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((1)/(2), (1)/(2), (1)/(2)) be the centre of the

Correct answer: 0 → p = → SP = ((1)/(2), - (1)/(2), - (1)/(2)) = (1)/(2)( ^ i - ^ j - ^ k ) → q = → SQ = (- (1)/(2), (1)/(2), - (1)/(2)) = (1)/(2)(- ^ i + ^ j - ^ k ) → r = → SR = (- (1)/(2), - (1)/(2), (1)/(2)) = (1)/(2)(- ^ i - ^ j + ^ k ) → t = → ST = ((1)/(2), (1)/(2), (1)/(2)) = (1)/(2)( ^ i + ^ j + ^ k ) |( → p × → q ) × ( → r × → t )| = |(1)/(4)| cases ^ i & ^ j & ^ k 1 & - 1 & - 1 - 1 & 1 & - 1 cases | × (1)/(4)| cases ^ i & ^ j & ^ k - 1 & - 1 & 1 1 & 1 & 1 cases || = (1)/(16)|(2 ^ i + 2 ^ j ) × (- 2 ^ i + 2 ^ j )| = | ^ k 2 | = (1)/(2)

JEE Advanced 2018Paper 2

Exam: JEE Advanced
Year: 2018
Paper: Paper 2
Subject: Mathematics
Chapter: Vector Algebra

JEE Advanced 2018 · Paper 2 · Mathematics · Vector Algebra

Consider the cube in the first octant with sides OP, OQ and OR of...

Consider the cube in the first octant with sides

OP, OQ and OR of length

, along the x-axis, y-axis and z-axis, respectively, where

O ( 0, 0, 0 )

is the origin. Let

S\left(\dfrac{1}{2}, \dfrac{1}{2}, \dfrac{1}{2}\right)

be the centre of the cube and

be the vertex of the cube opposite to the origin

such that

lies on the diagonal

. If

\overset{\to}{p} = \overset{\to}{SP}, \overset{\to}{q} = \overset{\to}{SQ}, \overset{\to}{r} = \overset{\to}{SR}

and

\overset{\to}{t} = \overset{\to}{ST}

, then the value of \left|\left(\overset{\to}{p} \times \overset{\to}{q}\right) \times \left(\overset{\to}{r} \times \overset{\to}{t}\right)\right|

is ______. 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,

Correct answer

Step-by-step solution

\overset{\to}{p} = \overset{\to}{SP} = \left(\dfrac{1}{2}, - \dfrac{1}{2}, - \dfrac{1}{2}\right) = \dfrac{1}{2}\left(\overset{^}{i} - \overset{^}{j} - \overset{^}{k}\right)

\overset{\to}{q} = \overset{\to}{SQ} = \left(- \dfrac{1}{2}, \dfrac{1}{2}, - \dfrac{1}{2}\right) = \dfrac{1}{2}\left(- \overset{^}{i} + \overset{^}{j} - \overset{^}{k}\right)

\overset{\to}{r} = \overset{\to}{SR} = \left(- \dfrac{1}{2}, - \dfrac{1}{2}, \dfrac{1}{2}\right) = \dfrac{1}{2}\left(- \overset{^}{i} - \overset{^}{j} + \overset{^}{k}\right)

\overset{\to}{t} = \overset{\to}{ST} = \left(\dfrac{1}{2}, \dfrac{1}{2}, \dfrac{1}{2}\right) = \dfrac{1}{2}\left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right)

\left|\left(\overset{\to}{p} \times \overset{\to}{q}\right) \times \left(\overset{\to}{r} \times \overset{\to}{t}\right)\right| = \left|\dfrac{1}{4}\left|

\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ 1 & - 1 & - 1 \\ - 1 & 1 & - 1\end{cases}

\right| \times \dfrac{1}{4}\left|

\begin{cases}\overset{^}{i} & \overset{^}{j} & \overset{^}{k} \\ - 1 & - 1 & 1 \\ 1 & 1 & 1\end{cases}

\right|\right|

<br >

= \dfrac{1}{16}\left|\left(2\overset{^}{i} + 2\overset{^}{j}\right) \times \left(- 2\overset{^}{i} + 2\overset{^}{j}\right)\right| = \left|\dfrac{\overset{^}{k}}{2}\right| = \dfrac{1}{2}

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