Cross Product (Vector Product) PYQs
Practice Class 12 CBSE Board Previous Year Questions (2008-2026)
Q1
2009 Board
00:00
Find the vector product $(\hat{i} + 2\hat{j} + 3\hat{k}) \times (2\hat{i} - \hat{j} + \hat{k})$. 2 Marks
Step 1: Set up Determinant
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & -1 & 1 \end{vmatrix}$.
Step 2: Expand Determinant
$= \hat{i}(2+3) - \hat{j}(1-6) + \hat{k}(-1-4)$.
$= 5\hat{i} + 5\hat{j} - 5\hat{k}$.
5i + 5j - 5k
Q2
2009 Board
00:00
Find the area of the parallelogram whose adjacent sides are represented by the vectors $\hat{i} + \hat{j}$ and $\hat{i} - \hat{j}$. 2 Marks
Step 1: Calculate Cross Product
$(\hat{i} + \hat{j}) \times (\hat{i} - \hat{j}) = \hat{i} \times \hat{i} - \hat{i} \times \hat{j} + \hat{j} \times \hat{i} - \hat{j} \times \hat{j}$.
$= 0 - \hat{k} - \hat{k} - 0 = -2\hat{k}$.
Step 2: Calculate Magnitude
Area $= |-2\hat{k}| = 2$ sq. units.
2 sq. units
Q3
2010 Board
00:00
Find the area of the triangle formed by the vectors $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$. 2 Marks
Step 1: Calculate Cross Product
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 1 \end{vmatrix} = \hat{i}(1+1) - \hat{j}(1-2) + \hat{k}(-1-2) = 2\hat{i} + \hat{j} - 3\hat{k}$.
Step 2: Calculate Magnitude
$|\vec{a} \times \vec{b}| = \sqrt{4+1+9} = \sqrt{14}$.
Step 3: Area of Triangle Formula
Area $= \frac{1}{2} |\vec{a} \times \vec{b}| = \frac{\sqrt{14}}{2}$ sq. units.
√14/2 sq. units
Q4
2011 Board
00:00
Find a unit vector perpendicular to both the vectors $\hat{i} + \hat{j}$ and $\hat{i} - \hat{j}$. 3 Marks
Step 1: Cross Product
$\vec{c} = (\hat{i} + \hat{j}) \times (\hat{i} - \hat{j}) = -2\hat{k}$.
Step 2: Find Unit Vector
$\hat{n} = \frac{-2\hat{k}}{|-2\hat{k}|} = -\hat{k}$ (or $\hat{k}$).
±k
Q5
2011 Board
00:00
Find the area of the triangle whose vertices are $(1,0,0), (0,1,0), (0,0,1)$. 4 Marks
Step 1: Position Vectors
$\vec{OA} = \hat{i}, \vec{OB} = \hat{j}, \vec{OC} = \hat{k}$.
Step 2: Find Side Vectors
$\vec{AB} = \hat{j} - \hat{i}, \vec{AC} = \hat{k} - \hat{i}$.
Step 3: Cross Product
$\vec{AB} \times \vec{AC} = (\hat{j}-\hat{i}) \times (\hat{k}-\hat{i}) = \hat{j} \times \hat{k} - \hat{j} \times \hat{i} - \hat{i} \times \hat{k} + \hat{i} \times \hat{i}$.
$= \hat{i} + \hat{k} + \hat{j} + 0 = \hat{i} + \hat{j} + \hat{k}$.
Step 4: Area
Area $= \frac{1}{2} |\hat{i} + \hat{j} + \hat{k}| = \frac{\sqrt{3}}{2}$ sq. units.
√3/2 sq. units
Q6
2014 Board
00:00
Show that $(\vec{a} + \vec{b}) \times \vec{c} = \vec{a} \times \vec{c} + \vec{b} \times \vec{c}$. 3 Marks
Uses the distributive property of the cross product over addition.
Proof involves expanding $(\vec{a}+\vec{b}) \times \vec{c}$ using component form or properties of determinants.
Hence Proved
Q7
2019 Board
00:00
Using vectors, find the area of the triangle whose vertices are $(1,2,3), (2,3,1), (3,1,2)$. 4 Marks
$\vec{AB} = \hat{i} + \hat{j} - 2\hat{k}$, $\vec{AC} = 2\hat{i} - \hat{j} - \hat{k}$.
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \\ 2 & -1 & -1 \end{vmatrix} = -3\hat{i} - 3\hat{j} - 3\hat{k}$.
Area $= \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2}\sqrt{9+9+9} = \frac{3\sqrt{3}}{2}$.
3√3/2 sq. units
Q8
2018 Board
00:00
Using vector product, find the area of the parallelogram formed by vectors $\hat{i} + 2\hat{j} + \hat{k}$ and $2\hat{i} - \hat{j} + \hat{k}$. 2 Marks
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ 2 & -1 & 1 \end{vmatrix} = 3\hat{i} + \hat{j} - 5\hat{k}$.
Area $= \sqrt{9+1+25} = \sqrt{35}$.
√35 sq. units
Q9
2017 Board
00:00
Find $(\hat{i} + 2\hat{j} + 3\hat{k}) \times (2\hat{i} + \hat{j} - \hat{k})$. 2 Marks
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 1 & -1 \end{vmatrix} = -5\hat{i} + 7\hat{j} - 3\hat{k}$.
