Anand Classes provides detailed NCERT Solutions for Exercise 10.2 (Set-1) of Chapter 10 Vector Algebra for Class 12 Mathematics, helping students understand concepts like scalar multiplication, vector addition, and properties of vectors through clear, step-by-step explanations. These notes strictly follow the latest NCERT and CBSE syllabus, making them perfect for revision, concept strengthening, and scoring high in board examinations. Click the print button to download study material and notes.
NCERT Solutions for Exercise 10.2 of Chapter 10 Vector Algebra for Class 12 Mathematics
NCERT Question.1 : Compute the magnitude of the following vectors
$$
\vec{a}= \hat{i}+\hat{j}+\hat{k},\;\;\;\vec{b}= 2\hat{i}-7\hat{j}-3\hat{k},\;\;\;
\vec{c}= \frac{1}{\sqrt{3}}\hat{i}+\frac{1}{\sqrt{3}}\hat{j}-\frac{1}{\sqrt{3}}\hat{k}
$$
Solution
Using the magnitude formula for vector
$$\vec{v}= x\hat{i}+y\hat{j}+z\hat{k}$$
$$|\vec{v}|=\sqrt{x^{2}+y^{2}+z^{2}}$$
For $\vec{a}$:
$$|\vec{a}|=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}$$
For $\vec{b}$:
$$|\vec{b}|=\sqrt{(2)^{2}+(-7)^{2}+(-3)^{2}}$$
$$|\vec{b}|=\sqrt{4+49+9}=\sqrt{62}$$
For $\vec{c}$:
$$|\vec{c}|=\sqrt{\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(-\frac{1}{\sqrt{3}}\right)^{2}}$$
$$|\vec{c}|=\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=\sqrt{1}=1$$
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NCERT Question.2 : Write two different vectors having the same magnitude.
Solution :
An infinite number of such vectors are possible. One example is:
$$
\vec{a}= \hat{i}+2\hat{j}-3\hat{k}
$$
$$
\vec{b}= 2\hat{i}-\hat{j}-3\hat{k}
$$
Magnitude of $\vec{a}$:
$$
|\vec{a}|= \sqrt{1^{2}+(-2)^{2}+(-3)^{2}}
$$
$$
|\vec{a}|=\sqrt{1+4+9}=\sqrt{14}
$$
Magnitude of $\vec{b}$:
$$
|\vec{b}|= \sqrt{(2)^{2}+(-1)^{2}+(-3)^{2}}
$$
$$
|\vec{b}|=\sqrt{4+1+9}=\sqrt{14}
$$
Thus,
$$
|\vec{a}| = |\vec{b}|
$$
but
$$
\vec{a} \ne \vec{b}
$$
Remark. In this way, we can construct an infinite number of possible answers.
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NCERT Question.3 : Write two different vectors having the same direction.
Solution
Let
$$\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k} \tag{i}$$
and
$$\vec{b} = 2(\hat{i} + 2\hat{j} + 3\hat{k}) \tag{ii}$$
From (i),
$$\vec{b} = 2\vec{a}$$
Thus,
$$\vec{b} = m\vec{a} \quad \text{where } m = 2 > 0.$$
Direction of Vectors
Since $\vec{b} = 2\vec{a}$, the vectors $\vec{a}$ and $\vec{b}$ have the same direction.
But
$$\vec{b} \ne \vec{a}$$
because
$$|\vec{b}| = |2||\vec{a}| = 2|\vec{a}| .$$
Hence
$$|\vec{b} | \ne |\vec{a}|.$$
Remark
In this manner, we can construct infinitely many possible answers.
An infinite number of such vectors are possible.
In terms of direction cosines
$$
\vec{a}= \hat{i}+\hat{j}+\hat{k}
$$
$$
\vec{b}= 2\hat{i}+2\hat{j}+2\hat{k}
$$
Direction Cosines of $\vec{a}$
$$
\ell=\frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{1}{\sqrt{3}}
$$
$$
m=\frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{1}{\sqrt{3}}
$$
$$
n=\frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}=\frac{1}{\sqrt{3}}
$$
Direction Cosines of $\vec{b}$
$$
\ell=\frac{2}{\sqrt{2^{2}+2^{2}+2^{2}}}=\frac{2}{\sqrt{12}}=\frac{1}{\sqrt{3}}
$$
$$
m=\frac{2}{\sqrt{2^{2}+2^{2}+2^{2}}}=\frac{1}{\sqrt{3}}
$$
$$
n=\frac{2}{\sqrt{2^{2}+2^{2}+2^{2}}}=\frac{1}{\sqrt{3}}
$$
This shows that both vectors $\vec{a}$ and $\vec{b}$ have the same direction, even though
$$
\vec{a}\ne \vec{b}
$$
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NCERT Question.4 : Find the values of $x$ and $y$ so that the vectors $2\hat{i} + 3\hat{j}$ and $x\hat{i} + y\hat{j}$ are equal.
Solution
Since the vectors are equal
$$x\hat{i} + y\hat{j} = 2\hat{i} + 3\hat{j}$$
Comparing coefficients of $\hat{i}$ and $\hat{j}$ on both sides, we have
$$ x = 2, \quad y = 3 $$
Final Result
$$\boxed{x = 2,\ y = 3}$$
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NCERT Question.5 : Find the scalar and vector components of the vector with initial point $(2, 1)$ and terminal point $(-5, 7)$.
Solution
Let the initial point be $P(2, 1)$ and the terminal point be $Q(-5, 7)$.
The position vectors are
$$\overrightarrow{OP} = 2\hat{i} + \hat{j}$$
$$\overrightarrow{OQ} = -5\hat{i} + 7\hat{j}$$
Now,
$$\overrightarrow{PQ} = \overrightarrow{OQ} – \overrightarrow{OP}$$
$$\overrightarrow{PQ} =(-5\hat{i} + 7\hat{j})-(2\hat{i} + \hat{j})$$
So,
$$\overrightarrow{PQ} = (-5 – 2)\hat{i} + (7 – 1)\hat{j}$$
$$\overrightarrow{PQ} = -7\hat{i} + 6\hat{j}$$
Thus,
- Scalar components: $-7$ and $6$
- Vector components: $-7\hat{i}$ and $6\hat{j}$
Final Result
$$\boxed{\overrightarrow{PQ} = -7\hat{i} + 6\hat{j}}$$
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NCERT Question.6 : Find the sum of the vectors
$\vec{a} = \hat{i} – 2\hat{j} + \hat{k},\;\;\;
\vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k},\;\;\;
\vec{c} = \hat{i} – 6\hat{j} – 7\hat{k}$
Solution
Given Vectors are :
$\vec{a} = \hat{i} – 2\hat{j} + \hat{k},\;\;\;
\vec{b} = -2\hat{i} + 4\hat{j} + 5\hat{k},\;\;\;
\vec{c} = \hat{i} – 6\hat{j} – 7\hat{k}$
$$\vec{a} + \vec{b} + \vec{c} =(\hat{i} – 2\hat{j} + \hat{k})+(-2\hat{i} + 4\hat{j} + 5\hat{k})+(\hat{i} – 6\hat{j} – 7\hat{k})$$
Add the corresponding $\hat{i}, \hat{j} $ and $\hat{k}$ components.
$$\vec{a} + \vec{b} + \vec{c} = (1 – 2 + 1)\hat{i} + (-2 + 4 – 6)\hat{j} + (1 + 5 – 7)\hat{k}$$
Simplifying:
$$\vec{a} + \vec{b} + \vec{c} = 0\hat{i} – 4\hat{j} – 1\hat{k}=-4\hat{j} – \hat{k}$$
Thus,
Final Result
$$\boxed{-4\hat{j} – \hat{k}}$$
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NCERT Question.7 : Find the unit vector in the direction of the vector $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$
Solution
Given
$$\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$$
First find the magnitude of $\vec{a}$
$$|\vec{a}| = \sqrt{1^2 + 1^2 + 2^2}$$
$$|\vec{a}| = \sqrt{6}$$
The unit vector is
$$\hat{a} = \frac{\vec{a}}{|\vec{a}|}$$
Thus
$$\hat{a} = \frac{\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{6}}$$
$$\hat{a} = \frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k}$$
Final Result
$$\boxed{\frac{1}{\sqrt{6}}\hat{i} + \frac{1}{\sqrt{6}}\hat{j} + \frac{2}{\sqrt{6}}\hat{k}}$$
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NCERT Question.8 : Find the unit vector in the direction of vector $\overrightarrow{PQ}$, where $P(1,2,3)$ and $Q(4,5,6)$.
Solution
The points are
$P(1,2,3)$ and $Q(4,5,6)$.
The vector
$$\overrightarrow{PQ} = \overrightarrow{OQ} – \overrightarrow{OP}$$
Compute components:
$$\overrightarrow{PQ} = (4-1)\hat{i} + (5-2)\hat{j} + (6-3)\hat{k}$$
$$\overrightarrow{PQ} = 3\hat{i} + 3\hat{j} + 3\hat{k}$$
Next compute its magnitude:
$$|\overrightarrow{PQ}| = \sqrt{3^2 + 3^2 + 3^2}$$
$$|\overrightarrow{PQ}| = \sqrt{27} = 3\sqrt{3}$$
Unit vector along $ \overrightarrow{PQ} $:
$$\widehat{PQ} = \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}$$
$$\widehat{PQ} = \frac{3\hat{i} + 3\hat{j} + 3\hat{k}}{3\sqrt{3}}$$
$$\widehat{PQ} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}$$
Final Result
$$\boxed{\frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} + \frac{1}{\sqrt{3}}\hat{k}}$$
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NCERT Question.9 : For the vectors
$\vec{a} = 2\hat{i} – \hat{j} + 2\hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} – \hat{k}$.
Find the unit vector in the direction of $\vec{a} + \vec{b}$.
Solution
Given
$$\vec{a}=2\hat{i}-\hat{j}+2\hat{k}$$
$$\vec{b}=-\hat{i}+\hat{j}-\hat{k}$$
Add the vectors:
$$\vec{a}+\vec{b}=(2-1)\hat{i}+(-1+1)\hat{j}+(2-1)\hat{k}$$
$$\vec{a}+\vec{b}=\hat{i}+0\hat{j}+\hat{k}$$
$$\vec{a}+\vec{b}=\hat{i}+\hat{k}$$
Magnitude:
$$|\vec{a}+\vec{b}|=\sqrt{1^2+1^2}=\sqrt{2}$$
Unit vector:
$$\widehat{a+b}=\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}$$
$$\widehat{a+b}=\frac{\hat{i}+\hat{k}}{\sqrt{2}}$$
Final Result
$$\boxed{\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{k}}$$
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NCERT Question.10 : Find a vector in the direction of the vector
$\vec{a} = 5\hat{i} – \hat{j} + 2\hat{k}$, which has a magnitude of 8 units.
Solution
Given:
$$\vec{a} = 5\hat{i} – \hat{j} + 2\hat{k}$$
Magnitude of $\vec{a}$:
$$|\vec{a}| = \sqrt{5^2 + (-1)^2 + 2^2}$$
$$|\vec{a}| = \sqrt{25 + 1 + 4}$$
$$|\vec{a}| = \sqrt{30}$$
Unit vector in the direction of $\vec{a}$:
$$\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{5\hat{i} – \hat{j} + 2\hat{k}}{\sqrt{30}}$$
A vector parallel to $\vec{a}$ with magnitude 8 is:
$$8\hat{a} = 8 \cdot \frac{5\hat{i} – \hat{j} + 2\hat{k}}{\sqrt{30}}$$
Simplifying:
$$8\hat{a} = \frac{40}{\sqrt{30}}\hat{i} – \frac{8}{\sqrt{30}}\hat{j} + \frac{16}{\sqrt{30}}\hat{k}$$
Final Result
$$\boxed{\frac{40}{\sqrt{30}}\hat{i} – \frac{8}{\sqrt{30}}\hat{j} + \frac{16}{\sqrt{30}}\hat{k}}$$
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