane passes through the point , will satisfy the above equation.

That implies

Now, substituting the value of in the equation above we get the final equation of the plane;

** is the required equation of the plane. **

** Answer: **

Here and

and and

Hence, using the relation , we get

or ..............(1)

where, is some real number.

Taking , we get

or

or .............(2)

Given that the plane passes through the point , it must satisfy (2), i.e.,

or

Putting the values of in (1), we get

or

or

** which is the required vector equation of the plane. **

** Answer: **

The equation of the plane through the intersection of the given two planes, and is given in Cartesian form as;

or ..................(1)

So, the direction ratios of (1) plane are which are .

Then, the plane in equation (1) is perpendicular to whose direction ratios are .

As planes are perpendicular then,

we get,

or

or

Then we will substitute the values of in the equation (1), we get

or

** This is the required equation of the plane. **

** Question:12 ** Find the angle between the planes whose vector equations are and .

** Answer: **

Given two vector equations of plane

and .

Here, and

The formula for finding the angle between two planes,

.............................(1)

and

Now, we can substitute the values in the angle formula (1) to get,

or

or

7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

** Answer: **

Two planes

whose direction ratios are and whose direction ratios are ,

are said to ** Parallel: **

If,

and ** Perpendicular: **

If,

And the angle between is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

** Parallel check: **

Clearly, the given planes are ** NOT ** parallel.

** Perpendicular check: **

.

Clearly, the given planes are ** NOT ** perpendicular.

Then find the angle between them,

2x + y + 3z – 2 = 0 and x – 2y + 5 = 0

** Answer: **

Two planes

whose direction ratios are and whose direction ratios are ,

are said to ** Parallel: **

If,

and ** Perpendicular: **

If,

And the angle between is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

** Perpendicular check: **

.

** Thus, the given planes are perpendicular to each other. **

2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0

** Answer: **

Two planes

whose direction ratios are and whose direction ratios are ,

are said to ** Parallel: **

If,

and ** Perpendicular: **

If,

And the angle between is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

** Parallel check: **

** Thus, the given planes are parallel as **

2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0

** Answer: **

Two planes

whose direction ratios are and whose direction ratios are ,

are said to ** Parallel: **

If,

and ** Perpendicular: **

If,

And the angle between is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

** Parallel check: **

Therefore

** Thus, the given planes are parallel to each other. **

4x + 8y + z – 8 = 0 and y + z – 4 = 0

** Answer: **

Two planes

whose direction ratios are and whose direction ratios are ,

are said to ** Parallel: **

If,

and ** Perpendicular: **

If,

And the angle between is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

** Parallel check: **

Clearly, the given planes are ** NOT ** parallel as .

** Perpendicular check: **

.

Clearly, the given planes are ** NOT ** perpendicular.

Then finding the angle between them,

** Question:14 ** In the following cases, find the distance of each of the given points from the corresponding given plane

** Answer: **

We know that the distance between a point and a plane is given by,

.......................(1)

So, calculating for each case;

** (a) ** Point and Plane

Therefore,

** (b) ** Point and Plane

Therefore,

** (c) ** Point and Plane

Therefore,

** (d) ** Point and Plane

Therefore,

**NCERT class 12 three dimensional geometry ncert solutions - Miscellaneous Exercise **

** Answer: **

We can assume the line joining the origin, be OA where and the point and PQ be the line joining the points and .

Then the direction ratios of the line OA will be and that of line PQ will be

So to check whether line OA is perpendicular to line PQ then,

Applying the relation we know,

** Therefore OA is perpendicular to line PQ. **

** Answer: **

Given that are the direction cosines of two mutually perpendicular lines.

Therefore, we have the relation:

** .........................(1) **

** .............(2) **

Now, let us assume be the new direction cosines of the lines which are perpendicular to the line with direction cosines.

Therefore we have,

Or,

......(3)

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

where, ** ........................(4) **

Then we know that,

So, from the equation (1) and (2) we have,

Therefore, ** ..(5) **

Now, we will substitute the values from the equation (4) and (5) in equation (3), to get

Therefore we have the direction cosines of the required line as;

** Question:3 ** Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.

** Answer: **

Given direction ratios and .

Thus the angle between the lines A is given by;

a

** Thus, the angle between the lines is **

** Question:4 ** Find the equation of a line parallel to x-axis and passing through the origin.

** Answer: **

Equation of a line parallel to the x-axis and passing through the origin is itself ** x-axis ** .

So, let A be a point on the x-axis.

Therefore, the coordinates of A are given by , where .

