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Question: vectors <img src="https://cdn.pureessence.tech/canvas_66.png?top_left_x=0&top_left_y=1143&width=300&...

vectors and with magnitude 2, 3 & 4 respectively are coplanar. A unit vector is perpendicular to all of them. If (a⃗×b⃗)×(c⃗×d⃗)( \vec { a } \times \vec { b } ) \times ( \vec { c } \times \vec { d } )
=i^6−j^3+k^3= \frac { \hat { \mathrm { i } } } { 6 } - \frac { \hat { \mathrm { j } } } { 3 } + \frac { \hat { \mathrm { k } } } { 3 } and the angle between a⃗\vec { a } and b→\overrightarrow { \mathrm { b } } is 300, then is equal to

A

53\frac { 5 } { 3 }

B

59\frac { 5 } { 9 }

C

512\frac { 5 } { 12 }

D

518\frac { 5 } { 18 }

Answer

518\frac { 5 } { 18 }

Explanation

Solution

= ∣a→∣| \overrightarrow { \mathrm { a } } | ∣b→∣| \overrightarrow { \mathrm { b } } | sin 300 n^\hat { n } =

since is perpendicular to a⃗&b⃗\vec { a } \& \vec { b } Ž = ±

\ = ± 3 Now (a⃗×b⃗)×(c⃗×d⃗)( \vec { a } \times \vec { b } ) \times ( \vec { c } \times \vec { d } )= ±

Ž ± 3

Or ±

Ž , ,