-5i + 7j - 3k
Q10
2013 Board
00:00
Find $(2\hat{i} - \hat{j} + \hat{k}) \times (\hat{i} + \hat{j} - \hat{k})$. 2 Marks
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} = 0\hat{i} + 3\hat{j} + 3\hat{k}$.
3j + 3k
Q11
2014 Board
00:00
Find the area of the parallelogram whose adjacent sides are $\hat{i} + \hat{j} + \hat{k}$ and $2\hat{i} - \hat{j} + \hat{k}$. 2 Marks
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & -1 & 1 \end{vmatrix} = 2\hat{i} + \hat{j} - 3\hat{k}$.
Area $= \sqrt{4+1+9} = \sqrt{14}$.
√14 sq. units
Q12
2015 Board
00:00
Find a unit vector perpendicular to both $\hat{i} + \hat{j} + \hat{k}$ and $2\hat{i} - \hat{j} + \hat{k}$. 3 Marks
$\vec{a} \times \vec{b} = 2\hat{i} + \hat{j} - 3\hat{k}$.
$\hat{n} = \frac{1}{\sqrt{14}}(2\hat{i} + \hat{j} - 3\hat{k})$.
(1/√14)(2i + j - 3k)
Q13
2026 Board
00:00
If |a| = 8, |b| = 3 and |$\vec{a}$ × b| = 12, then value of |$\vec{a}·b$| is ______. 1 Mark
Solution will be updated soon.
Q14
2025 Board
00:00
The unit vector perpendicular to the vectors $\hat{i}$−$\hat{j}$ and $\hat{i}+\hat{j}$ is ______. 1 Mark
Solution will be updated soon.
Q15
2024 Board
00:00
Let $\vec{a}$ be any vector such that |a| = a. The value of |$\vec{a}×\hat{i}$|²+|$\vec{a}×\hat{j}$|²+|$\vec{a}×\hat{k}$|² is ______. 1 Mark
Solution will be updated soon.
Q16
2024 Board
00:00
Find $\vec{a}$ vector of magnitude 4 units perpendicular to each of the vectors $2\hat{i}$−$\hat{j}+\hat{k}$ and $\hat{i}+\hat{j}$−$\hat{k}$ and hence verify your answer. 3 Marks
Solution will be updated soon.
Q17
2024 Board
00:00
Given $\vec{a}$ = $2\hat{i}$−$\hat{j}+\hat{k}$, $\vec{b}$ = $3\hat{i}$−$\hat{k}$ and $\vec{c}$ = $2\hat{i}+\hat{j}$−$2\hat{k}$. Find $\vec{a}$ vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ and $\vec{c}·\vec{d}$ = 3. 3 Marks
Solution will be updated soon.
Q18
2016 Board
00:00
Find λ and μ, if ($\hat{i}+3\hat{j}$+$9\hat{k}$)×($3\hat{i}$−λ$\hat{j}+$μ$\hat{k}$) = 0. 1 Mark
Solution will be updated soon.
Q19
2016 Board
00:00
Write the number of vectors of unit length perpendicular to both the vectors $\vec{a}=2\hat{i}$+$\hat{j}+2\hat{k}$ and $\vec{b}=\hat{j}+\hat{k}$. 1 Mark
Solution will be updated soon.
Q20
2019 Board
00:00
If |a|=2, |b|=7 and $\vec{a}×\vec{b}=3\hat{i}$+$2\hat{j}+6\hat{k}$, find the angle between $\vec{a}$ and b. 1 Mark
Solution will be updated soon.
Q21
2019 Board
00:00
Find |$\vec{a}×b$|, if $\vec{a}=2\hat{i}$+$\hat{j}+5\hat{k}$ and $\vec{b}=3\hat{i}$+$5\hat{j}$–$2\hat{k}$. 1 Mark
Solution will be updated soon.
Q22
2018 Board
00:00
If θ is the angle between two vectors $\hat{i}$–$2\hat{j}+3\hat{k}$ and $3\hat{i}$–$2\hat{j}+\hat{k}$, find sin θ. 1 Mark
Solution will be updated soon.
Q23
2018 Board
00:00
Let $\vec{a}=4\hat{i}$+$5\hat{j}$–$\hat{k}$, $\vec{b}=\hat{i}$–$4\hat{j}+5\hat{k}$ and $\vec{c}=3\hat{i}$+$\hat{j}$–$\hat{k}$. Find $\vec{a}$ vector $\vec{d}$ which is perpendicular to both $\vec{c}$ and $\vec{b}$ and $\vec{d}·\vec{a}=21$. 1 Mark
Solution will be updated soon.
Q24
2017 Board
00:00
Using vectors, find the area of the ΔABC, whose vertices are A(1,2,5), B(2,−1,4) and C(4,5,−1). 1 Mark
Solution will be updated soon.
Q25
2017 Board
00:00
Show that the points A, B, C with position vectors $2\hat{i}$–$\hat{j}+\hat{k}$, $\hat{i}$–$5\hat{j}$–$5\hat{k}$ and $5\hat{i}$–$4\hat{j}$–$4\hat{k}$ respectively, are the vertices of $\vec{a}$ $right-angled$ triangle. Hence find the area of the triangle. 1 Mark
Solution will be updated soon.
Q26
2016 Board
00:00
If $\vec{a}×\vec{b}=\vec{c}×\vec{d}$ and $\vec{a}×\vec{c}=\vec{b}×d$, then show that a−$\vec{d}$ is parallel to b−c, where a≠$\vec{d}$ and b≠c. 1 Mark
Solution will be updated soon.