Anand Classes presents Vector Algebra NCERT Solutions Miscellaneous Exercise Chapter-10 Class 12 Math Notes PDF Free Download (Set-2) to help students strengthen their understanding of vectors and master every type of question asked in the board exams. These solutions follow the latest NCERT guidelines and provide clear explanations, step-by-step methods, and exam-oriented problem-solving techniques to improve accuracy and conceptual clarity. Click the print button to download study material and notes.
NCERT Question.10 : The two adjacent sides of a parallelogram are
($2\hat{i}-4\hat{j}+5\hat{k})$ and ($\hat{i}-2\hat{j}-3\hat{k})$.
Find the unit vector parallel to its diagonal and also find its area.
Solution
Given vectors
$$
\vec{a}=2\hat{i}-4\hat{j}+5\hat{k},\qquad
\vec{b}=\hat{i}-2\hat{j}-3\hat{k}
$$
Diagonals of the parallelogram are $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$
One Diagonal of the parallelogram
$$
\vec{a}+\vec{b}=(2+1)\hat{i}+(-4-2)\hat{j}+(5-3)\hat{k}
$$
$$
\vec{a}+\vec{b}=3\hat{i}-6\hat{j}+2\hat{k}
$$
Unit vector parallel to diagonal
Magnitude:
$$
\lvert \vec{a}+\vec{b} \rvert =\sqrt{3^2+(-6)^2+2^2}
$$
$$
\lvert \vec{a}+\vec{b} \rvert =\sqrt{9+36+4}=\sqrt{49}=7
$$
Unit vector:
$$
\hat{u}=\frac{3\hat{i}-6\hat{j}+2\hat{k}}{7}
$$
$$
\hat{u}=\frac{3}{7}\hat{i}-\frac{6}{7}\hat{j}+\frac{2}{7}\hat{k}
$$
Another Diagonal of the parallelogram
$$
\vec{a}-\vec{b}=(2-1)\hat{i}+(-4+2)\hat{j}+(5+3)\hat{k}
$$
$$
\vec{a}-\vec{b}=\hat{i}-2\hat{j}+8\hat{k}
$$
Unit vector parallel to diagonal
Magnitude:
$$
\lvert \vec{a}-\vec{b} \rvert =\sqrt{1^2+(-2)^2+8^2}
$$
$$
\lvert \vec{a}-\vec{b} \rvert =\sqrt{1+4+64}=\sqrt{69}
$$
Unit vector:
$$
\hat{v}=\frac{\hat{i}-2\hat{j}+8\hat{k}}{{\sqrt{69}}}
$$
$$
\hat{v}=\frac{1}{\sqrt{69}}\hat{i}-\frac{2}{\sqrt{69}}\hat{j}+\frac{8}{\sqrt{69}}\hat{k}
$$
Area of the parallelogram
$$
\text{Area}=\lvert \vec{a}\times \vec{b} \rvert
$$
Compute cross product:
$$
\vec{a}\times \vec{b}=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \
2 & -4 & 5 \
1 & -2 & -3
\end{vmatrix}
$$
$$
\vec{a}\times \vec{b}
=\hat{i}(12+10)-\hat{j}(-6-5)+\hat{k}(-4+4)
$$
$$
\vec{a}\times \vec{b}
=22\hat{i}+11\hat{j}
$$
Magnitude:
$$
\lvert \vec{a}\times \vec{b} \rvert
=\sqrt{22^2+11^2}
$$
$$
\lvert \vec{a}\times \vec{b} \rvert=\sqrt{484+121}=\sqrt{605}=11\sqrt{5}
$$
Area of parallelogram
$$
\boxed{\sqrt{605}=11\sqrt{5}}
$$
Final Answer
- Unit vector parallel to diagonal:
$$
\boxed{\frac{3}{7}\hat{i}-\frac{6}{7}\hat{j}+\frac{2}{7}\hat{k}}
$$- $$
\boxed{\frac{1}{\sqrt{69}}\hat{i}-\frac{2}{\sqrt{69}}\hat{j}+\frac{8}{\sqrt{69}}\hat{k}}$$
- $$
- Area of parallelogram:
$$
\boxed{\sqrt{605}=11\sqrt{5}}
$$
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NCERT Question.11 : Show that the direction cosines of a vector equally inclined to the axes $OX, OY$, and $OZ$ are
$$\pm\left(\frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}},\ \frac{1}{\sqrt{3}}\right)$$
Solution
Assume that the vector is equally inclined to all three coordinate axes.
