Vector Algebra helps students understand direction, magnitude, and spatial reasoning, and Chapter 10: Vector Algebra is one of the most conceptual and scoring chapters in Class 12 Maths. This chapter introduces vectors as mathematical tools to represent physical quantities and focuses on vector operations such as addition, scalar multiplication, dot product, and cross product.
NCERT Solutions for Class 12 Maths Chapter 10 – Vector Algebra are prepared strictly according to the CBSE syllabus and board exam pattern. These solutions include important CBSE board questions asked between 2018 and 2025 and are explained in a clear, step-by-step manner using simple language and proper mathematical presentation. This helps students build strong conceptual clarity, practise efficiently, and score well in board examinations.
NCERT Solutions For Class 12 Maths Chapter 10 Vector Algebra
Q.
Find∣a×b∣,ifa=i−7j+7kandb=3i−2j+2k.
Q.
Find the sum of the vectorsa=i−2j+k,b=−2i+4j+5kandc=i−6j−7k.
Q.
Find the value of x for which x(i+j+k)isaunitvector.
Q.
Find the area of the triangle with vertices A (1, 1, 2), B (2, 3, 5) and C (1, 5, 5).
Q.
Show that the points A, B and C with position vectors, a=3i−4j−4k,b=2i−j+kand c=i−3j−5k respectively form the vertices of a right angled triangle.
Q.
Show that the vectori+j+kis equally inclined to the axes OX, OY and OZ.
Q.
Find the direction cosines of the vector joining the points A(1, 2, –3) and B(–1, – 2, 1) directed A to B.
Q.
Find the unit vector in the direction of the vectorPQ, where P and Q are the points (1,2,3) and (4,5,6)respectively.
Q.
Find the unit vector in the direction of the vectora=i+j+2k.
Q.
Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (– 5, 7).
Q.
Ifaunitvectoramakesangles3πwithi,4πwithj andanacuteangleθwithk,thenfindθandhence,thecomponentsofa.
Q.
Find the values of x and y so that the vectors2i+3jandxi+yjare equal.
Q.
Write two different vectors having same direction.
Q.
Compute the magnitude of the following vectors:a=i+j+k;b=2i−7j−3k;c=31i+31j−31k;
Q.
Findλandμif(2i+6j+27k)×(i+λj+μk)=0.
Q.
Classify the following as scalar and vector quantities.
(i) time period
(ii) distance
(iii) force
(iv) velocity
(v) work done
Q.
Classify the following measures as scalars and vectors.
(i) 10 kg
(ii) 2 meters north-west
(iii) 40°
(iv) 40 watt
(v) 10-19 coulomb
(vi) 20 m/s2
Q.
Represent graphically a displacement of 40 km, 30° east of north.
Q.
Showthat(a−b)×(a+b)=2(a×b)
Q.
Findavectorofmagnitude5units,andparalleltotheresultantofthevectorsa=2i+3j−kandb=i−2j+k.
Q. Find \( \lvert \vec{a}\times\vec{b}\rvert \), if \( \vec{a}=\hat{i}-7\hat{j}+7\hat{k} \) and \( \vec{b}=3\hat{i}-2\hat{j}+2\hat{k} \).
A. Given vectors are \( \vec{a}=\hat{i}-7\hat{j}+7\hat{k} \) and \( \vec{b}=3\hat{i}-2\hat{j}+2\hat{k} \).
Then,
\[
\vec{a}\times\vec{b}=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k}\\
1 & -7 & 7\\
3 & -2 & 2
\end{vmatrix}
=(-14+14)\hat{i}+(2-21)\hat{j}+(-2+21)\hat{k}
=-19\hat{j}+19\hat{k}
\]
\[
\therefore\ \lvert \vec{a}\times\vec{b}\rvert=\lvert -19\hat{j}+19\hat{k}\rvert
=\sqrt{(-19)^2+(19)^2}=19\sqrt{2}
\]
Subject: Mathematics | Chapter: Vector Algebra | Difficulty Level: Medium
Q. Find the sum of the vectors \( \vec{a}=\hat{i}-2\hat{j}+\hat{k} \), \( \vec{b}=-2\hat{i}+4\hat{j}+5\hat{k} \) and \( \vec{c}=\hat{i}-6\hat{j}-7\hat{k} \).
A. The given vectors are \( \vec{a}=\hat{i}-2\hat{j}+\hat{k} \), \( \vec{b}=-2\hat{i}+4\hat{j}+5\hat{k} \) and \( \vec{c}=\hat{i}-6\hat{j}-7\hat{k} \).
