CBSE Class 12 Maths Chapter 10: Vector Algebra NCERT Solutions

The scalar product, also known as the 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. This product results in a scalar quantity.

The formula is expressed as:

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

Where \(|\overrightarrow{a}|\) 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, \(\overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| \cos \theta\), when \(\theta = 90^{\circ}\), \(\cos 90^{\circ} = 0\). Therefore, the scalar product becomes zero.

The condition for perpendicularity is:

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

This means that if the dot product of two non-zero vectors is zero, they are perpendicular to each other.

The vectors \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) are the standard unit vectors along the positive x-axis, y-axis, and z-axis, respectively. Since they are mutually perpendicular, the dot product of any two distinct vectors is zero, and the dot product of a vector with itself (which is the square of its magnitude) is 1, as their magnitudes are 1.

The dot products are:

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

And:

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

The vector product, also known as the cross product, of two vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\) is a vector quantity. Its magnitude is equal to the product of the magnitudes of the two vectors and the sine of the angle \(\theta\) between them. The direction of the resulting vector is perpendicular to the plane containing \(\overrightarrow{a}\) and \(\overrightarrow{b}\), following the right-hand rule.

The formula is:

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

Where \(\hat{n}\) is a unit vector perpendicular to both \(\overrightarrow{a}\) and \(\overrightarrow{b}\).

The cross product of any vector with itself is zero, as the angle between them is \(0^{\circ}\) and \(\sin 0^{\circ} = 0\). For the standard basis vectors \(\hat{i}\), \(\hat{j}\), \(\hat{k}\), the cross products follow a cyclic order:

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

The cross products of distinct basis vectors are:

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

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

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

The magnitude of the cross product of two vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\) gives the area of the parallelogram formed by these vectors when they are considered as adjacent sides. A triangle formed by these vectors as adjacent sides has half the area of this parallelogram.

Therefore, the area of the triangle is:

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

The area of a quadrilateral can be calculated using its diagonals represented by vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\). The formula involves the cross product of the diagonal vectors.

The area of the quadrilateral is given by:

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

The magnitude of the vector (cross) product of two vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\) directly gives the area of the parallelogram formed by these vectors when they represent the adjacent sides originating from the same vertex.

The area of the parallelogram is:

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

Work done (W) by a constant force \(\overrightarrow{F}\) on an object that undergoes a displacement \(\overrightarrow{d}\) is defined as the scalar product of the force vector and the displacement vector. It represents the component of the force acting in the direction of the displacement, multiplied by the magnitude of the displacement.

The formula for work done is:

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

Where \(\theta\) is the angle between the force vector \(\overrightarrow{F}\) and the displacement vector \(\overrightarrow{d}\).

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

The moment of the force is:

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

The magnitude of the moment is \(|\overrightarrow{r}| |\overrightarrow{F}| \sin \theta\), and its direction is perpendicular to the plane containing \(\overrightarrow{r}\) and \(\overrightarrow{F}\).

Vector addition can be visualized using the parallelogram law or the triangle law of vector addition. If \(\overrightarrow{a}\) and \(\overrightarrow{b}\) represent two vectors originating from the same point O, say \(\overrightarrow{OA} = \overrightarrow{a}\) and \(\overrightarrow{OB} = \overrightarrow{b}\), then their sum \(\overrightarrow{a} + \overrightarrow{b}\) is represented by the diagonal of the parallelogram formed by \(\overrightarrow{a}\) and \(\overrightarrow{b}\) originating from O. Alternatively, if we place the tail of \(\overrightarrow{b}\) at the head of \(\overrightarrow{a}\), the resultant vector \(\overrightarrow{OC}\) from the tail of \(\overrightarrow{a}\) to the head of \(\overrightarrow{b}\) represents the sum.

If \(\overrightarrow{OA} = \overrightarrow{a}\) and \(\overrightarrow{OB} = \overrightarrow{b}\), then the sum is:

$$\backslash overrightarrow\{OC\}\; =\; \backslash overrightarrow\{OA\}\; +\; \backslash overrightarrow\{OB\}\; =\; \backslash overrightarrow\{a\}\; +\; \backslash overrightarrow\{b\}$$

The position vector of a point represents its location relative to an origin (usually denoted as O). If \(\overrightarrow{OA}\) is the position vector of point A and \(\overrightarrow{OB}\) is the position vector of point B, then the vector representing the displacement from point A to point B (\(\overrightarrow{AB}\)) can be found by subtracting the position vector of the initial point (A) from the position vector of the terminal point (B).

The displacement vector \(\overrightarrow{AB}\) is calculated as:

$$\backslash overrightarrow\{AB\}\; =\; \backslash overrightarrow\{OB\}\; -\; \backslash overrightarrow\{OA\}$$

Here, O represents the origin.

The section formula allows us to find the position vector of a point that divides a line segment internally in a specific ratio. Let \(\overrightarrow{OA} = \overrightarrow{a}\) and \(\overrightarrow{OB} = \overrightarrow{b}\) be the position vectors of points A and B, respectively. If point P divides the line segment AB internally in the ratio m:n, its position vector \(\overrightarrow{OP}\) is given by the weighted average of the position vectors of A and B.

The formula is:

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

If P is the midpoint of AB, then the ratio m:n is 1:1. In this case, the formula simplifies to:

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

A unit vector is a vector with a magnitude of 1. To find a unit vector in the direction of any 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 \(|\overrightarrow{r}| = \sqrt{x^2 + y^2 + z^2}\).

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. Let these angles be \(\alpha\), \(\beta\), and \(\gamma\), respectively. The direction cosines are denoted by \(\cos \alpha\), \(\cos \beta\), and \(\cos \gamma\).

For the vector \(\overrightarrow{r} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\), its magnitude is \(|\overrightarrow{r}| = \sqrt{a_1^2 + a_2^2 + a_3^2}\). The direction cosines are calculated as:

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

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

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

It is important to note that the sum of the squares of the direction cosines is always 1: \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\).