Let the angle made with each axis be $\alpha$.
The direction cosines are
$$l=\cos\alpha,\quad m=\cos\alpha,\quad n=\cos\alpha.$$
Using the identity for direction cosines:
$$l^2+m^2+n^2=1.$$
Substitute the values:
$$\cos^2\alpha+\cos^2\alpha+\cos^2\alpha=1$$
$$3\cos^2\alpha=1$$
$$\cos^2\alpha=\frac{1}{3}$$
$$\cos\alpha=\pm\frac{1}{\sqrt{3}}.$$
Hence ehe direction cosines are
$$l=m=n=\pm\frac{1}{\sqrt{3}}.$$
Final Result
$$\boxed{\pm\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)}$$
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NCERT Question.12 : Let
$$\vec{a}=\hat{i}+4\hat{j}+2\hat{k},\qquad
\vec{b}=3\hat{i}-2\hat{j}+7\hat{k},\qquad
\vec{c}=2\hat{i}-\hat{j}+4\hat{k}.$$
Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ and satisfies $\vec{c}\cdot\vec{d}=15$.
Solution
Let $\vec{d}=d_{1}\hat{i}+d_{2}\hat{j}+d_{3}\hat{k}$.
Perpendicularity to $\vec{a}$ and $\vec{b}$ gives
$$\vec{d}\cdot\vec{a}=(d_{1}\hat{i}+d_{2}\hat{j}+d_{3}\hat{k})\cdot (\hat{i}+4\hat{j}+2\hat{k})=0$$
$$\vec{d}\cdot\vec{a}=d_{1}+4d_{2}+2d_{3}=0 \tag{1}$$
$$\vec{d}\cdot\vec{b}=(d_{1}\hat{i}+d_{2}\hat{j}+d_{3}\hat{k})\cdot (3\hat{i}-2\hat{j}+7\hat{k})=0$$
$$\vec{d}\cdot\vec{b}=3d_{1}-2d_{2}+7d_{3}=0 \tag{2}$$
The condition with $\vec{c}$ is
$$\vec{c}\cdot\vec{d}=(2\hat{i}-\hat{j}+4\hat{k}) \cdot (d_{1}\hat{i}+d_{2}\hat{j}+d_{3}\hat{k})=15$$
$$\vec{c}\cdot\vec{d}=2d_{1}-d_{2}+4d_{3}=15 \tag{3}$$
Solve the linear system (1), (2), (3). Eliminating variables yields
$$d_{1}=\frac{160}{3},\qquad d_{2}=-\frac{5}{3},\qquad d_{3}=-\frac{70}{3}.$$
Thus
$$\boxed{\displaystyle \vec{d}=\frac{160}{3}\hat{i}-\frac{5}{3}\hat{j}-\frac{70}{3}\hat{k}
=\frac{1}{3}\bigl(160\hat{i}-5\hat{j}-70\hat{k}\bigr).}$$
Quick verification
$$\vec{d}\cdot\vec{a}
=\tfrac{160}{3}+4\bigl(-\tfrac{5}{3}\bigr)+2\bigl(-\tfrac{70}{3}\bigr)
=\tfrac{160-20-140}{3}=0$$
$$\vec{d}\cdot\vec{b}
=3\bigl(\tfrac{160}{3}\bigr)-2\bigl(-\tfrac{5}{3}\bigr)+7\bigl(-\tfrac{70}{3}\bigr)
=\tfrac{160+10-490}{3}=0$$
$$\vec{c}\cdot\vec{d}
=2\bigl(\tfrac{160}{3}\bigr)-\bigl(-\tfrac{5}{3}\bigr)+4\bigl(-\tfrac{70}{3}\bigr)
=\tfrac{320+5-280}{3}=\tfrac{45}{3}=15$$
All conditions are satisfied.
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NCERT Question.13 :The scalar product of the vector $\hat{i}+\hat{j}+\hat{k}$ with a unit vector along the sum of the vectors $2\hat{i}+4\hat{j}-5\hat{k}$ and $\lambda\hat{i}+2\hat{j}+3\hat{k}$ is equal to $1$. Find the value of $\lambda$.
