NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra

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NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra covers important concepts of the difference between a scalar and a vector quantity, the properties of these quantities, and the operations of vectors. There are two types of physical quantities, scalars and vectors. The scalar quantity has only magnitude, whereas the vector quantity has both magnitude and direction. Vector algebra studies the algebra of vector quantities.

The chapters Vectors and Three Dimensional Geometry holds a weightage of 14 marks in the CBSE Class 12 Examination. The questions asked in the examination test the concepts of types of vectors (equal, zero, unit, parallel and collinear vectors), position vector, negative of a vector, addition of vectors, multiplication of a vector by a scalar, and vector (cross) product of vectors.

Download PDF: NCERT Solutions for Class 12 Mathematics Chapter 10

NCERT Solutions for Class 12 Mathematics Chapter 10 Vector Algebra

Important Topics in Class 12 Mathematics Chapter 10 Vector Algebra

A vector has both magnitudes and direction. It is represented by an arrow that shows the direction (→) and its length shows the magnitude.

The vector is denoted as $\overrightarrow V$and its magnitude is represented as \|V\|.

Addition of Vectors: Let us consider there are two vectors P and Q, then the sum of these vectors can be when the tail of vector Q meets the head of vector A. During this addition, the magnitude and direction of the vectors should not change.

The vector addition follows these important laws: Commutative Law: P + Q = Q + P Associative Law: P + (Q + R) = (P + Q) + R

Subtraction of Vectors: In the subtraction of vectors, the direction of one vector is reversed and then the addition is performed on both the vectors.

It can be denoted as P – Q = P + (-Q)

Multiplication of Vectors: If k is a scalar quantity and is multiplied by vector A, then scalar multiplication is given by kA.

If k is positive then the direction of the vector kA is the direction of vector A, but if the value of k is negative, then the direction of vector kA is opposite of the direction of vector A. The magnitude of the vector kA can be calculated by \|kA\|.

Dot Product: The dot product is a scalar product. It is represented using a dot (.) between the two vectors.

Suppose P and Q are two given vectors, then the dot product for both the vectors is given through P.Q = \|P\| \|Q\| cosθ.

Cross Product: Denoted by a multiplication sign (x) between two vectors, the cross product is a binary vector operation that is defined in a three-dimensional system.

It is represented as P x Q = \|P\| \|Q\| sinθ

NCERT Solutions For Class 12 Maths Chapter 10 Exercises

The detailed solutions for all the NCERT Solutions for Chapter 10 Vector Algebra in different exercises are as follows:

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CBSE CLASS XII Related Questions

1.
The integrating factor of the differential equation $ \frac{dy}{dx} + y = \frac{1 + y}{x} $ is:

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2.
The probability that a student buys a colouring book is 0.7, and a box of colours is 0.2. The probability that she buys a colouring book, given that she buys a box of colours, is 0.3. Find:
(i) The probability that she buys both the colouring book and the box of colours.
(ii) The probability that she buys a box of colours given she buys the colouring book.

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3.
Solve the following linear programming problem graphically: Maximise $ Z = x + 2y $ Subject to the constraints: \[ x - y \geq 0 \] \[ x - 2y \geq -2 \] \[ x \geq 0, \, y \geq 0 \]

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4.
Let $f'(x) = 3(x^2 + 2x) - \frac{4}{x^3} + 5$, $f(1) = 0$. Then, $f(x)$ is:

$x^3 + 3x^2 + \frac{2}{x^2} + 5x + 11$
$x^3 + 3x^2 + \frac{2}{x^2} + 5x - 11$
$x^3 + 3x^2 - \frac{2}{x^2} + 5x - 11$
$x^3 - 3x^2 - \frac{2}{x^2} + 5x - 11$

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5.
A furniture workshop produces three types of furniture: chairs, tables, and beds each day. On a particular day, the total number of furniture pieces produced is 45. It was also found that the production of beds exceeds that of chairs by 8, while the total production of beds and chairs together is twice the production of tables. Determine the units produced of each type of furniture, using the matrix method.

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6.
Evaluate : \[ I = \int_0^{\frac{\pi}{4}} \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} \]

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NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra

NCERT Solutions for Class 12 Mathematics Chapter 10 Vector Algebra

Important Topics in Class 12 Mathematics Chapter 10 Vector Algebra

NCERT Solutions For Class 12 Maths Chapter 10 Exercises

CBSE CLASS XII Related Questions

1.
The integrating factor of the differential equation \( \frac{dy}{dx} + y = \frac{1 + y}{x} \) is:

3.
Solve the following linear programming problem graphically: Maximise \( Z = x + 2y \) Subject to the constraints: \[ x - y \geq 0 \] \[ x - 2y \geq -2 \] \[ x \geq 0, \, y \geq 0 \]

4.
Let $f'(x) = 3(x^2 + 2x) - \frac{4}{x^3} + 5$, $f(1) = 0$. Then, $f(x)$ is:

6.
Evaluate : \[ I = \int_0^{\frac{\pi}{4}} \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} \]

Similar Mathematics Concepts

Comments

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NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra

NCERT Solutions for Class 12 Mathematics Chapter 10 Vector Algebra

Important Topics in Class 12 Mathematics Chapter 10 Vector Algebra

NCERT Solutions For Class 12 Maths Chapter 10 Exercises

CBSE CLASS XII Related Questions

1.The integrating factor of the differential equation \( \frac{dy}{dx} + y = \frac{1 + y}{x} \) is:

3.Solve the following linear programming problem graphically: Maximise \( Z = x + 2y \) Subject to the constraints: \[ x - y \geq 0 \] \[ x - 2y \geq -2 \] \[ x \geq 0, \, y \geq 0 \]

4.Let $f'(x) = 3(x^2 + 2x) - \frac{4}{x^3} + 5$, $f(1) = 0$. Then, $f(x)$ is:

6. Evaluate : \[ I = \int_0^{\frac{\pi}{4}} \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} \]

Similar Mathematics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
The integrating factor of the differential equation \( \frac{dy}{dx} + y = \frac{1 + y}{x} \) is:

3.
Solve the following linear programming problem graphically: Maximise \( Z = x + 2y \) Subject to the constraints: \[ x - y \geq 0 \] \[ x - 2y \geq -2 \] \[ x \geq 0, \, y \geq 0 \]

4.
Let $f'(x) = 3(x^2 + 2x) - \frac{4}{x^3} + 5$, $f(1) = 0$. Then, $f(x)$ is:

6.
Evaluate : \[ I = \int_0^{\frac{\pi}{4}} \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} \]