CBSE Class 12 Maths Chapter 4 Vector Algebra NCERT Solutions

The scalar product (or dot product) of two vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\) is defined as the product of their magnitudes and the cosine of the angle \(\theta\) between them. Mathematically, it is expressed as:

$$\backslash overrightarrow\{a\}\; \backslash cdot\; \backslash overrightarrow\{b\}\; =\; |\backslash overrightarrow\{a\}|\; |\backslash overrightarrow\{b\}|\; \backslash cos\; \backslash theta$$

Here, \(|\overrightarrow{a}|\) and \(|\overrightarrow{b}|\) represent the magnitudes of vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\), respectively.

Two non-zero vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\) are said to be perpendicular (or orthogonal) if the angle \(\theta\) between them is \(90^{\circ}\) or \(\frac{\pi}{2}\) radians.

Using the formula for the scalar product, \(\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| \cos \theta\), if \(\theta = 90^{\circ}\), then \(\cos 90^{\circ} = 0\).

Therefore, the condition for perpendicularity is:

$$\backslash overrightarrow\{a\}\; \backslash cdot\; \backslash overrightarrow\{b\}\; =\; 0$$

The vectors \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) are the standard unit vectors along the positive x, y, and z axes, respectively. They are mutually perpendicular and have a magnitude of 1.

The dot product of a unit vector with itself is 1:

$$\backslash hat\{i\}\; \backslash cdot\; \backslash hat\{i\}\; =\; \backslash hat\{j\}\; \backslash cdot\; \backslash hat\{j\}\; =\; \backslash hat\{k\}\; \backslash cdot\; \backslash hat\{k\}\; =\; 1$$

Since they are mutually perpendicular, the dot product of any two distinct vectors is 0:

$$\backslash hat\{i\}\; \backslash cdot\; \backslash hat\{j\}\; =\; \backslash hat\{j\}\; \backslash cdot\; \backslash hat\{i\}\; =\; 0$$

$$\backslash hat\{j\}\; \backslash cdot\; \backslash hat\{k\}\; =\; \backslash hat\{k\}\; \backslash cdot\; \backslash hat\{j\}\; =\; 0$$

$$\backslash hat\{k\}\; \backslash cdot\; \backslash hat\{i\}\; =\; \backslash hat\{i\}\; \backslash cdot\; \backslash hat\{k\}\; =\; 0$$

The vector product (or cross product) of two vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\) is a vector whose magnitude is the product of the magnitudes of \(\overrightarrow{a}\) and \(\overrightarrow{b}\) and the sine of the angle \(\theta\) between them, and whose direction is perpendicular to the plane containing \(\overrightarrow{a}\) and \(\overrightarrow{b}\).

It is expressed as:

$$\backslash overrightarrow\{a\}\; \backslash times\; \backslash overrightarrow\{b\}\; =\; |\backslash overrightarrow\{a\}|\; |\backslash overrightarrow\{b\}|\; \backslash sin\; \backslash theta\; \backslash cdot\; \backslash hat\{n\}$$

where \(\hat{n}\) is a unit vector perpendicular to the plane containing \(\overrightarrow{a}\) and \(\overrightarrow{b}\), such that \(\overrightarrow{a}\), \(\overrightarrow{b}\), and \(\hat{n}\) form a right-handed system.

The cross product of the standard unit vectors \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) follows a cyclic rule:

The cross product of a unit vector with itself is the zero vector:

$$\backslash hat\{i\}\; \backslash times\; \backslash hat\{i\}\; =\; \backslash hat\{j\}\; \backslash times\; \backslash hat\{j\}\; =\; \backslash hat\{k\}\; \backslash times\; \backslash hat\{k\}\; =\; \backslash overrightarrow\{0\}$$

