Get detailed solutions for NCERT Class 12 Maths Exercise 10.3 focusing on the Scalar (Dot) Product of vectors. Learn to calculate the angle between two vectors ($\theta$), find the projection of one vector onto another, determine conditions for perpendicularity ($\vec{a} \cdot \vec{b} = 0$), and prove vector properties involving magnitudes and orthogonality. Essential for mastering Chapter 10 vector operations.

Ex. 10.1

Ex. 10.2

Ex. 10.3

Ex. 10.4

Ex. 10 Misc

This exercise focuses on the scalar product (or dot product) of two vectors, defined by $\mathbf{\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta}$ or $\mathbf{\vec{a} \cdot \vec{b} = a_x b_x + a_y b_y + a_z b_z}$.

image 180 Rbse Solutions Class 12 Maths Exercise 10.3 | Scalar Product | Dot Product

image 181 Rbse Solutions Class 12 Maths Exercise 10.3 | Scalar Product | Dot Product

1. Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and $2$, respectively, having $\vec{a} \cdot \vec{b} = \sqrt{6}$.

The angle $\theta$ between $\vec{a}$ and $\vec{b}$ is given by:

$$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$$

Given: $|\vec{a}| = \sqrt{3}$, $|\vec{b}| = 2$, and $\vec{a} \cdot \vec{b} = \sqrt{6}$.

$$\cos \theta = \frac{\sqrt{6}}{\sqrt{3} \cdot 2} = \frac{\sqrt{3} \cdot \sqrt{2}}{\sqrt{3} \cdot 2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$$

Since $\cos \theta = \frac{1}{\sqrt{2}}$, the angle is $\mathbf{\theta = 45^\circ \text{ or } \frac{\pi}{4}}$.

2. Find the angle between the vectors $\vec{a} = \hat{i} – 2\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} – 2\hat{j} + \hat{k}$.

Calculate the dot product $\vec{a} \cdot \vec{b}$:$$\vec{a} \cdot \vec{b} = (1)(3) + (-2)(-2) + (3)(1) = 3 + 4 + 3 = 10$$
Calculate the magnitudes:$$|\vec{a}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$$$$|\vec{b}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$$
Find the angle $\theta$:$$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{10}{\sqrt{14} \cdot \sqrt{14}} = \frac{10}{14} = \frac{5}{7}$$The angle is $\mathbf{\theta = \cos^{-1}\left(\frac{5}{7}\right)}$.

3. Find the projection of the vector $\vec{a} = \hat{i} – \hat{j}$ on the vector $\vec{b} = \hat{i} + \hat{j}$.

The projection of $\vec{a}$ on $\vec{b}$ is given by $\text{Proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.

Calculate the dot product $\vec{a} \cdot \vec{b}$:$$\vec{a} \cdot \vec{b} = (1)(1) + (-1)(1) = 1 – 1 = 0$$
Calculate the magnitude $|\vec{b}|$:$$|\vec{b}| = \sqrt{1^2 + 1^2} = \sqrt{2}$$
Projection:$$\text{Proj}_{\vec{b}} \vec{a} = \frac{0}{\sqrt{2}} = \mathbf{0}$$(Note: Since the projection is 0, the vectors $\vec{a}$ and $\vec{b}$ are perpendicular).

4. Find the projection of the vector $\vec{a} = \hat{i} + 3\hat{j} + 7\hat{k}$ on the vector $\vec{b} = 7\hat{i} – \hat{j} + 8\hat{k}$.

Calculate the dot product $\vec{a} \cdot \vec{b}$:$$\vec{a} \cdot \vec{b} = (1)(7) + (3)(-1) + (7)(8) = 7 – 3 + 56 = 60$$
Calculate the magnitude $|\vec{b}|$:$$|\vec{b}| = \sqrt{7^2 + (-1)^2 + 8^2} = \sqrt{49 + 1 + 64} = \sqrt{114}$$
Projection:$$\text{Proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \mathbf{\frac{60}{\sqrt{114}}}$$

5. Show that each of the given three vectors is a unit vector and that they are mutually perpendicular.

Let $\vec{a} = \frac{1}{7}(2\hat{i} + 3\hat{j} + 6\hat{k})$, $\vec{b} = \frac{1}{7}(3\hat{i} – 6\hat{j} + 2\hat{k})$, $\vec{c} = \frac{1}{7}(6\hat{i} + 2\hat{j} – 3\hat{k})$.

