Anand Classes offers comprehensive NCERT Solutions for Exercise 10.3 of Chapter 10 Vector Algebra for Class 12 Mathematics, helping students understand the dot product, scalar projection, and related vector concepts through clear, step-by-step explanations. These Set-2 notes are prepared according to the latest NCERT and CBSE guidelines, making them ideal for revision, doubt clearing, and effective board exam preparation. Click the print button to download study material and notes.
NCERT Solutions for Exercise 10.3 of Chapter 10 Vector Algebra for Class 12 Mathematics
NCERT Question.10 : If $\vec{a} = 2\hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = -\hat{i} + 2\hat{j} + \hat{k}$ and $\vec{c} = 3\hat{i} + \hat{j}$ are such that $(\vec{a} + \lambda \vec{b})$ is perpendicular to $\vec{c}$. Ffind the value of $\lambda$.
Solution
Since $\vec{a} + \lambda \vec{b}$ is perpendicular to $\vec{c}$, their dot product is zero:
$$
(\vec{a} + \lambda \vec{b}) \cdot \vec{c} = 0
$$
Expand:
$$
\vec{a} \cdot \vec{c} + \lambda \vec{b} \cdot \vec{c} = 0
$$
Compute each dot product.
1. Calculate $\vec{a} \cdot \vec{c}$
$$
(\vec{a} \cdot \vec{c})=(2\hat{i} + 2\hat{j} + 3\hat{k}) \cdot (3\hat{i} + \hat{j})
$$
$$
(\vec{a} \cdot \vec{c})= 2\cdot 3 + 2\cdot 1 + 3\cdot 0
$$
$$
(\vec{a} \cdot \vec{c})= 6 + 2 = 8
$$
2. Calculate $\vec{b} \cdot \vec{c}$
$$
(\vec{b} \cdot \vec{c})=(-\hat{i} + 2\hat{j} + \hat{k}) \cdot (3\hat{i} + \hat{j})
$$
$$
(\vec{b} \cdot \vec{c})= -1\cdot 3 + 2\cdot 1 + 1\cdot 0
$$
$$
(\vec{b} \cdot \vec{c})= -3 + 2 = -1
$$
Substitute into the perpendicularity condition:
$$
\vec{a} \cdot \vec{c} + \lambda \vec{b} \cdot \vec{c} = 0
$$
$$
8 + \lambda(-1) = 0
$$
$$
8 – \lambda = 0
$$
$$
\lambda = 8
$$
Final Result
$$\boxed{\lambda = 8}$$
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NCERT Question.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 $\vec b$.
Solution
Let
$$\vec u = |\vec a|\vec b + |\vec b|\vec a$$
and
$$\vec v = |\vec a|\vec b – |\vec b|\vec a.$$
To show perpendicularity, we compute the dot product:
$$\vec u \cdot \vec v = (|\vec a|\vec b + |\vec b|\vec a)\cdot (|\vec a|\vec b – |\vec b|\vec a).$$
Expanding:
$$\vec u \cdot \vec v = |\vec a|^2(\vec b \cdot \vec b) – |\vec a||\vec b|(\vec b \cdot \vec a) + |\vec b||\vec a|(\vec a \cdot \vec b) – |\vec b|^2(\vec a \cdot \vec a).$$
Since
$$\vec a \cdot \vec b = \vec b \cdot \vec a,$$
the middle terms cancel out.
Thus:
$$\vec u \cdot \vec v = |\vec a|^2|\vec b|^2 – |\vec b|^2|\vec a|^2.$$
So:
$$\vec u \cdot \vec v = 0.$$
Hence the vectors are perpendicular.
Final Result
$$\boxed{(|\vec a|\vec b + |\vec b|\vec a) \;\perp \; (|\vec a|\vec b – |\vec b|\vec a)}$$
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NCERT Question.12 : If $\vec a \cdot \vec a = 0$ and $\vec a \cdot \vec b = 0$ then what can be concluded about the vector $\vec a$?
