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 O0- 0- 0 is the origin- Let S 12- 12- 12 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 p- - S-P- q- - S-Q- r- - S-R- and t- - S-T- then the value of 2-p-q- r-t- is -

p⃗ = S⃗P⃗ = (12, 12, 12 ) = 12 (î ĵ k̂) q⃗ = S⃗Q⃗ = ( 12, 12, 12 ) = 12 ( î + ĵ k̂) r⃗ = S⃗R⃗ = ( 12, 12, 12 ) = 12 ( î ĵ + k̂) ...

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Byju's Answer

Standard XII

Mathematics

Volume of Parallelopiped

Consider the ...

Question

Consider the cube in the first octant with sides $O P, O Q$ and $O R$ 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 $O T$ . If $\to p = \to S P, \to q = \to S Q, \to r = \to S R$ and $\to t = \to S T$ , then the value of $2 | (\to p \times \to q) \times (\to r \times \to t) |$ is ________.

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Solution

$\to p = \to S P = (\frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}) = \frac{1}{2} (^i -^j -^k)$

$\to q = \to S Q = (- \frac{1}{2}, \frac{1}{2}, - \frac{1}{2}) = \frac{1}{2} (-^i +^j -^k)$

$\to r = \to S R = (- \frac{1}{2}, - \frac{1}{2}, \frac{1}{2}) = \frac{1}{2} (-^i -^j +^k)$

$\to t = \to S T = (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}) = \frac{1}{2} (^i +^j +^k)$

$\Rightarrow | (\to p \times \to q) \times (\to r \times \to t) = \frac{1}{4} ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 1 & - 1 & 1 - 1 & 1 & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ \times \frac{1}{4} ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k - 1 & - 1 & 1 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$

$= \frac{1}{16} | (2^i + 2^j) \times (- 2^i + 2^j) | = ∣ ∣ ∣ \frac{^k}{2} ∣ ∣ ∣ = \frac{1}{2} = 0.50$ .
$∴ 2 | (\to p \times \to q) \times (\to r \times \to t) | = 1$

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