(i) Unit Vector Check ($|\vec{v}| = 1$)

$$|\vec{a}|^2 = \left(\frac{1}{7}\right)^2 (2^2 + 3^2 + 6^2) = \frac{1}{49} (4 + 9 + 36) = \frac{49}{49} = 1$$

Thus, $|\vec{a}| = 1$.

$$|\vec{b}|^2 = \left(\frac{1}{7}\right)^2 (3^2 + (-6)^2 + 2^2) = \frac{1}{49} (9 + 36 + 4) = \frac{49}{49} = 1$$

Thus, $|\vec{b}| = 1$.

$$|\vec{c}|^2 = \left(\frac{1}{7}\right)^2 (6^2 + 2^2 + (-3)^2) = \frac{1}{49} (36 + 4 + 9) = \frac{49}{49} = 1$$

Thus, $|\vec{c}| = 1$.They are all unit vectors.

(ii) Mutually Perpendicular Check ($\vec{v}_1 \cdot \vec{v}_2 = 0$)

$$\vec{a} \cdot \vec{b} = \frac{1}{49} [(2)(3) + (3)(-6) + (6)(2)] = \frac{1}{49} [6 – 18 + 12] = \frac{0}{49} = 0$$

$$\vec{b} \cdot \vec{c} = \frac{1}{49} [(3)(6) + (-6)(2) + (2)(-3)] = \frac{1}{49} [18 – 12 – 6] = \frac{0}{49} = 0$$

$$\vec{a} \cdot \vec{c} = \frac{1}{49} [(2)(6) + (3)(2) + (6)(-3)] = \frac{1}{49} [12 + 6 – 18] = \frac{0}{49} = 0$$

Since their dot products are all zero, they are mutually perpendicular (orthogonal).

6. Find $|\vec{a}|$, if $(\vec{a} + \vec{b}) \cdot (\vec{a} – \vec{b}) = 8$ and $|\vec{b}| = 8|\vec{a}|$.

Simplify the dot product:$$(\vec{a} + \vec{b}) \cdot (\vec{a} – \vec{b}) = \vec{a} \cdot \vec{a} – \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} – \vec{b} \cdot \vec{b}$$Since $\vec{a} \cdot \vec{a} = |\vec{a}|^2$, $\vec{b} \cdot \vec{b} = |\vec{b}|^2$, and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$:$$(\vec{a} + \vec{b}) \cdot (\vec{a} – \vec{b}) = |\vec{a}|^2 – |\vec{b}|^2$$
Substitute the given values:$$|\vec{a}|^2 – |\vec{b}|^2 = 8$$Substitute $|\vec{b}| = 8|\vec{a}|$ into the equation:$$|\vec{a}|^2 – (8|\vec{a}|)^2 = 8$$$$|\vec{a}|^2 – 64|\vec{a}|^2 = 8$$$$-63|\vec{a}|^2 = 8$$There appears to be a typo in the original question, as magnitude squared cannot be negative. Assuming the expression was $(\vec{b} + \vec{a}) \cdot (\vec{b} – \vec{a})$ or the relationship was $|\vec{a}|=8|\vec{b}|$ to yield a positive result.

Assuming the relationship was $|\vec{b}| = \frac{1}{8}|\vec{a}|$:

$$|\vec{a}|^2 – \left(\frac{1}{8}|\vec{a}|\right)^2 = 8$$

$$|\vec{a}|^2 – \frac{1}{64}|\vec{a}|^2 = 8$$

$$\frac{63}{64}|\vec{a}|^2 = 8 \implies |\vec{a}|^2 = \frac{8 \cdot 64}{63} = \frac{512}{63}$$

$$|\vec{a}| = \sqrt{\frac{512}{63}} = \frac{16\sqrt{2}}{\sqrt{63}} = \frac{16\sqrt{2}}{3\sqrt{7}}$$

Assuming the intended relationship was $|\vec{a}| = 2|\vec{b}|$ and the difference of squares gives $3|\vec{b}|^2=8$ (a common typo in this problem set):

$$|\vec{a}|^2 – |\vec{b}|^2 = 8$$

$$(2|\vec{b}|)^2 – |\vec{b}|^2 = 8$$

$$4|\vec{b}|^2 – |\vec{b}|^2 = 8$$

$$3|\vec{b}|^2 = 8 \implies |\vec{b}|^2 = 8/3$$

$$|\vec{a}| = 2|\vec{b}| = 2\sqrt{8/3} = \mathbf{4\sqrt{2/3}}$$

Based on the provided text, the computation leads to:

$$-63|\vec{a}|^2 = 8 \implies |\vec{a}| = \mathbf{\sqrt{-\frac{8}{63}}}$$

Since a magnitude must be real, we conclude there is an error in the problem statement.

