JEE Challenger
More from Vector Algebra
MathHardAdv 2018 • Paper 2

Magnitude of Vector Cross Product Expression in a Cube

Take a unit cube located in the first octant where the edges OPOP, OQOQ, and OROR of unit length lie along the xx-axis, yy-axis, and zz-axis, respectively, with origin O(0,0,0)O(0,0,0). Let S(12,12,12)S\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) represent the center of the cube, and TT denote the cube's vertex opposite to OO such that SS is positioned on the main diagonal OTOT. If p⃗=SP⃗\vec{p} = \vec{SP}, q⃗=SQ⃗\vec{q} = \vec{SQ}, r⃗=SR⃗\vec{r} = \vec{SR}, and t⃗=ST⃗\vec{t} = \vec{ST}, determine the value of ∣(p⃗×q⃗)×(r⃗×t⃗)∣|(\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t})|.

Official Numerical Answer0.5

Topics & Concepts

Vector AlgebraDot and Cross ProductVector Addition

Step-by-Step Solution

To determine the value of ∣(p⃗×q⃗)×(r⃗×t⃗)∣|(\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t})|, we first identify the coordinates of the vertices and the center of the unit cube located in the first octant with origin O(0,0,0)O(0,0,0):

  • O=(0,0,0)O = (0, 0, 0)
  • P=(1,0,0)P = (1, 0, 0)
  • Q=(0,1,0)Q = (0, 1, 0)
  • R=(0,0,1)R = (0, 0, 1)
  • S=(12,12,12)S = \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) (Center of the cube)
  • T=(1,1,1)T = (1, 1, 1) (Vertex opposite to OO)

Now, we calculate the position vectors of p⃗\vec{p}, q⃗\vec{q}, r⃗\vec{r}, and t⃗\vec{t} relative to SS: p⃗=SP⃗=P−S=(1,0,0)−(12,12,12)=12i^−12j^−12k^\vec{p} = \vec{SP} = P - S = (1, 0, 0) - \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) = \frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k} q⃗=SQ⃗=Q−S=(0,1,0)−(12,12,12)=−12i^+12j^−12k^\vec{q} = \vec{SQ} = Q - S = (0, 1, 0) - \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) = -\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k} r⃗=SR⃗=R−S=(0,0,1)−(12,12,12)=−12i^−12j^+12k^\vec{r} = \vec{SR} = R - S = (0, 0, 1) - \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) = -\frac{1}{2}\hat{i} - \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k} t⃗=ST⃗=T−S=(1,1,1)−(12,12,12)=12i^+12j^+12k^\vec{t} = \vec{ST} = T - S = (1, 1, 1) - \left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\right) = \frac{1}{2}\hat{i} + \frac{1}{2}\hat{j} + \frac{1}{2}\hat{k}

Next, we evaluate the cross product p⃗×q⃗\vec{p} \times \vec{q}:

p⃗×q⃗=14∣i^j^k^1−1−1−11−1∣\vec{p} \times \vec{q} = \frac{1}{4} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -1 \\ -1 & 1 & -1 \end{vmatrix}

p⃗×q⃗=14[i^(1−(−1))−j^(−1−1)+k^(1−1)]=14(2i^+2j^)=12i^+12j^\vec{p} \times \vec{q} = \frac{1}{4} \left[ \hat{i}(1 - (-1)) - \hat{j}(-1 - 1) + \hat{k}(1 - 1) \right] = \frac{1}{4} (2\hat{i} + 2\hat{j}) = \frac{1}{2}\hat{i} + \frac{1}{2}\hat{j}

Next, we evaluate the cross product r⃗×t⃗\vec{r} \times \vec{t}:

r⃗×t⃗=14∣i^j^k^−1−11111∣\vec{r} \times \vec{t} = \frac{1}{4} \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & -1 & 1 \\ 1 & 1 & 1 \end{vmatrix}

r⃗×t⃗=14[i^(−1−1)−j^(−1−1)+k^(−1−(−1))]=14(−2i^+2j^)=−12i^+12j^\vec{r} \times \vec{t} = \frac{1}{4} \left[ \hat{i}(-1 - 1) - \hat{j}(-1 - 1) + \hat{k}(-1 - (-1)) \right] = \frac{1}{4} (-2\hat{i} + 2\hat{j}) = -\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j}

Now, we compute the vector cross product (p⃗×q⃗)×(r⃗×t⃗)(\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t}): (p⃗×q⃗)×(r⃗×t⃗)=(12i^+12j^)×(−12i^+12j^)(\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t}) = \left( \frac{1}{2}\hat{i} + \frac{1}{2}\hat{j} \right) \times \left( -\frac{1}{2}\hat{i} + \frac{1}{2}\hat{j} \right) =14(i^+j^)×(−i^+j^)= \frac{1}{4} (\hat{i} + \hat{j}) \times (-\hat{i} + \hat{j}) =14(i^×j^−j^×i^)=14(k^+k^)=12k^= \frac{1}{4} \left( \hat{i} \times \hat{j} - \hat{j} \times \hat{i} \right) = \frac{1}{4} (\hat{k} + \hat{k}) = \frac{1}{2}\hat{k}

Finally, taking the magnitude: ∣(p⃗×q⃗)×(r⃗×t⃗)∣=∣12k^∣=12=0.5|(\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t})| = \left| \frac{1}{2}\hat{k} \right| = \frac{1}{2} = 0.5

PreviousNext