Question Statement We are tasked with finding the volume of a parallelepiped given its three edge vectors. The volume of a parallelepiped can be calculated using the scalar triple product of the vectors representing the edges. The vectors for each part of the problem are as follows:
Part (i)
U = 3 i ^ + 0 j ^ + 2 k ^ \mathbf{U} = 3 \hat{i} + 0 \hat{j} + 2 \hat{k} U = 3 i ^ + 0 j ^ + 2 k ^ V = i ^ + 2 j ^ + k ^ \mathbf{V} = \hat{i} + 2 \hat{j} + \hat{k} V = i ^ + 2 j ^ + k ^ W = 0 i ^ − j ^ + 4 k ^ \mathbf{W} = 0 \hat{i} - \hat{j} + 4 \hat{k} W = 0 i ^ − j ^ + 4 k ^ Part (ii)
a = i ^ − 4 j ^ − k ^ \mathbf{a} = \hat{i} - 4 \hat{j} - \hat{k} a = i ^ − 4 j ^ − k ^ b = i ^ − j ^ − 2 k ^ \mathbf{b} = \hat{i} - \hat{j} - 2 \hat{k} b = i ^ − j ^ − 2 k ^ c = 2 i ^ − 3 j ^ + k ^ \mathbf{c} = 2 \hat{i} - 3 \hat{j} + \hat{k} c = 2 i ^ − 3 j ^ + k ^ Part (iii)
a ^ = i ^ − 2 j ^ + 3 k ^ \hat{\mathbf{a}} = \hat{i} - 2 \hat{j} + 3 \hat{k} a ^ = i ^ − 2 j ^ + 3 k ^ b = 2 i ^ − j ^ − k ^ \mathbf{b} = 2 \hat{i} - \hat{j} - \hat{k} b = 2 i ^ − j ^ − k ^ c ^ = 0 i ^ − j ^ + k ^ \hat{\mathbf{c}} = 0 \hat{i} - \hat{j} + \hat{k} c ^ = 0 i ^ − j ^ + k ^ Background and Explanation To find the volume of a parallelepiped given three vectors, we use the formula:
Volume = ∣ U ⋅ ( V × W ) ∣ \text{Volume} = | \mathbf{U} \cdot (\mathbf{V} \times \mathbf{W}) | Volume = ∣ U ⋅ ( V × W ) ∣ This involves finding the scalar triple product of the three vectors. The cross product of two vectors gives a vector perpendicular to both, and the dot product of this result with the third vector gives a scalar value. The magnitude of this scalar gives the volume.
Solution Part (i)
Given vectors:
U = 3 i ^ + 0 j ^ + 2 k ^ \mathbf{U} = 3 \hat{i} + 0 \hat{j} + 2 \hat{k} U = 3 i ^ + 0 j ^ + 2 k ^ V = i ^ + 2 j ^ + k ^ \mathbf{V} = \hat{i} + 2 \hat{j} + \hat{k} V = i ^ + 2 j ^ + k ^ W = 0 i ^ − j ^ + 4 k ^ \mathbf{W} = 0 \hat{i} - \hat{j} + 4 \hat{k} W = 0 i ^ − j ^ + 4 k ^
We begin by finding the determinant of the matrix formed by these vectors:
Volume = ∣ 3 0 21 2 10 − 1 4 ∣ \text{Volume} = \left| \begin{array}{ccc}
3 & 0 & 2
1 & 2 & 1
0 & -1 & 4
\end{array} \right| Volume = 3 0 21 2 10 − 1 4 Now expand this determinant using the first row:
= 3 ( ∣ 2 1 − 1 4 ∣ ) − 0 + 2 ( ∣ 1 20 − 1 ∣ ) = 3 \left( \begin{vmatrix} 2 & 1 -1 & 4 \end{vmatrix} \right) - 0 + 2 \left( \begin{vmatrix} 1 & 2 0 & -1 \end{vmatrix} \right) = 3 ( 2 1 − 1 4 ) − 0 + 2 ( 1 20 − 1 ) Evaluating the 2x2 determinants:
∣ 2 1 − 1 4 ∣ = 2 ( 4 ) − 1 ( − 1 ) = 8 + 1 = 9 \begin{vmatrix} 2 & 1 -1 & 4 \end{vmatrix} = 2(4) - 1(-1) = 8 + 1 = 9 2 1 − 1 4 = 2 ( 4 ) − 1 ( − 1 ) = 8 + 1 = 9 ∣ 1 20 − 1 ∣ = 1 ( − 1 ) − 2 ( 0 ) = − 1 \begin{vmatrix} 1 & 2 0 & -1 \end{vmatrix} = 1(-1) - 2(0) = -1 1 20 − 1 = 1 ( − 1 ) − 2 ( 0 ) = − 1 Thus, the volume is:
= 3 ( 9 ) + 2 ( − 1 ) = 27 − 2 = 25 cubic units = 3(9) + 2(-1) = 27 - 2 = 25 \text{ cubic units} = 3 ( 9 ) + 2 ( − 1 ) = 27 − 2 = 25 cubic units Part (ii)
Given vectors:
a = i ^ − 4 j ^ − k ^ \mathbf{a} = \hat{i} - 4 \hat{j} - \hat{k} a = i ^ − 4 j ^ − k ^ b = i ^ − j ^ − 2 k ^ \mathbf{b} = \hat{i} - \hat{j} - 2 \hat{k} b = i ^ − j ^ − 2 k ^ c = 2 i ^ − 3 j ^ + k ^ \mathbf{c} = 2 \hat{i} - 3 \hat{j} + \hat{k} c = 2 i ^ − 3 j ^ + k ^
We now find the determinant of the matrix formed by these vectors:
Volume = ∣ 1 − 4 − 11 − 1 − 22 − 3 1 ∣ \text{Volume} = \left| \begin{array}{ccc}
1 & -4 & -1
1 & -1 & -2
2 & -3 & 1
\end{array} \right| Volume = 1 − 4 − 11 − 1 − 22 − 3 1 Expand using the first row:
= 1 ( ∣ − 1 − 2 − 3 1 ∣ ) − ( − 4 ) ( ∣ 1 − 22 1 ∣ ) + ( − 1 ) ( ∣ 1 − 12 − 3 ∣ ) = 1 \left( \begin{vmatrix} -1 & -2 -3 & 1 \end{vmatrix} \right) - (-4) \left( \begin{vmatrix} 1 & -2 2 & 1 \end{vmatrix} \right) + (-1) \left( \begin{vmatrix} 1 & -1 2 & -3 \end{vmatrix} \right) = 1 ( − 1 − 2 − 3 1 ) − ( − 4 ) ( 1 − 22 1 ) + ( − 1 ) ( 1 − 12 − 3 ) Evaluating the 2x2 determinants:
∣ − 1 − 2 − 3 1 ∣ = ( − 1 ) ( 1 ) − ( − 2 ) ( − 3 ) = − 1 − 6 = − 7 \begin{vmatrix} -1 & -2 -3 & 1 \end{vmatrix} = (-1)(1) - (-2)(-3) = -1 - 6 = -7 − 1 − 2 − 3 1 = ( − 1 ) ( 1 ) − ( − 2 ) ( − 3 ) = − 1 − 6 = − 7 ∣ 1 − 22 1 ∣ = 1 ( 1 ) − ( − 2 ) ( 2 ) = 1 + 4 = 5 \begin{vmatrix} 1 & -2 2 & 1 \end{vmatrix} = 1(1) - (-2)(2) = 1 + 4 = 5 1 − 22 1 = 1 ( 1 ) − ( − 2 ) ( 2 ) = 1 + 4 = 5 ∣ 1 − 12 − 3 ∣ = 1 ( − 3 ) − ( − 1 ) ( 2 ) = − 3 + 2 = − 1 \begin{vmatrix} 1 & -1 2 & -3 \end{vmatrix} = 1(-3) - (-1)(2) = -3 + 2 = -1 1 − 12 − 3 = 1 ( − 3 ) − ( − 1 ) ( 2 ) = − 3 + 2 = − 1 Thus, the volume is:
= 1 ( − 7 ) + 4 ( 5 ) − 1 ( − 1 ) = − 7 + 20 + 1 = 14 cubic units = 1(-7) + 4(5) - 1(-1) = -7 + 20 + 1 = 14 \text{ cubic units} = 1 ( − 7 ) + 4 ( 5 ) − 1 ( − 1 ) = − 7 + 20 + 1 = 14 cubic units Part (iii)
Given vectors:
a ^ = i ^ − 2 j ^ + 3 k ^ \hat{\mathbf{a}} = \hat{i} - 2 \hat{j} + 3 \hat{k} a ^ = i ^ − 2 j ^ + 3 k ^ b = 2 i ^ − j ^ − k ^ \mathbf{b} = 2 \hat{i} - \hat{j} - \hat{k} b = 2 i ^ − j ^ − k ^ c ^ = 0 i ^ − j ^ + k ^ \hat{\mathbf{c}} = 0 \hat{i} - \hat{j} + \hat{k} c ^ = 0 i ^ − j ^ + k ^
We find the determinant of the matrix formed by these vectors:
Volume = ∣ 1 − 2 32 − 1 − 10 1 1 ∣ \text{Volume} = \left| \begin{array}{ccc}
1 & -2 & 3
2 & -1 & -1
0 & 1 & 1
\end{array} \right| Volume = 1 − 2 32 − 1 − 10 1 1 Expand using the first row:
= 1 ( ∣ − 1 − 11 1 ∣ ) − ( − 2 ) ( ∣ 2 − 10 1 ∣ ) + 3 ( ∣ 2 − 10 1 ∣ ) = 1 \left( \begin{vmatrix} -1 & -1 1 & 1 \end{vmatrix} \right) - (-2) \left( \begin{vmatrix} 2 & -1 0 & 1 \end{vmatrix} \right) + 3 \left( \begin{vmatrix} 2 & -1 0 & 1 \end{vmatrix} \right) = 1 ( − 1 − 11 1 ) − ( − 2 ) ( 2 − 10 1 ) + 3 ( 2 − 10 1 ) Evaluating the 2x2 determinants:
∣ − 1 − 11 1 ∣ = ( − 1 ) ( 1 ) − ( − 1 ) ( 1 ) = − 1 + 1 = 0 \begin{vmatrix} -1 & -1 1 & 1 \end{vmatrix} = (-1)(1) - (-1)(1) = -1 + 1 = 0 − 1 − 11 1 = ( − 1 ) ( 1 ) − ( − 1 ) ( 1 ) = − 1 + 1 = 0 ∣ 2 − 10 1 ∣ = 2 ( 1 ) − ( − 1 ) ( 0 ) = 2 \begin{vmatrix} 2 & -1 0 & 1 \end{vmatrix} = 2(1) - (-1)(0) = 2 2 − 10 1 = 2 ( 1 ) − ( − 1 ) ( 0 ) = 2 Thus, the volume is:
= 1 ( 0 ) + 2 ( 2 ) + 3 ( 2 ) = 0 + 4 + 6 = 10 cubic units = 1(0) + 2(2) + 3(2) = 0 + 4 + 6 = 10 \text{ cubic units} = 1 ( 0 ) + 2 ( 2 ) + 3 ( 2 ) = 0 + 4 + 6 = 10 cubic units
Scalar Triple Product for volume:
Volume = ∣ U ⋅ ( V × W ) ∣ \text{Volume} = | \mathbf{U} \cdot (\mathbf{V} \times \mathbf{W}) | Volume = ∣ U ⋅ ( V × W ) ∣
Determinant of a 3x3 matrix to compute the scalar triple product.
Summary of Steps
Write down the vectors for each part of the problem.
Set up the determinant using the components of the vectors.
Expand the determinant along the first row (or another row for convenience).
Compute the 2x2 determinants.
Combine the results to obtain the volume.
Ensure the final volume is in cubic units.