The cross product of different unit vectors follows the cyclic order \(\hat{i} \rightarrow \hat{j} \rightarrow \hat{k} \rightarrow \hat{i}\):

$$\backslash hat\{i\}\; \backslash times\; \backslash hat\{j\}\; =\; \backslash hat\{k\},\; \backslash quad\; \backslash hat\{j\}\; \backslash times\; \backslash hat\{i\}\; =\; -\backslash hat\{k\}$$

$$\backslash hat\{j\}\; \backslash times\; \backslash hat\{k\}\; =\; \backslash hat\{i\},\; \backslash quad\; \backslash hat\{k\}\; \backslash times\; \backslash hat\{j\}\; =\; -\backslash hat\{i\}$$

$$\backslash hat\{k\}\; \backslash times\; \backslash hat\{i\}\; =\; \backslash hat\{j\},\; \backslash quad\; \backslash hat\{i\}\; \backslash times\; \backslash hat\{k\}\; =\; -\backslash hat\{j\}$$

If two adjacent sides of a triangle are represented by vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\), the area of the triangle is half the magnitude of the vector product of these two vectors.

Area = $\backslash frac\{1\}\{2\}\; |\backslash overrightarrow\{a\}\; \backslash times\; \backslash overrightarrow\{b\}|$

This is because the magnitude \(|\overrightarrow{a} \times \overrightarrow{b}|\) represents the area of the parallelogram formed by \(\overrightarrow{a}\) and \(\overrightarrow{b}\) as adjacent sides, and a triangle formed by these sides is exactly half of that parallelogram.

If the two diagonals of a quadrilateral are given by vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\), the area of the quadrilateral can be calculated using the formula:

Area = $\backslash frac\{1\}\{2\}\; |\; \backslash overrightarrow\{a\}\; \backslash times\; \backslash overrightarrow\{b\}\; |$

This formula holds true regardless of whether the quadrilateral is convex or concave.

The area of a parallelogram is determined by the magnitude of the vector product of its adjacent sides.

If \(\overrightarrow{a}\) and \(\overrightarrow{b}\) represent the adjacent sides of a parallelogram, its area is given by:

Area = $|\backslash overrightarrow\{a\}\; \backslash times\; \backslash overrightarrow\{b\}|$

This is because \(|\overrightarrow{a} \times \overrightarrow{b}| = |\overrightarrow{a}| |\overrightarrow{b}| \sin \theta\), which is the base times the height of the parallelogram.

Work done is a scalar quantity defined as the product of the force and the displacement in the direction of the force. When force and displacement are represented by vectors, the work done is calculated using the scalar product (dot product).

Work Done, $W\; =\; \backslash overrightarrow\{F\}\; \backslash cdot\; \backslash overrightarrow\{d\}$

This can also be expressed in terms of magnitudes and the angle \(\theta\) between the force and displacement vectors:

$W\; =\; |\backslash overrightarrow\{F\}|\; |\backslash overrightarrow\{d\}|\; \backslash cos\; \backslash theta$

The moment of a force about a point (also known as torque) is a measure of the turning effect of the force. It is calculated as the vector product (cross product) of the position vector \(\overrightarrow{r}\) (from the point to the point of application of the force) and the force vector \(\overrightarrow{F}\).

Moment = $\backslash overrightarrow\{r\}\; \backslash times\; \backslash overrightarrow\{F\}$

The direction of the moment vector is perpendicular to the plane containing \(\overrightarrow{r}\) and \(\overrightarrow{F}\), following the right-hand rule.

This statement describes the addition of vectors using the parallelogram law of vector addition. If \(\overrightarrow{a}\) and \(\overrightarrow{b}\) represent two adjacent sides of a parallelogram originating from the same point O (i.e., \(\overrightarrow{OA} = \overrightarrow{a}\) and \(\overrightarrow{OB} = \overrightarrow{b}\)), then their sum \(\overrightarrow{a} + \overrightarrow{b}\) is represented by the diagonal \(\overrightarrow{OC}\) of the parallelogram passing through O.

Alternatively, using the triangle law, if \(\overrightarrow{OA} = \overrightarrow{a}\) and \(\overrightarrow{AB} = \overrightarrow{b}\), then the resultant vector \(\overrightarrow{OB}\) is the sum \(\overrightarrow{a} + \overrightarrow{b}\).

This formula represents the vector joining point A to point B. If \(\overrightarrow{OA}\) is the position vector of point A and \(\overrightarrow{OB}\) is the position vector of point B (both with respect to the origin O), then the vector \(\overrightarrow{AB}\) is found by subtracting the position vector of the initial point (A) from the position vector of the terminal point (B).

This is derived from the triangle law of vector addition: \(\overrightarrow{OA} + \overrightarrow{AB} = \overrightarrow{OB}\), which rearranges to \(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}\).

This describes the section formula for position vectors. It provides a way to find the position vector of a point that divides a line segment internally in a given ratio.

For internal division of the line segment joining points A and B (with position vectors \(\overrightarrow{OA} = \overrightarrow{a}\) and \(\overrightarrow{OB} = \overrightarrow{b}\)) in the ratio m:n, the position vector of the dividing point P is:

$$\backslash overrightarrow\{OP\}\; =\; \backslash frac\{m\backslash overrightarrow\{OB\}\; +\; n\backslash overrightarrow\{OA\}\}\{m+n\}$$

When P is the midpoint, the ratio is 1:1 (m=1, n=1), simplifying the formula to:

$$\backslash overrightarrow\{OP\}\; =\; \backslash frac\{1\backslash cdot\backslash overrightarrow\{OB\}\; +\; 1\backslash cdot\backslash overrightarrow\{OA\}\}\{1+1\}\; =\; \backslash frac\{\backslash overrightarrow\{OB\}\; +\; \backslash overrightarrow\{OA\}\}\{2\}\; =\; \backslash frac\{\backslash overrightarrow\{a\}\; +\; \backslash overrightarrow\{b\}\}\{2\}$$

A unit vector is a vector with a magnitude of 1. To find the unit vector in the direction of a given vector \(\overrightarrow{r}\), we divide the vector by its magnitude.

The magnitude of \(\overrightarrow{r} = x\hat{i} + y\hat{j} + z\hat{k}\) is $|\backslash overrightarrow\{r\}|\; =\; \backslash sqrt\{x^2\; +\; y^2\; +\; z^2\}$.

Therefore, the unit vector \(\hat{r}\) in the direction of \(\overrightarrow{r}\) is:

$$\backslash hat\{r\}\; =\; \backslash frac\{\backslash overrightarrow\{r\}\}\{|\backslash overrightarrow\{r\}|\}\; =\; \backslash frac\{x\backslash hat\{i\}\; +\; y\backslash hat\{j\}\; +\; z\backslash hat\{k\}\}\{\backslash sqrt\{x^2\; +\; y^2\; +\; z^2\}\}$$

The direction cosines of a vector are the cosines of the angles that the vector makes with the positive x, y, and z axes, respectively. These angles are denoted by \(\alpha\), \(\beta\), and \(\gamma\).

For a vector \(\overrightarrow{r} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\), the magnitude is $|\backslash overrightarrow\{r\}|\; =\; \backslash sqrt\{a\_1^2\; +\; a\_2^2\; +\; a\_3^2\}$.

The direction cosines are calculated as:

$$\backslash cos\; \backslash alpha\; =\; \backslash frac\{a\_1\}\{|\backslash overrightarrow\{r\}|\}\; =\; \backslash frac\{a\_1\}\{\backslash sqrt\{a\_1^2\; +\; a\_2^2\; +\; a\_3^2\}\}$$

$$\backslash cos\; \backslash beta\; =\; \backslash frac\{a\_2\}\{|\backslash overrightarrow\{r\}|\}\; =\; \backslash frac\{a\_2\}\{\backslash sqrt\{a\_1^2\; +\; a\_2^2\; +\; a\_3^2\}\}$$

$$\backslash cos\; \backslash gamma\; =\; \backslash frac\{a\_3\}\{|\backslash overrightarrow\{r\}|\}\; =\; \backslash frac\{a\_3\}\{\backslash sqrt\{a\_1^2\; +\; a\_2^2\; +\; a\_3^2\}\}$$

These values indicate the direction of the vector in 3D space. A key property is that the sum of the squares of the direction cosines is always 1: \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\).