Then,
\[
\vec{a}+\vec{b}+\vec{c}
=(\hat{i}-2\hat{j}+\hat{k})+(-2\hat{i}+4\hat{j}+5\hat{k})+(\hat{i}-6\hat{j}-7\hat{k})
=-4\hat{j}-\hat{k}
\]
Subject: Mathematics | Chapter: Vector Algebra | Difficulty Level: Medium
Q. Find the value of \(x\) for which \(x(\hat{i}+\hat{j}+\hat{k})\) is a unit vector.
A. \(x(\hat{i}+\hat{j}+\hat{k})\) is a unit vector, so
\[
\lvert x(\hat{i}+\hat{j}+\hat{k})\rvert=1
\Rightarrow \lvert x\rvert\ \lvert \hat{i}+\hat{j}+\hat{k}\rvert=1
\Rightarrow \lvert x\rvert\sqrt{1^2+1^2+1^2}=1
\Rightarrow \lvert x\rvert\sqrt{3}=1
\Rightarrow x=\pm\frac{1}{\sqrt{3}}
\]
Subject: Mathematics | Chapter: Vector Algebra | Difficulty Level: Medium
Q. Find the area of the triangle with vertices \(A(1,1,2)\), \(B(2,3,5)\) and \(C(1,5,5)\).
A. The vertices of triangle \(ABC\) are \(A(1,1,2)\), \(B(2,3,5)\) and \(C(1,5,5)\).
The adjacent sides \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) are:
\[
\overrightarrow{AB}=(2-1)\hat{i}+(3-1)\hat{j}+(5-2)\hat{k}=\hat{i}+2\hat{j}+3\hat{k}
\]
\[
\overrightarrow{BC}=(1-2)\hat{i}+(5-3)\hat{j}+(5-5)\hat{k}=-\hat{i}+2\hat{j}
\]
\[
\overrightarrow{AB}\times\overrightarrow{BC}=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k}\\
1 & 2 & 3\\
-1 & 2 & 0
\end{vmatrix}
=(0-6)\hat{i}-(0+3)\hat{j}+(2+2)\hat{k}
=-6\hat{i}-3\hat{j}+4\hat{k}
\]
\[
\text{Area of } \triangle ABC=\frac{1}{2}\lvert \overrightarrow{AB}\times\overrightarrow{BC}\rvert
=\frac{1}{2}\sqrt{(-6)^2+(-3)^2+4^2}
=\frac{1}{2}\sqrt{36+9+16}
=\frac{1}{2}\sqrt{61}\ \text{square units}
\]
Subject: Mathematics | Chapter: Vector Algebra | Difficulty Level: Medium
Q. Show that the points \(A, B\) and \(C\) with position vectors \(\vec{a}=3\hat{i}-4\hat{j}-4\hat{k}\), \(\vec{b}=2\hat{i}-\hat{j}+\hat{k}\) and \(\vec{c}=\hat{i}-3\hat{j}-5\hat{k}\) respectively form the vertices of a right angled triangle.
A. Position vectors of points \(A, B\) and \(C\) are:
\(\vec{a}=3\hat{i}-4\hat{j}-4\hat{k},\ \vec{b}=2\hat{i}-\hat{j}+\hat{k},\ \vec{c}=\hat{i}-3\hat{j}-5\hat{k}\)
\[
\overrightarrow{AB}=\vec{b}-\vec{a}=(2\hat{i}-\hat{j}+\hat{k})-(3\hat{i}-4\hat{j}-4\hat{k})
=-\hat{i}+3\hat{j}+5\hat{k}
\]
\[
\overrightarrow{BC}=\vec{c}-\vec{b}=(\hat{i}-3\hat{j}-5\hat{k})-(2\hat{i}-\hat{j}+\hat{k})
=-\hat{i}-2\hat{j}-6\hat{k}
\]
\[
\overrightarrow{CA}=\vec{a}-\vec{c}=(3\hat{i}-4\hat{j}-4\hat{k})-(\hat{i}-3\hat{j}-5\hat{k})
=2\hat{i}-\hat{j}+\hat{k}
\]
So,
\[
\lvert \overrightarrow{AB}\rvert^2=(-1)^2+3^2+5^2=35,\quad
\lvert \overrightarrow{BC}\rvert^2=(-1)^2+(-2)^2+(-6)^2=41,\quad
\lvert \overrightarrow{CA}\rvert^2=2^2+(-1)^2+1^2=6
\]
\[
\lvert \overrightarrow{AB}\rvert^2+\lvert \overrightarrow{CA}\rvert^2=35+6=41=\lvert \overrightarrow{BC}\rvert^2
\]
Hence, \(ABC\) is a right-angled triangle.
Subject: Mathematics | Chapter: Vector Algebra | Difficulty Level: Medium
Q. Show that the vector \(\hat{i}+\hat{j}+\hat{k}\) is equally inclined to the axes \(OX\), \(OY\) and \(OZ\).
A. The given vector is \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\).
\[
\lvert \vec{a}\rvert=\sqrt{1^2+1^2+1^2}=\sqrt{3}
\]
The direction cosines of \(\vec{a}\) are:
\[
\left(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)
\]
Let \(\alpha,\beta,\gamma\) be the angles made by \(\vec{a}\) with the positive directions of \(x,y,z\) axes. Then,
\[
\cos\alpha=\frac{1}{\sqrt{3}},\ \cos\beta=\frac{1}{\sqrt{3}},\ \cos\gamma=\frac{1}{\sqrt{3}}
\]
Hence, the vector is equally inclined to the axes \(OX\), \(OY\) and \(OZ\).
