Consider the cube in the first octant with sides OP, OQ and OR of length 1, along the x-axis, y-axis and z-axis

Question

Consider the cube in the first octant with sides OP, OQ and OR of length 1, along the x-axis, y-axis and z-axis, respectively, where O(0,0,0) is the origin. Let $S (\frac{1}{2}, \frac{1}{2}, \frac{1}{2})$ be the centre of the cube and T be the vertex of the cube opposite to the origin O such that S lies on the diagonal OT. If $\vec{p} = \vec{S P}, \vec{q} = \vec{S Q}, \vec{r} = \vec{S R}$ and $\vec{t} = \vec{S T}$ , then the value of $2$ $| (\vec{p} \times \vec{q}) \times (\vec{r} \times \vec{t}) |$ is _______

Smart Achievers · Accepted Answer

(1)

$Point S (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}), Point T (1, 1, 1)$

$\vec{p} = \vec{S P} = \frac{\hat{i} - \hat{j} - \hat{k}}{2}, \vec{q} = \vec{S Q} = \frac{- \hat{i} + \hat{j} - \hat{k}}{2}, \vec{r} = \vec{S R} = \frac{- \hat{i} - \hat{j} + \hat{k}}{2}, \vec{t} = \vec{S T} = \frac{\hat{i} + \hat{j} + \hat{k}}{2}$

$Now,$ $\vec{p} \times \vec{q} = \frac{\hat{i} + \hat{j}}{2}$ $\vec{r} \times \vec{t} = \frac{- \hat{i} + \hat{j}}{2}$

$Now,$ $(\vec{p} \times \vec{q}) \cdot (\vec{r} \times \vec{t}) = \frac{\hat{k}}{2}$

Solution

Want Us to Email you About Special Offers & Updates?

Solution

Ans. (1) Point S(12,12,12), Point T(1,1,1) p→=SP→=i^-j^-k^2, q→=SQ→=-i^+j^-k^2, r→=SR→=-i^-j^+k^2, t→=ST→=i^+j^+k^2 Now, p→×q→=i^+j^2 r→×t→=-i^+j^2 Now, (p→×q→)·(r→×t→)=k^2

Want Us to Email you About Special Offers & Updates?