Now, the direction ratios of OA are

So, the equation of OA is given by,

or

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

** Answer: **

Direction ratios of AB are

and Direction ratios of CD are

So, it can be noticed that,

Therefore, AB is parallel to CD.

** Thus, we can easily say the angle between AB and CD which is either . **

** Question:6 ** If the lines and are perpendicular, find the value of k.

** Answer: **

Given both lines are perpendicular so we have the relation;

For the two lines whose direction ratios are known,

We have the direction ratios of the lines, and are and respectively.

Therefore applying the formula,

or

** For, the lines are perpendicular. **

** Question:7 ** Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane

** Answer: **

Given that the plane is passing through the point so, the position vector of the point A is and perpendicular to the plane whose direction ratios are and the normal vector is

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

, where

** Question:8 ** Find the equation of the plane passing through (a, b, c) and parallel to the plane .

** Answer: **

Given that the plane is passing through and is parallel to the plane

So, we have

The position vector of the point is,

and any plane which is parallel to the plane, is of the form,

. ** .......................(1) **

Therefore the equation we get,

Or,

So, now substituting the value of in equation (1), we get

** .................(2) **

** So, this is the required equation of the plane . **

Now, substituting in equation (2), we get

Or,

** Question:9 ** Find the shortest distance between lines and .

** Answer: **

Given lines are;

and

So, we can find the shortest distance between two lines and by the formula,

** ...........................(1) **

Now, we have from the comparisons of the given equations of lines.

** **

** **

So,

and

Now, substituting all values in equation (3) we get,

** Hence 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: **

We know that the equation of the line that passes through the points and is given by the relation;

and the line passing through the points,

And any point on the line is of the form .

So, the equation of the YZ plane is

Since the line passes through YZ- plane,

we have then,

or and

So, 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: **

We know that the equation of the line that passes through the points and is given by the relation;

and the line passing through the points,

And any point on the line is of the form .

So, the equation of ZX plane is

Since the line passes through YZ- plane,

we have then,

or and

So, therefore the required point is

** Answer: **

We know that the equation of the line that passes through the points and is given by the relation;

and the line passing through the points, .

And any point on the line is of the form.

This point lies on the plane,

or .

Hence, the coordinates of the required point are or .

** Answer: **

Given

two planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

the normal vectors of these plane are

Since the normal vector of the required plane is perpendicular to the normal vector of given planes, the required plane's normal vector will be :

Now, as we know

the equation of a plane in vector form is :

Now Since this plane passes through the point (-1,3,2)

Hence the equation of the plane is

** Question:14 ** If the points (1, 1, p) and (– 3, 0, 1) be equidistant from the plane then find the value of p.

** Answer: **

Given that the points and are equidistant from the plane

So we can write the position vector through the point is

Similarly, the position vector through the point is

The equation of the given plane is

and We know that the perpendicular distance between a point whose position vector is and the plane, and

Therefore, the distance between the point and the given plane is

** nbsp; .........................(1) **

Similarly, the distance between the point , and the given plane is

** .........................(2) **

** And it is given that the distance between the required plane and the points, and is equal. **

** therefore we have, **

** or or **

** Answer: **

So, the given planes are:

and

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

** ..............(1) **

Its direction ratios are and = 0

The required plane is parallel to the x-axis.

Therefore, its normal is perpendicular to the x-axis.

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

Substituting in equation (1), we obtain

So, the Cartesian equation is

** Answer: **

We have the coordinates of the points and respectively.

Therefore, the direction ratios of OP are

And we know that the equation of the plane passing through the point is

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

Here, the direction ratios of normal are and and the point P is .

Thus, the equation of the required plane is

** Answer: **

The equation of the plane passing through the line of intersection of the given plane in

,,,,,,,,,,,,,(1)

The plane in equation (1) is perpendicular to the plane, Therefore

Substituting in equation (1), we obtain

.......................(4)

So, this is the vector equation of the required plane.

The Cartesian equation of this plane can be obtained by substituting in equation (1).

Therefore we get the answer

** Question:18 ** Find the distance of the point (– 1, – 5, – 10) from the point of intersection of the line and the plane .

** Answer: **

Given,

Equation of a line :

Equation of the plane

Let's first find out the point of intersection of line and plane.

putting the value of into the equation of a plane from the equation from line

Now, from the equation, any point p in line is

So the point of intersection is

SO, Now,

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

Hence the required distance is 13.

** Question:19 ** Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes and .

** Answer: **

Given

A point through which line passes

two plane

And

it can be seen that normals of the planes are

since the line is parallel to both planes, its parallel vector will be perpendicular to normals of both planes.