7. Evaluate the product $(3\vec{a} – 5\vec{b}) \cdot (2\vec{a} + 7\vec{b})$.

Use the distributive property of the dot product:

$$(3\vec{a} – 5\vec{b}) \cdot (2\vec{a} + 7\vec{b}) = 3\vec{a} \cdot 2\vec{a} + 3\vec{a} \cdot 7\vec{b} – 5\vec{b} \cdot 2\vec{a} – 5\vec{b} \cdot 7\vec{b}$$

$$= 6(\vec{a} \cdot \vec{a}) + 21(\vec{a} \cdot \vec{b}) – 10(\vec{b} \cdot \vec{a}) – 35(\vec{b} \cdot \vec{b})$$

Since $\vec{a} \cdot \vec{a} = |\vec{a}|^2$, $\vec{b} \cdot \vec{b} = |\vec{b}|^2$, and $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$:

$$= 6|\vec{a}|^2 + 21(\vec{a} \cdot \vec{b}) – 10(\vec{a} \cdot \vec{b}) – 35|\vec{b}|^2$$

$$= \mathbf{6|\vec{a}|^2 + 11(\vec{a} \cdot \vec{b}) – 35|\vec{b}|^2}$$

8. Find the magnitude of two vectors $\vec{a}$ and $\vec{b}$, having the same magnitude and such that the angle between them is $60^\circ$ and their scalar product is $\frac{1}{2}$.

Given: $|\vec{a}| = |\vec{b}|$, $\theta = 60^\circ$, $\vec{a} \cdot \vec{b} = \frac{1}{2}$.

The dot product formula is $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$.

Let $|\vec{a}| = |\vec{b}| = x$.

$$\frac{1}{2} = x \cdot x \cdot \cos(60^\circ)$$

$$\frac{1}{2} = x^2 \cdot \frac{1}{2}$$

$$x^2 = 1 \implies x = 1$$

The magnitude of both vectors is $\mathbf{1}$.

9. Find $|\vec{x}|$, if for a unit vector $\vec{a}$, $(\vec{x} – \vec{a}) \cdot (\vec{x} + \vec{a}) = 12$.

Given: $|\vec{a}| = 1$ (unit vector).

$$(\vec{x} – \vec{a}) \cdot (\vec{x} + \vec{a}) = |\vec{x}|^2 – |\vec{a}|^2$$

Substitute the given value and $|\vec{a}| = 1$:

$$12 = |\vec{x}|^2 – 1^2$$

$$12 = |\vec{x}|^2 – 1$$

$$|\vec{x}|^2 = 13 \implies |\vec{x}| = \mathbf{\sqrt{13}}$$

10. If $\vec{a} = 2\hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = -\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{c} = 3\hat{i} + \lambda\hat{j}$ are such that $\vec{a} + \lambda\vec{b}$ is perpendicular to $\vec{c}$, then find the value of $\lambda$.

Two vectors $\vec{x}$ and $\vec{y}$ are perpendicular if $\vec{x} \cdot \vec{y} = 0$.

Here, $\vec{x} = \vec{a} + \lambda\vec{b}$ and $\vec{y} = \vec{c}$.

Calculate $\vec{a} + \lambda\vec{b}$:$$\vec{a} + \lambda\vec{b} = (2\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(-\hat{i} + 2\hat{j} + \hat{k})$$$$\vec{a} + \lambda\vec{b} = (2 – \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k}$$
Set the dot product to zero:$$(\vec{a} + \lambda\vec{b}) \cdot \vec{c} = 0$$$$[(2 – \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k}] \cdot [3\hat{i} + \lambda\hat{j} + 0\hat{k}] = 0$$$$3(2 – \lambda) + \lambda(2 + 2\lambda) + 0(3 + \lambda) = 0$$
Solve for $\lambda$:$$6 – 3\lambda + 2\lambda + 2\lambda^2 = 0$$$$2\lambda^2 – \lambda + 6 = 0$$Calculate the discriminant $\Delta = b^2 – 4ac = (-1)^2 – 4(2)(6) = 1 – 48 = -47$.Since the discriminant is negative, the quadratic equation has no real roots. There appears to be a typo in the original question.

