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Question

The volume of the parallelepiped formed by

\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}

\vec{b} = 2\hat{i} + \hat{j} + 3\hat{k}

, and

\vec{c} = \hat{i} + \hat{j} + \hat{k}

is:

Solution Path

Scalar triple product via 3x3 determinant = 4. Volume = |4| = 4.

01Question Setup

1/4

Find the volume of parallelepiped with edges

\vec{a} = 2\hat{i} - 3\hat{j}

\vec{b} = \hat{i} + \hat{j} - \hat{k}

\vec{c} = 3\hat{i} - \hat{k}

Volume

= ;?

02Determinant Setup

2/4

Volume

= |[\vec{a};\vec{b};\vec{c}]| = |\vec{a} \cdot (\vec{b} \times \vec{c})|

. Set up 3x3 determinant.

\begin{vmatrix} 2 & -3 & 0 \\ 1 & 1 & -1 \\ 3 & 0 & -1 \end{vmatrix}

03Expand DeterminantKEY INSIGHT

3/4

= 2(-1-0) + 3(-1+3) + 0 = -2 + 6 = 4

. Wait - recompute:

2(1 \cdot (-1) - (-1) \cdot 0) - (-3)(1 \cdot (-1) - (-1) \cdot 3) + 0 = 2(-1) + 3(-1+3) = -2 + 6 = 4

[\vec{a};\vec{b};\vec{c}] = 4

04Final Answer

4/4

Volume

= |4| = 4

\boxed{4}

Concepts from this question1 concepts unlocked

★

Scalar Triple Product and Volume

STANDARD

The scalar triple product [a b c] = a.(b x c) gives the signed volume of the parallelepiped formed by the three vectors. The absolute value gives the actual volume.

V = |\vec{a} \cdot (\vec{b} \times \vec{c})|

Volume of parallelepipeds and tetrahedra (1/6 of parallelepiped) appear regularly in JEE. The scalar triple product is also the determinant of the 3x3 matrix formed by the component vectors.

Volume of parallelepipedVolume of tetrahedronCoplanarity test

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