Solution
Given:
$$\vec a \cdot \vec a = 0$$
and
$$\vec a \cdot \vec b = 0$$
which means $\vec a \; \perp \;\vec b $
Now,
$$\vec a \cdot \vec a = |\vec a|^2.$$
Thus,
$$|\vec a|^2 = 0.$$
This gives:
$$|\vec a| = 0.$$
A vector whose magnitude is zero is the zero vector:
$$\vec a = \vec 0.$$
Now consider:
$$\vec a \cdot \vec b = 0.$$
Since $\vec a = \vec 0$,
$$\vec 0 \cdot \vec b = 0,$$
which is true for any vector $\vec b$.
Thus, the conclusion about $\vec b $ is that it can be any vector, as long as it is perpendicular to $\vec a $. The only constraint is that $\vec b $ must be perpendicular to $\vec a $.
Hence, $\vec b$ can be any vector because the zero vector is perpendicular to every vector.
Final Result
$$\boxed{\vec a = \vec 0 \text{ and } \vec b \text{ can be any vector}}$$
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NCERT Question.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$.
Solution
Given $|\vec a|=|\vec b|=|\vec c|=1$ and
$$\vec a + \vec b + \vec c = \vec 0.$$
Take dot product of the equation with $\vec a$:
$$\vec a\cdot(\vec a+\vec b+\vec c)=\vec a\cdot\vec 0 = 0.$$
Thus
$$\vec a\cdot\vec a + \vec a\cdot\vec b + \vec a\cdot\vec c = 0.$$
Since $|\vec a|^2=\vec a\cdot\vec a = 1$ we get
$$1 + \vec a\cdot\vec b + \vec a\cdot\vec c = 0\qquad(1)$$
Similarly dotting with $\vec b$ gives
$$1 + \vec b\cdot\vec c + \vec b\cdot\vec a = 0\qquad(2)$$
Dotting with $\vec c$ gives
$$1 + \vec c\cdot\vec a + \vec c\cdot\vec b = 0\qquad(3)$$
Add equations (1), (2) and (3):
$$3 + 2\bigl(\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a\bigr) = 0.$$
Hence
$$\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a = -\frac{3}{2}.$$
Final Result
$$\boxed{\;\vec a\cdot\vec b + \vec b\cdot\vec c + \vec c\cdot\vec a = -\frac{3}{2}\;}$$
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NCERT Question.14 : If either vector $ \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.
Solution
The dot product formula is
$$\vec a \cdot \vec b = |\vec a||\vec b|\cos\theta$$
where $|\vec a|$ and $|\vec b|$ are magnitudes and $\theta$ is the angle between the vectors.
If $ \vec a = \vec 0 $ then $|\vec a| = 0$ and therefore for any $ \vec b $
$$\vec a \cdot \vec b = 0\cdot|\vec b|\cos\theta = 0.$$
Similarly if $ \vec b = \vec 0 $ then $ \vec a \cdot \vec b = 0.$
For the converse suppose $ \vec a \cdot \vec b = 0 $. This implies either $|\vec a|=0$ or $|\vec b|=0$ or $\cos\theta = 0$. The last case means $\theta = 90^\circ$ and the vectors are perpendicular but not zero. Thus $ \vec a \cdot \vec b = 0 $ does not force either vector to be the zero vector.
A simple concrete example in $\mathbb R^2$ is
$$\vec a = (1,0)=\hat\imath,\qquad \vec b=(0,1)=\hat\jmath.$$
Their dot product is
$$\vec a\cdot\vec b = 1\cdot 0 + 0\cdot 1 = 0.$$
But $ \vec a \neq \vec 0 $ and $ \vec b \neq \vec 0 $. Hence the converse is false.
Final Result
$$\boxed{\text{If } \vec a\cdot\vec b=0 \text{ then } \vec a \text{ or } \vec b \text{ need not be } \vec 0\ ;\ \text{they may be perpendicular}}$$
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NCERT Question.15 : If the vertices $A, B, C$ of a triangle $ABC$ are $(1, 2, 3), (–1, 0, 0), (0, 1, 2)$, respectively, then find $\angle ABC$. [Here $\angle ABC$ is the angle between vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}$.
