Anand Classes provides accurate and well-explained NCERT Solutions for Exercise 10.3 of Chapter 10 Vector Algebra for Class 12 Mathematics, helping students master concepts such as the dot product, projection formulas, and applications of scalar products through clear, step-by-step solutions. These Set-1 notes strictly follow the latest NCERT and CBSE syllabus, making them highly useful for revision, concept building, and scoring well in board examinations. Click the print button to download study material and notes.
NCERT Question.1 : Find the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $|\vec{a}| = \sqrt{3}$ and $|\vec{b}| = 2$ and $\vec{a} \cdot \vec{b} = \sqrt{6}$
Solution :
Using the dot product formula:
$$
\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta
$$
Substitute the values:
$$
\cos\theta = \frac{\sqrt{6}}{\sqrt{3} \cdot 2}
$$
Simplify:
$$
\cos\theta = \frac{\sqrt{6}}{2\sqrt{3}}
= \frac{\sqrt{6}}{2\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}
= \frac{\sqrt{18}}{6}
= \frac{3\sqrt{2}}{6}
= \frac{\sqrt{2}}{2}
$$
Now find $\theta$:
$$
\theta = \cos^{-1}\left(\frac{\sqrt{2}}{2}\right)
$$
Thus,
$$
\theta = \frac{\pi}{4} \quad \text{or} \quad 45^\circ
$$
So, the angle between $\vec{a}$ and $\vec{b}$ is 45°.
NCERT Question.2 : Find the angle between the vectors
$\vec{a} = \hat{i} – 2\hat{j} + 3\hat{k}$ and
$\vec{b} = 3\hat{i} – 2\hat{j} + \hat{k}$
Solution
Given Vectors
$\vec{a} = \hat{i} – 2\hat{j} + 3\hat{k}$ and
$\vec{b} = 3\hat{i} – 2\hat{j} + \hat{k}$
Dot Product : To calculate the dot product, we multiply corresponding components and then add:
$$\vec{a} \cdot \vec{b} =( \hat{i} – 2\hat{j} + 3\hat{k})\cdot ( 3\hat{i} – 2\hat{j} + \hat{k})$$
$$\vec{a} \cdot \vec{b} = (1 \cdot 3) + (-2 \cdot -2) + (3 \cdot 1)$$
$$\vec{a} \cdot \vec{b}= 3 + 4 + 3 = 10$$
Magnitude of Vectors
Using $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$
$$|\vec{a}| = \sqrt{1^2 + (-2)^2 + 3^2}$$
$$|\vec{a}|= \sqrt{1 + 4 + 9} = \sqrt{14}$$
$$|\vec{b}| = \sqrt{3^2 + (-2)^2 + 1^2}$$
$$|\vec{b}|= \sqrt{9 + 4 + 1} = \sqrt{14}$$
Cosine of the Angle
$$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$$
$$\cos\theta= \frac{10}{\sqrt{14}\sqrt{14}}$$
$$\cos\theta= \frac{10}{14}$$
$$\cos\theta= \frac{5}{7}$$
Angle Between the Vectors
$$\theta = \cos^{-1}\left(\frac{5}{7}\right)$$
Final Result
$$\boxed{\theta = \cos^{-1}\left(\frac{5}{7}\right)}$$
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NCERT Question.3 : Find the projection of the vector
$\vec{a} = \hat{i} – \hat{j}\;$ on the vector $\vec{b} = \hat{i} + \hat{j}$
Solution
First, let’s find the dot product of given vectors.
Dot Product
$$\vec{a} \cdot \vec{b} =( \hat{i} – \hat{j})\cdot ( \hat{i} + \hat{j})$$
$$\vec{a} \cdot \vec{b} = (1 \cdot 1) + (-1 \cdot 1)$$
$$\vec{a} \cdot \vec{b}= 1 – 1 = 0$$
Magnitude of $\vec{b}$
$$|\vec{b}| = \sqrt{1^2 + 1^2} = \sqrt{2}$$
Projection of $\vec{a}$ on $\vec{b}$
Using
$$\text{Projection}_{\vec{b}}(\vec{a}) = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\vec{b}$$
$$\text{Projection}_{\vec{b}}(\vec{a}) = \frac{0}{2}(\hat{i} + \hat{j})$$
$$= 0(\hat{i} + \hat{j})$$
$$= \vec{0}$$
Final Result
$$\boxed{\vec{0}}$$
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NCERT Question.4 : Find the projection of the vector
$\vec{a} = \hat{i} + 3\hat{j} + 7\hat{k}\;$ on the vector $\vec{b} = 7\hat{i} – \hat{j} + 8\hat{k}$
Solution
First, let’s find the dot product of given vectors.
Dot Product
$$\vec{a} \cdot \vec{b} = (1 \cdot 7) + (3 \cdot -1) + (7 \cdot 8)$$
$$\vec{a} \cdot \vec{b}= 7 – 3 + 56 = 60$$
Magnitude of $\vec{b}$
$$|\vec{b}| = \sqrt{7^2 + (-1)^2 + 8^2}$$
$$|\vec{b}|= \sqrt{49 + 1 + 64} = \sqrt{114}$$
Projection of $\vec{a}$ on $\vec{b}$
Using
$$\text{Projection}_{\vec{b}}(\vec{a}) = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\vec{b}$$
Since
$$|\vec{b}|^2 = 114$$
$$\text{Projection}_{\vec{b}}(\vec{a}) = \frac{60}{114}\vec{b}$$
Thus,
$$\text{Projection}_{\vec{b}}(\vec{a}) = \frac{60}{114}(7\hat{i} – \hat{j} + 8\hat{k})$$
Final Result
$$\boxed{\text{Projection}_{\vec{b}}(\vec{a}) = \frac{60}{114}(7\hat{i} – \hat{j} + 8\hat{k})}$$
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NCERT Question.5 : Show that each of the given three vectors is a unit vector:
$ \frac{1}{7}(2\hat{i} + 3\hat{j} + 6\hat{k}),
\frac{1}{7}(3\hat{i} – 6\hat{j} + 2\hat{k}),
\frac{1}{7}(6\hat{i} + 2\hat{j} – 3\hat{k}) $
Also show that they are mutually perpendicular.
Solution
Let
$$ \vec{v_1} = \frac{1}{7}(2\hat{i} + 3\hat{j} + 6\hat{k}) $$
$$ \vec{v_2} = \frac{1}{7}(3\hat{i} – 6\hat{j} + 2\hat{k}) $$
$$ \vec{v_3} = \frac{1}{7}(6\hat{i} + 2\hat{j} – 3\hat{k}) $$
Checking if each vector is a unit vector
For $( \vec{v_1} )$:
$$ |\vec{v_1}| = \sqrt{\left(\frac{2}{7}\right)^2 + \left(\frac{3}{7}\right)^2 + \left(\frac{6}{7}\right)^2} $$
$$ |\vec{v_1}| = \sqrt{\frac{4 + 9 + 36}{49}} = \sqrt{\frac{49}{49}} = 1 $$
For $( \vec{v_2} )$:
$$ |\vec{v_2}| = \sqrt{\left(\frac{3}{7}\right)^2 + \left(\frac{-6}{7}\right)^2 + \left(\frac{2}{7}\right)^2} $$
$$ |\vec{v_2}| = \sqrt{\frac{9 + 36 + 4}{49}} = \sqrt{\frac{49}{49}} = 1 $$
For $( \vec{v_3} )$:
$$ |\vec{v_3}| = \sqrt{\left(\frac{6}{7}\right)^2 + \left(\frac{2}{7}\right)^2 + \left(\frac{-3}{7}\right)^2} $$
$$ |\vec{v_3}| = \sqrt{\frac{36 + 4 + 9}{49}} = \sqrt{\frac{49}{49}} = 1 $$
Thus all three vectors are unit vectors.
Showing they are mutually perpendicular
Vectors are perpendicular if their dot product is zero.
$( \vec{v_1} \cdot \vec{v_2} )$ :
$$ \vec{v_1} \cdot \vec{v_2} = \frac{2}{7}\frac{3}{7} + \frac{3}{7}\frac{-6}{7} + \frac{6}{7}\frac{2}{7} $$
$$ \vec{v_1} \cdot \vec{v_2}= \frac{6 – 18 + 12}{49} = 0 $$
$( \vec{v_1} \cdot \vec{v_3} ) :$
$$ \vec{v_1} \cdot \vec{v_3} = \frac{2}{7}\frac{6}{7} + \frac{3}{7}\frac{2}{7} + \frac{6}{7}\frac{-3}{7} $$
$$ \vec{v_1} \cdot \vec{v_3}= \frac{12 + 6 – 18}{49} = 0 $$
$( \vec{v_2} \cdot \vec{v_3} ) :$
$$ \vec{v_2} \cdot \vec{v_3} = \frac{3}{7}\frac{6}{7} + \frac{-6}{7}\frac{2}{7} + \frac{2}{7}\frac{-3}{7} $$
$$ \vec{v_2} \cdot \vec{v_3}= \frac{18 – 12 – 6}{49} = 0 $$
Final Result
All three vectors are unit vectors, and they are mutually perpendicular to each other.
