Scalar Triple Product PYQs
Practice Class 12 CBSE Board Previous Year Questions (2008-2026)
Q1
2010 Board
00:00
Show that the vectors $\hat{i} + \hat{j} + \hat{k}, 2\hat{i} + 3\hat{j} + 4\hat{k}, 3\hat{i} + 5\hat{j} + 7\hat{k}$ are coplanar. 4 Marks
Step 1: Form Determinant
Vectors are coplanar if their scalar triple product $[\vec{a}, \vec{b}, \vec{c}] = 0$.
$\begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 3 & 5 & 7 \end{vmatrix}$.
Step 2: Expand Determinant
$= 1(21-20) - 1(14-12) + 1(10-9)$.
$= 1(1) - 1(2) + 1(1) = 1 - 2 + 1 = 0$.
Step 3: Conclude
Since the scalar triple product is zero, the vectors are coplanar.
Vectors are coplanar
Q2
2013 Board
00:00
Prove that $\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{b} \cdot (\vec{c} \times \vec{a})$. 3 Marks
Scalar triple product is invariant under cyclic permutation.
$[\vec{a}, \vec{b}, \vec{c}] = [\vec{b}, \vec{c}, \vec{a}] = [\vec{c}, \vec{a}, \vec{b}]$.
This can be shown using the properties of determinants (interchanging two rows twice).
Hence Proved
Q3
2015 Board
00:00
Find the scalar triple product $[\vec{a}, \vec{b}, \vec{c}]$ for $\vec{a} = \hat{i} + \hat{j} + \hat{k}, \vec{b} = 2\hat{i} - \hat{j} + \hat{k}, \vec{c} = \hat{i} + 2\hat{j} - \hat{k}$. 2 Marks
$[\vec{a}, \vec{b}, \vec{c}] = \begin{vmatrix} 1 & 1 & 1 \\ 2 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix}$.
$= 1(1-2) - 1(-2-1) + 1(4+1) = -1 + 3 + 5 = 7$.
7
Q4
2017 Board
00:00
Show that the vectors $\hat{i} + \hat{j} + \hat{k}, 2\hat{i} + 3\hat{j} + 4\hat{k}, 3\hat{i} + 4\hat{j} + 5\hat{k}$ are coplanar. 4 Marks
$[\vec{a}, \vec{b}, \vec{c}] = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}$.
$= 1(15-16) - 1(10-12) + 1(8-9) = -1 + 2 - 1 = 0$.
Vectors are coplanar
Q5
2013 Board
00:00
If $\vec{a}, \vec{b}, \vec{c}$ are coplanar, show that $[\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}] = 0$. 4 Marks
Expression $= 2[\vec{a}, \vec{b}, \vec{c}]$.
Since $\vec{a}, \vec{b}, \vec{c}$ are coplanar, $[\vec{a}, \vec{b}, \vec{c}] = 0$.
Thus, $2(0) = 0$.
Hence Proved