So, a vector perpendicular to both these normal vector is

Now a line which passes through and parallels to is

So the required line is

** Questio ** n: ** 20 ** Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines:

** Answer: **

Given

Two straight lines in 3D whose direction cosines (3,-16,7) and (3,8,-5)

Now the two vectors which are parallel to the two lines are

and

As we know, a vector perpendicular to both vectors and is , so

A vector parallel to this vector is

Now as we know the vector equation of the line which passes through point p and parallel to vector d is

Here in our question, give point p = (1,2,-4) which means position vector of this point is

So, the required line is

** Question:21 ** Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then .

** Answer: **

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

The distance p of the plane from the origin is given by

Hence proved

** Question:22 ** Distance between the two planes: 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is

(A) 2 units (B) 4 units (C) 8 units (D)

** Answer: **

Given equations are

and

Now, it is clear from equation (i) and (ii) that given planes are parallel

We know that the distance between two parallel planes is given by

Put the values in this equation

we will get,

Therefore, the correct answer is (D)

** Question:23 ** The planes: 2x – y + 4z = 5 and 5x – 2.5y + 10z = 6 are

(A) Perpendicular (B) Parallel (C) intersect y-axis (D) passes through

** Answer: **

Given equations of planes are

and

Now, from equation (i) and (ii) it is clear that given planes are parallel to each other

Therefore, the correct answer is (B)

**If you are looking for exercises solutions for chapter 3d Geometry class 12 then they are listed below.**

- Three Dimensional Geometry Class 12 Exercise 11.1
- Three Dimensional Geometry Class 12 Exercise 11.2
- Three Dimensional Geometry Class 12 Exercise 11.3
- Three Dimensional Geometry Class 12 Miscellaneous Exercise

All the important topics are covered in the maths chapter 11 class 12 NCERT solutions.

A total of 36 questions in 3 exercises are given in this Maths chapter 11 class 12 solutions.

All these NCERT questions are solved and explained in the NCERT solutions for class 12 maths chapter 11 three dimensional geometry article to clear your doubts.

In this NCERT Solutions for Class 12 Maths Chapter 11 PDF Download, we deal with formulas like-

If l, m, n are the direction cosines of a line, then.

andDirection cosines of a line joining two pointsare, where

If l, m, n are the direction cosines and a, b, c are the direction of a line then-

**Also read,**

__NCERT Exemplar Class 12 Chemistry Solutions____NCERT Exemplar Class 12 Mathematics Solutions____NCERT Exemplar Class 12 Biology Solutions____NCERT Exemplar Class 12 Physics Solutions__

11.1 Introduction

11.2 Direction Cosines and Direction Ratios of a Line

11.2.1 Relation between the direction cosines of a line

11.2.2 Direction cosines of a line passing through two points

11.3 Equation of a Line in Space

11.3.1Equation of a line through a given point and parallel to a given vector b

11.3.2 Equation of a line passing through two given points

11.4 Angle between Two Lines

11.5 Shortest Distance between Two Lines

11.5.1 Distance between two skew lines

11.5.2 Distance between parallel lines

11.6 Plane

11.6.1 Equation of a plane in normal form

11.6.2 Equation of a plane perpendicular to a given vector and passing through a given point

11.6.3 Equation of a plane passing through three noncollinear points

11.6.4 Intercept form of the equation of a plane

11.6.5 Plane passing through the intersection of two given planes

11.7 Coplanarity of Two Lines

11.8 Angle between Two Planes

11.9 Distance of a Point from a Plane

11.10 Angle between a Line and a Plane

Chapter 1 | NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions |

Chapter 2 | NCERT solutions for class 12 maths chapter 2 Inverse Trigonometric Functions |

Chapter 3 | |

Chapter 4 | |

Chapter 5 | NCERT solutions for class 12 maths chapter 5 Continuity and Differentiability |

Chapter 6 | NCERT solutions for class 12 maths chapter 6 Application of Derivatives |

Chapter 7 | |

Chapter 8 | NCERT solutions for class 12 maths chapter 8 Application of Integrals |

Chapter 9 | NCERT solutions for class 12 maths chapter 9 Differential Equations |

Chapter 10 | NCERT solutions for class 12 maths chapter 10 Vector Algebra |

Chapter 11 | NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry |

Chapter 12 | NCERT solutions for class 12 maths chapter 12 Linear Programming |

Chapter 13 |

Chapter 11 of Class 12 Maths is titled "Three Dimensional Geometry." NCERT Solutions for this chapter provide step-by-step solutions to all the exercises and problems included in the textbook. Some of the key features of NCERT Solutions for Class 12 Maths Chapter 11 are:

Comprehensive Coverage: The ch 11 maths class 12 solutions cover all the important topics and concepts discussed in the chapter, including the coordinate axes and coordinate planes in three dimensions, distance between two points, section formula, direction cosines and direction ratios of a line, angle between two lines, equation of a line and a plane, and the distance of a point from a plane.

