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MathMedAdv 2013 • Paper 1

Volume of Parallelepiped from Diagonals of Parallelogram Base

Let PR⃗=3i^+j^−2k^\vec{PR} = 3\hat{i} + \hat{j} - 2\hat{k} and SQ⃗=i^−3j^−4k^\vec{SQ} = \hat{i} - 3\hat{j} - 4\hat{k} represent the diagonals of a parallelogram PQRSPQRS, and let PT⃗=i^+2j^+3k^\vec{PT} = \hat{i} + 2\hat{j} + 3\hat{k} be another vector. Then the volume of the parallelepiped formed by the vectors PT⃗\vec{PT}, PQ⃗\vec{PQ}, and PS⃗\vec{PS} is equal to:

Options

A

5

B

20

C

10

Correct
D

30

Topics & Concepts

Vector AlgebraScalar Triple ProductVector Addition

Step-by-Step Solution

To find the volume of the parallelepiped formed by the coterminous vectors PT⃗\vec{PT}, PQ⃗\vec{PQ}, and PS⃗\vec{PS}, we use the scalar triple product: V=∣[PT⃗,PQ⃗,PS⃗]∣=∣PT⃗⋅(PQ⃗×PS⃗)∣V = \left| \left[ \vec{PT}, \vec{PQ}, \vec{PS} \right] \right| = \left| \vec{PT} \cdot (\vec{PQ} \times \vec{PS}) \right|

Let the position vectors of the vertices of the parallelogram PQRSPQRS be defined with PP as the origin. Then: PQ⃗=q⃗\vec{PQ} = \vec{q} PS⃗=s⃗\vec{PS} = \vec{s}

Since PQRSPQRS is a parallelogram, the diagonals can be expressed in terms of the adjacent sides: PR⃗=PQ⃗+PS⃗=q⃗+s⃗\vec{PR} = \vec{PQ} + \vec{PS} = \vec{q} + \vec{s} SQ⃗=PQ⃗−PS⃗=q⃗−s⃗\vec{SQ} = \vec{PQ} - \vec{PS} = \vec{q} - \vec{s}

Now, let us compute the cross product of the diagonal vectors: PR⃗×SQ⃗=(q⃗+s⃗)×(q⃗−s⃗)\vec{PR} \times \vec{SQ} = (\vec{q} + \vec{s}) \times (\vec{q} - \vec{s}) PR⃗×SQ⃗=(q⃗×q⃗)−(q⃗×s⃗)+(s⃗×q⃗)−(s⃗×s⃗)\vec{PR} \times \vec{SQ} = (\vec{q} \times \vec{q}) - (\vec{q} \times \vec{s}) + (\vec{s} \times \vec{q}) - (\vec{s} \times \vec{s})

Using the anti-commutative property of the cross product (s⃗×q⃗=−q⃗×s⃗\vec{s} \times \vec{q} = -\vec{q} \times \vec{s}) and v⃗×v⃗=0⃗\vec{v} \times \vec{v} = \vec{0}: PR⃗×SQ⃗=−q⃗×s⃗−q⃗×s⃗=−2(q⃗×s⃗)=−2(PQ⃗×PS⃗)\vec{PR} \times \vec{SQ} = -\vec{q} \times \vec{s} - \vec{q} \times \vec{s} = -2(\vec{q} \times \vec{s}) = -2(\vec{PQ} \times \vec{PS})

Thus: PQ⃗×PS⃗=−12(PR⃗×SQ⃗)\vec{PQ} \times \vec{PS} = -\frac{1}{2} (\vec{PR} \times \vec{SQ})

Substituting this back into the volume formula gives: V=∣PT⃗⋅(−12(PR⃗×SQ⃗))∣=12∣[PT⃗,PR⃗,SQ⃗]∣V = \left| \vec{PT} \cdot \left( -\frac{1}{2} (\vec{PR} \times \vec{SQ}) \right) \right| = \frac{1}{2} \left| \left[ \vec{PT}, \vec{PR}, \vec{SQ} \right] \right|

Given the vectors: PT⃗=i^+2j^+3k^\vec{PT} = \hat{i} + 2\hat{j} + 3\hat{k} PR⃗=3i^+j^−2k^\vec{PR} = 3\hat{i} + \hat{j} - 2\hat{k} SQ⃗=i^−3j^−4k^\vec{SQ} = \hat{i} - 3\hat{j} - 4\hat{k}

We evaluate the determinant [PT⃗,PR⃗,SQ⃗]\left[ \vec{PT}, \vec{PR}, \vec{SQ} \right]:

[PT⃗,PR⃗,SQ⃗]=∣12331−21−3−4∣\left[ \vec{PT}, \vec{PR}, \vec{SQ} \right] = \begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & -2 \\ 1 & -3 & -4 \end{vmatrix}

Expanding along the first row:

[PT⃗,PR⃗,SQ⃗]=1⋅(1(−4)−(−2)(−3))−2⋅(3(−4)−(−2)(1))+3⋅(3(−3)−1(1))=1⋅(−4−6)−2⋅(−12+2)+3⋅(−9−1)=1(−10)−2(−10)+3(−10)=−10+20−30=−20\begin{aligned} \left[ \vec{PT}, \vec{PR}, \vec{SQ} \right] &= 1 \cdot (1(-4) - (-2)(-3)) - 2 \cdot (3(-4) - (-2)(1)) + 3 \cdot (3(-3) - 1(1)) \\ &= 1 \cdot (-4 - 6) - 2 \cdot (-12 + 2) + 3 \cdot (-9 - 1) \\ &= 1(-10) - 2(-10) + 3(-10) \\ &= -10 + 20 - 30 \\ &= -20 \end{aligned}

Hence, the volume of the parallelepiped is: V=12∣−20∣=10V = \frac{1}{2} |-20| = 10

Therefore, the correct option is C.

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