Assuming $\vec{c} = 3\hat{i} + \hat{j}$ (a common version of this question):

$$3(2 – \lambda) + 1(2 + 2\lambda) + 0 = 0$$

$$6 – 3\lambda + 2 + 2\lambda = 0$$

$$8 – \lambda = 0 \implies \mathbf{\lambda = 8}$$

11. Show that $|\vec{a}|\vec{b} + |\vec{b}|\vec{a}$ is perpendicular to $|\vec{a}|\vec{b} – |\vec{b}|\vec{a}$, for any two nonzero vectors $\vec{a}$ and $\vec{b}$.

We need to show that their dot product is zero. Use the identity $(\vec{x} + \vec{y}) \cdot (\vec{x} – \vec{y}) = |\vec{x}|^2 – |\vec{y}|^2$.

Let $\vec{x} = |\vec{a}|\vec{b}$ and $\vec{y} = |\vec{b}|\vec{a}$.

$$(|\vec{a}|\vec{b} + |\vec{b}|\vec{a}) \cdot (|\vec{a}|\vec{b} – |\vec{b}|\vec{a}) = |\vec{x}|^2 – |\vec{y}|^2$$

Calculate $|\vec{x}|^2$:

$$|\vec{x}|^2 = |(|\vec{a}|\vec{b})|^2 = (|\vec{a}|)^2 |\vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2$$

Calculate $|\vec{y}|^2$:

$$|\vec{y}|^2 = |(|\vec{b}|\vec{a})|^2 = (|\vec{b}|)^2 |\vec{a}|^2 = |\vec{a}|^2 |\vec{b}|^2$$

Substitute back:

$$\text{Dot Product} = |\vec{a}|^2 |\vec{b}|^2 – |\vec{a}|^2 |\vec{b}|^2 = 0$$

Since the dot product is zero, the two vectors are perpendicular.

12. If $\vec{a} \cdot \vec{a} = 0$, then what can be concluded about the vector $\vec{a}$?

The dot product of a vector with itself is its magnitude squared:

$$\vec{a} \cdot \vec{a} = |\vec{a}|^2$$

Given $\vec{a} \cdot \vec{a} = 0$:

$$|\vec{a}|^2 = 0 \implies |\vec{a}| = 0$$

A vector whose magnitude is zero must be the zero vector.

$$\mathbf{\vec{a} = \vec{0}}$$

13. If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, find the value of $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}$.

Given: $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$ and $\vec{a} + \vec{b} + \vec{c} = \vec{0}$.

Square the sum of the vectors:

$$|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{0}|^2 = 0$$

Expand the magnitude squared:

$$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$$

Substitute the magnitudes:

$$1^2 + 1^2 + 1^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$$

$$3 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$$

$$2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -3$$

$$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = \mathbf{-\frac{3}{2}}$$

14. If $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$, then $\vec{a} \cdot \vec{b} = 0$. But the converse need not be true. Justify your answer with an example.

The converse statement is: If $\vec{a} \cdot \vec{b} = 0$, then either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$.

Justification (Converse is False):

The dot product $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = 0$ implies that either $|\vec{a}| = 0$, $|\vec{b}| = 0$, or $\cos \theta = 0$.

If $\cos \theta = 0$, then $\theta = 90^\circ$ (or $\frac{\pi}{2}$).

This means that two non-zero vectors can have a zero dot product if they are perpendicular (orthogonal).

Example:

Let $\vec{a} = \hat{i}$ (where $\vec{a} \neq \vec{0}$)
Let $\vec{b} = \hat{j}$ (where $\vec{b} \neq \vec{0}$)
$\vec{a} \cdot \vec{b} = (1)(0) + (0)(1) + (0)(0) = 0$.Here, $\vec{a} \cdot \vec{b} = 0$, but neither $\vec{a}$ nor $\vec{b}$ is the zero vector.

15. If the vertices A, B, C of a triangle ABC are $(1, 2, 3)$, $(-1, 0, 0)$, $(0, 1, 2)$, respectively, then find $\angle ABC$. [$\angle ABC$ is the angle between the vectors $\vec{BA}$ and $\vec{BC}$].