$$ \boxed{\text{All three vectors are mutually perpendicular unit vectors}} $$
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NCERT Question.6 : Find $|\vec{a}|$ and $|\vec{b}|$ if
$(\vec{a} + \vec{b}) \cdot (\vec{a} – \vec{b}) = 8$ and $|\vec{a}| = 8|\vec{b}|.$
Solution
Given
$$(\vec{a} + \vec{b}) \cdot (\vec{a} – \vec{b}) = 8$$
Expand the dot product:
$$\vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{a} – \vec{a} \cdot \vec{b} – \vec{b} \cdot \vec{b} = 8$$
Since $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$, the middle terms cancel:
$$|\vec{a}|^2 – |\vec{b}|^2 = 8$$
Given
$$|\vec{a}| = 8|\vec{b}|$$
so
$$|\vec{a}|^2 = 64|\vec{b}|^2.$$
Substitute:
$$64|\vec{b}|^2 – |\vec{b}|^2 = 8$$
$$63|\vec{b}|^2 = 8$$
$$|\vec{b}|^2 = \frac{8}{63}$$
Thus,
$$|\vec{b}| = \frac{2\sqrt{2}}{3\sqrt{7}}$$
and
$$|\vec{a}| = 8|\vec{b}| = 8\sqrt{\frac{8}{63}}.$$
Simplify:
$$|\vec{a}| = \frac{16\sqrt{2}}{3\sqrt{7}}.$$
Final Result
$$\boxed{|\vec{b}| = \frac{2\sqrt{2}}{3\sqrt{7}} \quad,\quad |\vec{a}| = \frac{16\sqrt{2}}{3\sqrt{7}}}$$
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NCERT Question.7 : Evaluate the product
$$(3\vec{a} – 5\vec{b}) \cdot (2\vec{a} + 7\vec{b}).$$
Solution
Expand the dot product:
$$
(3\vec{a} – 5\vec{b}) \cdot (2\vec{a} + 7\vec{b})= 3\vec{a} \cdot 2\vec{a} + 3\vec{a} \cdot 7\vec{b} – 5\vec{b} \cdot 2\vec{a} – 5\vec{b} \cdot 7\vec{b}
$$
Apply dot product properties:
$$
(3\vec{a} – 5\vec{b}) \cdot (2\vec{a} + 7\vec{b})= 6|\vec{a}|^2 + 21|\vec{a}||\vec{b}| – 10|\vec{a}||\vec{b}| – 35|\vec{b}|^2
$$
Combine like terms:
$$
(3\vec{a} – 5\vec{b}) \cdot (2\vec{a} + 7\vec{b})= 6|\vec{a}|^2 + 11|\vec{a}||\vec{b}| – 35|\vec{b}|^2
$$
Final Result
$$\boxed{6|\vec{a}|^2 + 11|\vec{a}||\vec{b}| – 35|\vec{b}|^2}$$
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NCERT Question.8 : Find the magnitude of two vectors $\vec{a}$ and $\vec{b}$ having the same magnitude, such that the angle between them is $60^\circ$ and their scalar product is $\dfrac12$.
Solution
Using the scalar product formula:
$$
\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta
$$
Let the common magnitude be
$$
|\vec{a}| = |\vec{b}| = x
$$
Given:
$$
\vec{a} \cdot \vec{b} = \frac12
$$
and
$$
\cos 60^\circ = \frac12
$$
Substitute:
$$
\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta
$$
$$
\frac12 = x \cdot x \cdot \frac12
$$
$$
\frac12 = \frac{x^2}{2}
$$
Multiply both sides by 2:
$$
1 = x^2
$$
Take square root:
$$
x = 1
$$
Hence
$$|\vec{a}| = |\vec{b}| = 1$$
Final Result
$$\boxed{|\vec{a}| = |\vec{b}| = 1}$$
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NCERT Question.9 : Find $|\vec{x}|$ if for a unit vector $\vec{a}$,
$$(\vec{x} – \vec{a}) \cdot (\vec{x} + \vec{a}) = 12$$
Solution
Expand the dot product:
$$
(\vec{x} – \vec{a}) \cdot (\vec{x} + \vec{a})= \vec{x} \cdot \vec{x} + \vec{x} \cdot \vec{a} – \vec{a} \cdot \vec{x} – \vec{a} \cdot \vec{a}
$$
Since $\vec{x} \cdot \vec{a} = \vec{a} \cdot \vec{x}$, these two terms cancel:
$$
(\vec{x} – \vec{a}) \cdot (\vec{x} + \vec{a})=|\vec{x}|^2 – |\vec{a}|^2 = 12
$$
Given:
$$
|\vec{a}| = 1
$$
Substitute:
$$
|\vec{x}|^2 – 1 = 12
$$
$$
|\vec{x}|^2 = 13
$$
Take square root:
$$
|\vec{x}| = \sqrt{13}
$$
Final Result
$$\boxed{|\vec{x}| = \sqrt{13}}$$
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