Solution
Let
$$
\vec{a}=\hat{i}+\hat{j}+\hat{k},\qquad
\vec{b}=2\hat{i}+4\hat{j}-5\hat{k},\qquad
\vec{c}=\lambda\hat{i}+2\hat{j}+3\hat{k}.
$$
Form the sum
$$
\vec{d}=\vec{b}+\vec{c}=(2+\lambda)\hat{i}+6\hat{j}-2\hat{k}.
$$
The unit vector along $\vec{d}$ is
$$
\hat{d}=\frac{\vec{d}}{|\vec{d}|}
=\frac{(2+\lambda)\hat{i}+6\hat{j}-2\hat{k}}
{\sqrt{(2+\lambda)^2+6^2+(-2)^2}}
=\frac{(2+\lambda)\hat{i}+6\hat{j}-2\hat{k}}
{\sqrt{\lambda^2+4\lambda+44}}.
$$
Given the condition :
$$\vec{a}\cdot\hat{d}=1,$$
$$
\frac{\vec{a}\cdot\vec{d}}{|\vec{d}|}=1
$$
Compute
$$
\vec{a}\cdot\vec{d}=(1)(2+\lambda)+(1)(6)+(1)(-2)=\lambda+6.
$$
Thus,
$$
\frac{\lambda+6}{\sqrt{\lambda^2+4\lambda+44}}=1.
$$
Square both sides:
$$
(\lambda+6)^2=\lambda^2+4\lambda+44
$$
$$
\lambda^2+12\lambda+36=\lambda^2+4\lambda+44
$$
$$
8\lambda=8
$$
$$
\lambda=1.
$$
Final Answer
$$
\boxed{\lambda=1}
$$
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NCERT Question.14 : If $\vec a,\vec b,\vec c$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec a+\vec b+\vec c$ is equally inclined to $\vec a,\vec b,$ and $\vec c$.
Solution
Let the common magnitude be $p$, so
$$|\vec a|=|\vec b|=|\vec c|=p.$$
Mutual perpendicularity gives
$$\vec a\cdot\vec b=\vec b\cdot\vec c=\vec c\cdot\vec a=0.$$
Let the angles made by $\vec a+\vec b+\vec c$ with $\vec a,\vec b,\vec c$ be $\theta_1,\theta_2,\theta_3$ respectively. Then
$$\cos\theta_1=\frac{(\vec a+\vec b+\vec c)\cdot\vec a}{\lvert\vec a+\vec b+\vec c\rvert|\vec a|}.$$
Compute the numerator using orthogonality:
$$
(\vec a+\vec b+\vec c)\cdot\vec a
=\vec a\cdot\vec a+\vec b\cdot\vec a+\vec c\cdot\vec a
=|\vec a|^2+0+0=|\vec a|^2.
$$
Hence
$$\cos\theta_1=\frac{|\vec a|^2}{\lvert\vec a+\vec b+\vec c\rvert|\vec a|}
=\frac{|\vec a|}{\lvert\vec a+\vec b+\vec c\rvert}.$$
Similarly,
$$\cos\theta_2=\frac{(\vec a+\vec b+\vec c)\cdot\vec b}{\lvert\vec a+\vec b+\vec c\rvert|\vec b|}
=\frac{|\vec b|}{\lvert\vec a+\vec b+\vec c\rvert},$$
$$\cos\theta_3=\frac{(\vec a+\vec b+\vec c)\cdot\vec c}{\lvert\vec a+\vec b+\vec c\rvert|\vec c|}
=\frac{|\vec c|}{\lvert\vec a+\vec b+\vec c\rvert}.$$
Since $|\vec a|=|\vec b|=|\vec c|=p$, we get
$$\cos\theta_1=\cos\theta_2=\cos\theta_3=\frac{p}{\lvert\vec a+\vec b+\vec c\rvert},$$
so
$$\theta_1=\theta_2=\theta_3.$$
Therefore the vector $\vec a+\vec b+\vec c$ is equally inclined to $\vec a,\vec b,$ and $\vec c$.
Final conclusion
$$\boxed{\text{The angles } \theta_1,\theta_2,\theta_3\text{ are equal, so } \vec a+\vec b+\vec c \text{ is equally inclined to } \vec a,\vec b,\vec c.}$$
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NCERT Question.15 :Prove that
$$ (\vec a+\vec b)\cdot(\vec a+\vec b)=|\vec a|^2+|\vec b|^2 $$
if and only if $\vec a$ and $\vec b$ are perpendicular, given $\vec a\neq\vec 0,\ \vec b\neq\vec 0$.
Solution :
Expand the left hand side using bilinearity of the dot product:
$$
(\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.
$$
Using $\vec a\cdot\vec a=|\vec a|^2\;$, $\;\vec b\cdot\vec b=|\vec b|^2$ and commutativity $\vec a\cdot\vec b=\vec b\cdot\vec a$, the above expression becomes
$$
(\vec a+\vec b)\cdot(\vec a+\vec b)
=|\vec a|^2+2\vec a\cdot\vec b+|\vec b|^2.
$$
Now compare with the given expression $|\vec a|^2+|\vec b|^2$.
$$
(\vec a+\vec b)\cdot(\vec a+\vec b)=|\vec a|^2+|\vec b|^2,
$$
then from the expansion we get
$$
|\vec a|^2+2\vec a\cdot\vec b+|\vec b|^2=|\vec a|^2+|\vec b|^2,
$$
hence
$$
2\vec a\cdot\vec b=0\quad\Longrightarrow\quad \vec a\cdot\vec b=0.
$$
So $\vec a$ is perpendicular to $\vec b$.
(⇐) Conversely, if $\vec a\cdot\vec b=0$ then the expansion above reduces to
$$
(\vec a+\vec b)\cdot(\vec a+\vec b)=|\vec a|^2+|\vec b|^2,
$$
so the stated equality holds.
Conclusion
$$\boxed{(\vec a+\vec b)\cdot(\vec a+\vec b)=|\vec a|^2+|\vec b|^2\ \Longleftrightarrow\ \vec a\cdot\vec b=0,}$$
i.e. the equality holds exactly when $\vec a$ and $\vec b$ are perpendicular (nonzero condition only rules out the trivial zero cases).
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NCERT Question 16: Choose the correct answer
If $ \theta $ is the angle between two vectors $ \vec{a} $ and $ \vec{b} $, then
$ \vec{a} \cdot \vec{b} \ge 0 $ only when:
(A) $0 < \theta < \frac{\pi}{2}$
(B) $0 \le \theta \le \frac{\pi}{2}$
(C) $0 < \theta < \pi$
(D) $0 \le \theta \le \pi$
Solution
Let $ \theta $ be the angle between two vectors $ \vec{a} $ and $ \vec{b} $.
We know that
$$ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta $$
Given:
$$ \vec{a} \cdot \vec{b} \ge 0 $$
So,
$$ \cos \theta \ge 0 $$
This inequality holds when:
$$ 0 \le \theta \le \frac{\pi}{2} $$
Thus, the correct answer is (B) $,0 \le \theta \le \frac{\pi}{2}$.
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NCERT Question 17: Choose the correct answer
Let $ \vec{a} $ and $ \vec{b} $ be two unit vectors and $ \theta $ is the angle between them. Then $ \vec{a} + \vec{b} $ is a unit vector if:
(A) $ \theta = \frac{\pi}{4} $
(B) $ \theta = \frac{\pi}{3} $
(C) $ \theta = \frac{\pi}{2} $
(D) $ \theta = \frac{2\pi}{3} $
Solution
Given two unit vectors $ \vec{a} $ and $ \vec{b} $:
$$ |\vec{a}| = |\vec{b}| = 1 $$
If $ \vec{a} + \vec{b} $ is a unit vector, then
$$ |\vec{a} + \vec{b}| = 1 $$
Now,
$$ |\vec{a} + \vec{b}|^2 = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) $$
$$ |\vec{a} + \vec{b}|^2= \vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} $$
$$|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2\vec{a} \cdot \vec{b} + |\vec{b}|^2 $$
$$|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + 2|\vec{a}| |\vec{b}|\cos\theta + |\vec{b}|^2 $$
Substituting $ |\vec{a}| = |\vec{b}| = 1 , |\vec{a} + \vec{b}| = 1$:
$$ 1 = 1 + 2 \cos\theta + 1 $$
$$ 1 = 2 + 2 \cos\theta $$
$$ 2 \cos\theta = -1 $$
$$ \cos\theta = -\frac{1}{2} $$
Thus,
$$ \theta = \frac{2\pi}{3} $$
So, the correct answer is (D) $ \theta = \dfrac{2\pi}{3} $.