We need to find the angle between $\vec{BA}$ and $\vec{BC}$. The vertex $B$ is the common initial point.

Let $A(1, 2, 3)$, $B(-1, 0, 0)$, $C(0, 1, 2)$.

Calculate $\vec{BA}$ and $\vec{BC}$:$$\vec{BA} = \vec{a} – \vec{b} = (1 – (-1))\hat{i} + (2 – 0)\hat{j} + (3 – 0)\hat{k} = 2\hat{i} + 2\hat{j} + 3\hat{k}$$$$\vec{BC} = \vec{c} – \vec{b} = (0 – (-1))\hat{i} + (1 – 0)\hat{j} + (2 – 0)\hat{k} = 1\hat{i} + 1\hat{j} + 2\hat{k}$$
Calculate the dot product $\vec{BA} \cdot \vec{BC}$:$$\vec{BA} \cdot \vec{BC} = (2)(1) + (2)(1) + (3)(2) = 2 + 2 + 6 = 10$$
Calculate the magnitudes:$$|\vec{BA}| = \sqrt{2^2 + 2^2 + 3^2} = \sqrt{4 + 4 + 9} = \sqrt{17}$$$$|\vec{BC}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$$
Find the angle $\theta$:$$\cos \theta = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}| |\vec{BC}|} = \frac{10}{\sqrt{17} \cdot \sqrt{6}} = \frac{10}{\sqrt{102}}$$$$\angle ABC = \mathbf{\cos^{-1}\left(\frac{10}{\sqrt{102}}\right)}$$

16. Show that the points $A(1, 2, 7)$, $B(2, 6, 3)$ and $C(3, 10, -1)$ are collinear.

Three points $A, B, C$ are collinear if the vectors formed by any two pairs (e.g., $\vec{AB}$ and $\vec{BC}$) are collinear (i.e., $\vec{AB} = \lambda \vec{BC}$) and share a common point.

Calculate $\vec{AB}$ and $\vec{BC}$:$$\vec{AB} = \vec{b} – \vec{a} = (2-1)\hat{i} + (6-2)\hat{j} + (3-7)\hat{k} = \hat{i} + 4\hat{j} – 4\hat{k}$$$$\vec{BC} = \vec{c} – \vec{b} = (3-2)\hat{i} + (10-6)\hat{j} + (-1-3)\hat{k} = \hat{i} + 4\hat{j} – 4\hat{k}$$
Check for collinearity:Since $\vec{AB} = 1 \cdot \vec{BC}$, the vectors are collinear ($\lambda = 1$).Since $B$ is a common point, the points $\mathbf{A, B, C \text{ are collinear}}$.

17. Show that the vectors $\vec{a} = 2\hat{i} – \hat{j} + \hat{k}$, $\vec{b} = \hat{i} – 3\hat{j} – 5\hat{k}$ and $\vec{c} = 3\hat{i} – 4\hat{j} – 4\hat{k}$ form the vertices of a right angled triangle.

Let $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of the vertices $A, B, C$. The side vectors are $\vec{AB}, \vec{BC}, \vec{CA}$. The triangle is right-angled if the dot product of two side vectors is zero.

Side Vectors:$$\vec{AB} = \vec{b} – \vec{a} = (1-2)\hat{i} + (-3-(-1))\hat{j} + (-5-1)\hat{k} = -\hat{i} – 2\hat{j} – 6\hat{k}$$$$\vec{BC} = \vec{c} – \vec{b} = (3-1)\hat{i} + (-4-(-3))\hat{j} + (-4-(-5))\hat{k} = 2\hat{i} – \hat{j} + \hat{k}$$$$\vec{CA} = \vec{a} – \vec{c} = (2-3)\hat{i} + (-1-(-4))\hat{j} + (1-(-4))\hat{k} = -\hat{i} + 3\hat{j} + 5\hat{k}$$
Check Dot Products:$$\vec{AB} \cdot \vec{BC} = (-1)(2) + (-2)(-1) + (-6)(1) = -2 + 2 – 6 = -6$$$$\vec{BC} \cdot \vec{CA} = (2)(-1) + (-1)(3) + (1)(5) = -2 – 3 + 5 = 0$$Since $\vec{BC} \cdot \vec{CA} = 0$, $\vec{BC}$ is perpendicular to $\vec{CA}$.Therefore, the triangle $ABC$ is a right-angled triangle at C.