Subject: Mathematics | Chapter: Vector Algebra | Difficulty Level: Medium
Q. Find the direction cosines of the vector joining the points \(A(1,2,-3)\) and \(B(-1,-2,1)\) directed \(A\) to \(B\).
A. The given points are \(A(1,2,-3)\) and \(B(-1,-2,1)\).
\[
\overrightarrow{AB}=(-1-1)\hat{i}+(-2-2)\hat{j}+(1+3)\hat{k}=-2\hat{i}-4\hat{j}+4\hat{k}
\]
\[
\lvert \overrightarrow{AB}\rvert=\sqrt{(-2)^2+(-4)^2+4^2}=\sqrt{4+16+16}=\sqrt{36}=6
\]
\[
\text{Direction cosines of } \overrightarrow{AB}=
\left(\frac{-2}{6},\frac{-4}{6},\frac{4}{6}\right)=\left(-\frac{1}{3},-\frac{2}{3},\frac{2}{3}\right)
\]
Subject: Mathematics | Chapter: Vector Algebra | Difficulty Level: Medium
Q. Find the unit vector in the direction of the vector \(\overrightarrow{PQ}\), where \(P\) and \(Q\) are the points \((1,2,3)\) and \((4,5,6)\) respectively.
A. Here,
\[
\overrightarrow{OP}=\hat{i}+2\hat{j}+3\hat{k},\quad
\overrightarrow{OQ}=4\hat{i}+5\hat{j}+6\hat{k}
\]
\[
\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}
=(4\hat{i}+5\hat{j}+6\hat{k})-(\hat{i}+2\hat{j}+3\hat{k})
=3\hat{i}+3\hat{j}+3\hat{k}
\]
\[
\text{Unit vector of } \overrightarrow{PQ}=
\frac{\overrightarrow{PQ}}{\lvert \overrightarrow{PQ}\rvert}
=\frac{3\hat{i}+3\hat{j}+3\hat{k}}{\sqrt{3^2+3^2+3^2}}
=\frac{3\hat{i}+3\hat{j}+3\hat{k}}{3\sqrt{3}}
=\frac{\hat{i}}{\sqrt{3}}+\frac{\hat{j}}{\sqrt{3}}+\frac{\hat{k}}{\sqrt{3}}
\]
Subject: Mathematics | Chapter: Vector Algebra | Difficulty Level: Medium
Q. Find the unit vector in the direction of the vector \(\vec{a}=\hat{i}+\hat{j}+2\hat{k}\).
A. The unit vector \(\hat{a}\) in the direction of \(\vec{a}\) is:
\[
\hat{a}=\frac{\vec{a}}{\lvert \vec{a}\rvert}
=\frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{1^2+1^2+2^2}}
=\frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{6}}
=\frac{1}{\sqrt{6}}\hat{i}+\frac{1}{\sqrt{6}}\hat{j}+\frac{2}{\sqrt{6}}\hat{k}
\]
Subject: Mathematics | Chapter: Vector Algebra | Difficulty Level: Medium
Q. Find the scalar and vector components of the vector with initial point \((2,1)\) and terminal point \((-5,7)\).
A. The vector with initial point \(P(2,1)\) and terminal point \(Q(-5,7)\) is:
\[
\overrightarrow{PQ}=(-5-2)\hat{i}+(7-1)\hat{j}=-7\hat{i}+6\hat{j}
\]
Hence, the required scalar components are \(-7\) and \(6\) while the vector components are \(-7\hat{i}\) and \(6\hat{j}\).
Subject: Mathematics | Chapter: Vector Algebra | Difficulty Level: Medium
FAQs: NCERT Solutions for Class 12 Maths Chapter 10 – Vector Algebra
Q1. Why is Vector Algebra important for Class 12 Maths?
Vector Algebra is essential for understanding three-dimensional geometry and physics concepts and carries good weightage in CBSE board exams.
Q2. How many marks are usually asked from Vector Algebra in board exams?
Generally, 5 to 6 marks worth of questions are asked from this chapter in the CBSE board exams.
Q3. Are NCERT Solutions enough to prepare this chapter?
Yes. NCERT textbook questions and NCERT Solutions are sufficient for CBSE board exams, as most questions are directly based on NCERT exercises and examples.
Q4. Which topics in Vector Algebra are most important for exams?
The most important topics include:
Q5. Why do students find Vector Algebra confusing?
Students often struggle with:
Practising diagrams and understanding the geometric meaning of vectors helps reduce confusion.
Q6. Is Vector Algebra useful for competitive exams?
Yes. Vector Algebra is widely used in JEE and other competitive exams, especially in questions related to 3D geometry and physics applications.