Solution
Compute the vectors from Vertx $B$ to the other vertices.
$$\overrightarrow{BA}=\overrightarrow{A}-\overrightarrow{B}=(1-(-1),\;2-0,\;3-0)=(2,\;2,\;3)$$
$$\overrightarrow{BC}=\overrightarrow{C}-\overrightarrow{B}=(0-(-1),\;1-0,\;2-0)=(1,\;1,\;2)$$
Dot product:
$$\overrightarrow{BA}\cdot\overrightarrow{BC}=(2)(1)+(2)(1)+(3)(2)=2+2+6=10$$
Magnitudes:
$$|\overrightarrow{BA}|=\sqrt{2^2+2^2+3^2}=\sqrt{4+4+9}=\sqrt{17}$$
$$|\overrightarrow{BC}|=\sqrt{1^2+1^2+2^2}=\sqrt{1+1+4}=\sqrt{6}$$
Cosine of the angle:
$$\cos\theta=\frac{\overrightarrow{BA}\cdot\overrightarrow{BC}}{|\overrightarrow{BA}|\;|\overrightarrow{BC}|}
=\frac{10}{\sqrt{17}\sqrt{6}}=\frac{10}{\sqrt{102}}$$
Therefore
$$\theta=\cos^{-1}\left(\frac{10}{\sqrt{102}}\right)$$
Numerical value
$$\theta\approx 8.0495^\circ\quad\text{(approximately)}$$
Final Result
$$\boxed{\theta=\cos^{-1}\left(\frac{10}{\sqrt{102}}\right)\approx 8.0495^\circ}$$
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NCERT Question.16 : Show that the points $A(1, 2, 7), B(2, 6, 3)$ and $C(3, 10, -1)$ are collinear.
Solution :
Direction vector $\overrightarrow{AB}$
$$
\vec{AB}=(2-1,6-2,3-7)=(1,4,-4)
$$
Direction vector $\overrightarrow{BC}$
$$
\overrightarrow{BC}=(3-2,10-6,-1-3)=(1,4,-4)
$$
Conclusion
Since
$$\overrightarrow{AB}=\vec{BC}=(1,4,-4),$$
both vectors have the same direction.
Therefore, the points $(A), (B)$ and $(C)$ are collinear.
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NCERT Question.17 : Show that the vectors
$2\hat{i}-\hat{j} + \hat{k}, \hat{i} – 3\hat{j} – 5\hat{k}, 3\hat{i} – 4\hat{j} – 4\hat{k}$
form the vertices of a right-angled triangle.
Solution
Let
$$\vec{A}=2\hat{i}-\hat{j}+\hat{k},\quad \vec{B}=\hat{i}-3\hat{j}-5\hat{k},\quad \vec{C}=3\hat{i}-4\hat{j}-4\hat{k}.$$
Step 1: Compute magnitudes
$$
|\vec{A}|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{4+1+1}=\sqrt{6}
$$
$$
|\vec{B}|=\sqrt{1^2+(-3)^2+(-5)^2}=\sqrt{1+9+25}=\sqrt{35}
$$
$$
|\vec{C}|=\sqrt{3^2+(-4)^2+(-4)^2}=\sqrt{9+16+16}=\sqrt{41}
$$
Step 2: Check Pythagoras relation
$$
|\vec{A}|^2+|\vec{B}|^2=6+35=41=|\vec{C}|^2
$$
Conclusion
Since
$$|\vec{A}|^2+|\vec{B}|^2=|\vec{C}|^2$$
the three vectors satisfy the Pythagoras theorem. Therefore, the vectors represent the vertices of a right-angled triangle.
Final Result
$$\boxed{\text{The three given vectors form a right-angled triangle.}}$$
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NCERT Question.18 : If $\vec{a}$ is a nonzero vector of magnitude $a$ and $\lambda$ a nonzero scalar, then $\lambda\vec{a}$ is a unit vector if:
(A) $\lambda = 1$
(B) $\lambda = -1$
(C) $a = |\lambda|$
(D) $a = \frac{1}{|\lambda|}$
Solution
For $\lambda\vec{a}$ to be a unit vector:
$$|\lambda\vec{a}|=1$$
Magnitude property:
$$|\lambda\vec{a}|=|\lambda||\vec{a}|$$
Given $|\vec{a}|=a$, we have:
$$|\lambda|a=1$$
Solve for (a):
$$a=\frac{1}{|\lambda|}$$
Final Result
$$\boxed{a=\frac{1}{|\lambda|}}$$
So the correct option is (D).
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