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Question: The sine of the angle between the vectors \(\mathbf { a } = 3 \mathbf { i } + \mathbf { j } + \mathb...

The sine of the angle between the vectors a=3i+j+k,b=2i2j+k\mathbf { a } = 3 \mathbf { i } + \mathbf { j } + \mathbf { k } , \mathbf { b } = 2 \mathbf { i } - 2 \mathbf { j } + \mathbf { k } is

A

7499\sqrt { \frac { 74 } { 99 } }

B

2599\sqrt { \frac { 25 } { 99 } }

C

3799\sqrt { \frac { 37 } { 99 } }

D

541\frac { 5 } { \sqrt { 41 } }

Answer

7499\sqrt { \frac { 74 } { 99 } }

Explanation

Solution

a×b=ijk311221=3ij8k\mathbf { a } \times \mathbf { b } = \left| \begin{array} { c c c } \mathbf { i } & \mathbf { j } & \mathbf { k } \\ 3 & 1 & 1 \\ 2 & - 2 & 1 \end{array} \right| = 3 \mathbf { i } - \mathbf { j } - 8 \mathbf { k }; sinθ=a×bab=74119=7499\sin \theta = \frac { | \mathbf { a } \times \mathbf { b } | } { | \mathbf { a } | | \mathbf { b } | } = \frac { \sqrt { 74 } } { \sqrt { 11 } \cdot \sqrt { 9 } } = \sqrt { \frac { 74 } { 99 } }