Simple and Clear Explanation: The class 12 maths ch 11 question answer are written in a simple and clear language that makes it easy for students to understand even the most complex concepts.

Step-by-step Approach: The class 12 maths ch 11 question answer are provided in a step-by-step manner, making it easy for students to follow and learn.

Class 12 Maths Chapter 11 NCERT solutions are very easy for you to understand the concepts as they are explained in a step-by-step manner.

NCERT Class 12 Maths solutions chapter 11 will give you some new insight into the concepts.

Scoring good marks in the exam is now a reality with the help of these solutions of NCERT for class 12 maths chapter 11 three dimensional geometry as these questions are answered by the experts who know how best to answer the questions in the board exam.

You should solve the miscellaneous exercise also, to develop a grip on the concepts. Here, you will get solutions for miscellaneous exercise too.

Three dimensional geometry class 12 ncert solutions PDF Download will also be made available soon.

1. What is the weightage of the chapter three-dimensional geometry for the CBSE board exam?

The concepts of vector algebra and Three Dimensional Geometry can be used interchangeably. if two chapters are combined, vector algebra & three-dimensional geometry has a 17% weightage in the 12th board maths final exam. after getting command of these concepts it becomes easy for students to score well in the exam therefore NCERT Notes, NCERT syllabus, and NCERT textbooks are recommended.

2. Would it be accurate to say that three dimensional geometry class 12 ncert solutions is the most helpful study material for students during revision?

Undoubtedly, NCERT Solutions for Class 12 Maths Chapter 11 stand out as the top study material aiding students in effortlessly revising complex concepts. The solutions present a well-reasoned explanation to facilitate student learning. The team of experts at Careers360 has carefully crafted step-by-step solutions that encourage students to employ an analytical thinking approach. Moreover, these solutions can be cross-referenced to gain insights into alternative methods to solve textbook problems.

3. How the NCERT solutions are helpful in the CBSE board exam?

Only knowing the answer is not enough to score good marks in the exam. One should know how best to answer in order to get good marks. NCERT solutions are provided by the experts who know how best to write answer in the board exam in order to get good marks.

4. What are the important topics covered in the chapter of ncert solutions class 12 maths chapter 11?

Class 12 chapter 11 maths Maths covers the following important topics:

Introduction (11.1)

Direction cosines and direction ratios of a line (11.2)

Equation of a line in space (11.3)

Angle between two lines (11.4)

Shortest distance between two lines (11.5)

Plane (11.6)

Coplanarity of two lines (11.7)

Angle between two planes (11.8)

Distance of a point from a plane (11.9)

Angle between a line and a plane (11.10)

5. How class 12 math chapter 11 miscellaneous exercises are useful?

The ncert solutions class 12 maths chapter 11 miscellaneous is useful for students in several ways. These exercises include additional problems that are not part of the main textbook exercises but are relevant to the chapter's concepts.

Firstly, NCERT solutions for class 12 maths chapter 11 miscellaneous exercise help students to gain a deeper understanding of the concepts covered in Chapter 11.

Secondly, these ch 11 miscellaneous class 12 exercises help students to test their problem-solving abilities and improve their skills.

Thirdly, solving the miscellaneous exercise problems for 3d class 12 ncert solutions can help students to prepare for their exams.

Application Date:20 November,2023 - 19 December,2023

Application Date:20 November,2023 - 19 December,2023

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For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.

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Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

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Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

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Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

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The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

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Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production.

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A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

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Individuals who want to opt for a career as a Visual Communication Designer will work in the graphic design and arts industry. Every sector in the modern age is using visuals to connect with people, clients, or customers. This career involves art and technology and candidates who want to pursue their career as visual communication designer has a great scope of career opportunity.

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In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

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Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

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In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

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Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

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Advertising managers consult with the financial department to plan a marketing strategy schedule and cost estimates. We often see advertisements that attract us a lot, not every advertisement is just to promote a business but some of them provide a social message as well. There was an advertisement for a washing machine brand that implies a story that even a man can do household activities. And of course, how could we even forget those jingles which we often sing while working?

2 Jobs Available

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

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A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available

3 Jobs Available

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

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**Quality Assurance Manager Job Description:** A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes.

2 Jobs Available

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans.

2 Jobs Available

Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment.

2 Jobs Available

A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

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Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

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ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available

3 Jobs Available

An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.

2 Jobs Available

A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements.

2 Jobs Available

The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.

2 Jobs Available

A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy.

2 Jobs Available

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