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NCERT Question 18 : Choose the correct answer. The value of
$$ \hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j}) $$
is:
(A) 0 (B) −1 (C) 1 (D) 3
Solution
Evaluate each term using standard vector identities.
We know:
$$ \hat{i} \times \hat{k} = -\hat{j} $$
$$ \hat{i} \times \hat{j} = \hat{k} $$
$$ \hat{j} \times \hat{k} = \hat{i} $$
Now compute each dot product:
$$ \hat{i} \cdot (\hat{j} \times \hat{k}) = \hat{i} \cdot \hat{i} = 1 $$
$$ \hat{j} \cdot (\hat{i} \times \hat{k}) = \hat{j} \cdot (-\hat{j}) = -1 $$
$$ \hat{k} \cdot (\hat{i} \times \hat{j}) = \hat{k} \cdot \hat{k} = 1 $$
Hence :
$$ \hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j}) =1 – 1 + 1 = 1 $$
$$ \hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j}) = 1 $$
Thus, the correct answer is (C) 1.
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NCERT Question 19 : Choose the correct answer.
If $ \theta $ is the angle between any two vectors $ \vec{a} $ and $ \vec{b} $, then
$$ |\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}| $$
when $ \theta $ is equal to:
(A) $0$ (B) $\frac{\pi}{4}$ (C) $\frac{\pi}{2}$ (D) $\pi$
Solution
Let $ \theta $ be the angle between $ \vec{a} $ and $ \vec{b} $.
Since $ \vec{a} $ and $ \vec{b} $ are non-zero vectors, their magnitudes are positive.
We use the formulas:
$$ |\vec{a} \cdot \vec{b}| = |\vec{a}| |\vec{b}| \cos \theta $$
$$ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta $$
The given condition is:
$$ |\vec{a}| |\vec{b}| \cos \theta = |\vec{a}| |\vec{b}| \sin \theta $$
Dividing both sides by $|\vec{a}| |\vec{b}|$ (non-zero):
$$ \cos \theta = \sin \theta $$
Thus:
$$ \tan \theta = 1 $$
$$ \theta = \frac{\pi}{4} $$
Therefore, the correct answer is (B) $\dfrac{\pi}{4}$.
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NCERT Question.4 : If $\vec{a}$, $\vec{b}$, $\vec{c}$ are such that $|\vec{a}|=3$, $|\vec{b}|=4$, $|\vec{c}|=5$ and each of them is perpendicular to the sum of the other two, find $|\vec{a}+\vec{b}+\vec{c}|$.
Solution
The condition “each vector is perpendicular to the sum of the other two” gives the three dot product relations
$$\vec{a}\cdot(\vec{b}+\vec{c})=0,\qquad
\vec{b}\cdot(\vec{c}+\vec{a})=0,\qquad
\vec{c}\cdot(\vec{a}+\vec{b})=0.$$
Expand each
$$\vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}=0\quad(1),\qquad
\vec{b}\cdot\vec{c}+\vec{b}\cdot\vec{a}=0\quad(2),\qquad
\vec{c}\cdot\vec{a}+\vec{c}\cdot\vec{b}=0\quad(3).$$
Add (1), (2) and (3). Using commutativity of dot product we get
$$2\bigl(\vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}+\vec{b}\cdot\vec{c}\bigr)=0$$
so
$$\vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}+\vec{b}\cdot\vec{c}=0.$$
Now compute the square of the magnitude required
$$
|\vec{a}+\vec{b}+\vec{c}|^2
=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2
+2\bigl(\vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}+\vec{b}\cdot\vec{c}\bigr)$$
$$ |\vec{a}+\vec{b}+\vec{c}|^2=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2\cdot 0.
$$
Substitute the given magnitudes
$$|\vec{a}+\vec{b}+\vec{c}|^2=3^2+4^2+5^2=9+16+25=50.$$
Hence
$$|\vec{a}+\vec{b}+\vec{c}|=\sqrt{50}=5\sqrt{2}.$$
Final Result
$$\boxed{|\vec{a}+\vec{b}+\vec{c}|=5\sqrt{2},}$$
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