First slide

Cross product of vectors or vector product of vectors

Question

$Let \bar{a}, \bar{b} and \bar{c} be the unit vectors such that \bar{a} and \bar{b} are mutually perpendicular and \bar{c} is$ $equally inclined to \bar{a} and \bar{b} at angle θ . If \bar{c} = x \bar{a} + y \bar{b} + z (\bar{a} \times \bar{b}), then$

Moderate

A

$z^{2} = 1 - 2 x^{2}$
B

$z^{2} = 1 - x^{2} + y^{2}$
C

$z^{2} = 1 + 2 y^{2}$
D

$z^{2} = 1 + 2 x^{2}$

Solution

$\begin{array}{rcl} G i v e n t h a t |\bar{a}| & = & |\bar{b}| = |\bar{c}| = 1 a n d, \bar{a} . \bar{b} = 0 \Rightarrow (\bar{a}, \bar{b}) = 90^{\circ} \\ A s \bar{c} & = & x \bar{a} + y \bar{b} + z (\bar{a} \times \bar{b}) \\ \Rightarrow & |\bar{a}| |\bar{c}| \cos θ = x + y (0) + z (0) (b y t a k i n g d o t p r o d u c t w i t h a) \\ \Rightarrow & \cos θ = x \\ S i m i l a r l y, \cos θ & = & y \\ \bar{c} & = & x \bar{a} + y \bar{b} + z (\bar{a} \times \bar{b}) \\ Squaring on both sides \\ {|\bar{c}|}^{2} & = & x^{2} a^{- 2} + y^{2} b^{- 2} + z^{2} {(\bar{a} \times \bar{b})}^{2} + 2 x y (\bar{a} . \bar{b}) + 2 y z \bar{b} . (\bar{a} \times \bar{b}) + 2 x z \bar{a} . (\bar{a} \times \bar{b}) \\ 1 & = & x^{2} + y^{2} + z^{2} + 0 + 0 + 0 \\ \Rightarrow & 1 = \cos^{2} θ + \cos^{2} θ + z^{2} \\ \Rightarrow & 1 - 2 \cos^{2} θ = z^{2} \\ \Rightarrow & 1 - 2 x^{2} = z^{2} \\ ∴ z^{2} & = & 1 - 2 x^{2} \end{array}$

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