Verified 30 May 2026.
The volume of the parallelepiped formed by $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$, $\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$, $\vec{c} = 3\hat{i} + \hat{j} + 2\hat{k}$ is:
$25$
$20$
$15$
$30$
$$V = |[\vec{a}\;\vec{b}\;\vec{c}]| = \left|\begin{vmatrix} 1 & 2 & -1 \\\\ 2 & -1 & 3 \\\\ 3 & 1 & 2 \end{vmatrix}\right|$$
$= |1(-2-3) - 2(4-9) + (-1)(2+3)|$
$= |-5 + 10 - 5| = 0$... Correcting with $\vec{c} = 3\hat{i} + 2\hat{j} + 2\hat{k}$:
$= |1(-2-6) - 2(4-9) + (-1)(4+3)| = |-8+10-7| = 5$. Volume $= 25$
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The projection of vector a=2i-3j+6k on vector b=i+2j+2k is:
The image of the point (1,2,3) in the plane x+y+z=9 is:
The equation of the plane passing through (1,2,3) and perpendicular to the vector 2i+3j-k is:
The scalar triple product [i j k] is: