Let → x , → y and → z be three vectors each of magnitude 2 and the angle between each pair of them is (π)/(3) . If → a is a non - zero vector perpendicular to → x and → y × → z and → b is a non - zero vector

Correct answer: A. → b = ( → b . → z )( → z - → x ) → a is in direction of → x × ( → y × → z ) i.e., ( → x . → z ) → y - ( → x . → y ) → z ⇒ → a = λ _ 1 [2 × (1)/(2)( → y - → z )] → a = λ _ 1 ( → y - → z ) ........(i) Now → a . → y = λ _ 1 ( → y . → y - → y . → z ) = λ _ 1 (2 - 1) ⇒ λ _ 1 = → a . → y ........(ii) From (i) and (ii), → a = → a . → y ( → y - → z ) Similarly, → b = ( → b . → z )( → z - → x ) Now, → a . → b = ( → a . → y )( → b . → z )[( → y - → z ) . ( → z - → x )] = ( → a . → y )( → b . → z )[1 - 1 - 2 + 1] = - ( → a . → y )( → b . → z )

JEE Advanced 2014Paper 1

Exam: JEE Advanced
Year: 2014
Paper: Paper 1
Subject: Mathematics
Chapter: Vector Algebra

JEE Advanced 2014 · Paper 1 · Mathematics · Vector Algebra

Let → x , → y and → z be three vectors each of magnitude 2 and the...

Let

\overset{\to}{x}, \overset{\to}{y}

and

\overset{\to}{z}

be three vectors each of magnitude

\sqrt{2}

and the angle between each pair of them is

\dfrac{\pi}{3} .

\overset{\to}{a}

is a non - zero vector perpendicular to

\overset{\to}{x}

and

\overset{\to}{y} \times \overset{\to}{z}

and

\overset{\to}{b}

is a non - zero vector perpendicular to

\overset{\to}{y}

and

\overset{\to}{z} \times \overset{\to}{x}

, then

Options

A $\overset{\to}{b} = \left(\overset{\to}{b} . \overset{\to}{z}\right)\left(\overset{\to}{z} - \overset{\to}{x}\right)$
B $\overset{\to}{a} = \left(\overset{\to}{a} . \overset{\to}{y}\right)\left(\overset{\to}{y} - \overset{\to}{z}\right)$
C $\overset{\to}{a} . \overset{\to}{b} = - \left(\overset{\to}{a} . \overset{\to}{y}\right)\left(\overset{\to}{b} . \overset{\to}{z}\right)$
D $\overset{\to}{a} = \left(\overset{\to}{a} . \overset{\to}{y}\right)\left(\overset{\to}{z} - \overset{\to}{y}\right)$

Correct answer

A. $\overset{\to}{b} = \left(\overset{\to}{b} . \overset{\to}{z}\right)\left(\overset{\to}{z} - \overset{\to}{x}\right)$

Step-by-step solution

\overset{\to}{a}

is in direction of

\overset{\to}{x} \times \left(\overset{\to}{y} \times \overset{\to}{z}\right)

<br >i.e., \left(\overset{\to}{x} . \overset{\to}{z}\right)\overset{\to}{y} - \left(\overset{\to}{x} . \overset{\to}{y}\right)\overset{\to}{z}

⇒

\overset{\to}{a} = {\lambda}_{1}\left[2 \times \dfrac{1}{2}\left(\overset{\to}{y} - \overset{\to}{z}\right)\right]

\overset{\to}{a} = {\lambda}_{1}\left(\overset{\to}{y} - \overset{\to}{z}\right)

........(i)
Now

\overset{\to}{a} . \overset{\to}{y} = {\lambda}_{1}\left(\overset{\to}{y} . \overset{\to}{y} - \overset{\to}{y} . \overset{\to}{z}\right)

= {\lambda}_{1}\left(2 - 1\right) ⇒ {\lambda}_{1} = \overset{\to}{a} . \overset{\to}{y}

........(ii)
From (i) and (ii),

\overset{\to}{a} = \overset{\to}{a} . \overset{\to}{y}\left(\overset{\to}{y} - \overset{\to}{z}\right)

Similarly,

\overset{\to}{b} = \left(\overset{\to}{b} . \overset{\to}{z}\right)\left(\overset{\to}{z} - \overset{\to}{x}\right)

Now,

\overset{\to}{a} . \overset{\to}{b} = \left(\overset{\to}{a} . \overset{\to}{y}\right)\left(\overset{\to}{b} . \overset{\to}{z}\right)\left[\left(\overset{\to}{y} - \overset{\to}{z}\right) . \left(\overset{\to}{z} - \overset{\to}{x}\right)\right]

= \left(\overset{\to}{a} . \overset{\to}{y}\right)\left(\overset{\to}{b} . \overset{\to}{z}\right)\left[1 - 1 - 2 + 1\right]

= - \left(\overset{\to}{a} . \overset{\to}{y}\right)\left(\overset{\to}{b} . \overset{\to}{z}\right)

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