Question:

If \(\vec{a} = \alpha\hat{i} + \beta\hat{j} + 3\hat{k}\), \(\vec{b} = -\beta\hat{i} - \alpha\hat{j} - \hat{k}\) and \(\vec{c} = \hat{i} - 2\hat{j} - \hat{k}\) such that \(\vec{a}\cdot\vec{b} = 1\) and \(\vec{b}\cdot\vec{c} = -3\), then \(\dfrac{1}{3}\left((\vec{a}\times\vec{b})\cdot\vec{c}\right)\) is equal to _____________.

Show Hint

Use the two dot-product conditions to find \(\alpha, \beta\), then evaluate the scalar triple product as a determinant.
Updated On: Nov 4, 2026
Show Solution
collegedunia
Verified By Collegedunia

Correct Answer: 2

Solution and Explanation

\(\vec{a}\cdot\vec{b} = -\alpha\beta - \alpha\beta - 3 = 1\Rightarrow \alpha\beta = -2\).
\(\vec{b}\cdot\vec{c} = -\beta + 2\alpha + 1 = -3\Rightarrow \beta = 2\alpha + 4\).
\(\alpha(2\alpha + 4) = -2\Rightarrow 2\alpha^2 + 4\alpha + 2 = 0\Rightarrow (\alpha+1)^2 = 0\Rightarrow \alpha = -1,\ \beta = 2\).
\(\vec{a} = (-1, 2, 3)\), \(\vec{b} = (-2, 1, -1)\), \(\vec{c} = (1, -2, -1)\).
\((\vec{a}\times\vec{b})\cdot\vec{c} = \begin{vmatrix} -1 & 2 & 3 \\ -2 & 1 & -1 \\ 1 & -2 & -1 \end{vmatrix} = -1(-1 - 2) - 2(2 + 1) + 3(4 - 1) = 3 - 6 + 9 = 6\).
\(\frac13\times 6 = 2\).
Answer: 2
Was this answer helpful?
0
0

Top JEE Main Vector Algebra Questions

View More Questions