Question

Two vectors $\alpha$with Let the magnitude3s of the vectors be represented by A, B and R respectively. Which of the following relations is not correct?

• A.
• B.
• C.
• D. $R \sin \beta = A \sin ( \alpha + \beta )$

Answer 1

Answer: (A)

Answer 2

Let OP and OQ represent two vectors $\overrightarrow { \mathrm { A } }$ and $( \alpha + \beta )$.

Using the parallelogram method of vector additions, Resultant vector,

SN is normal to OP and PM is normal to OS.

From the geometry of the figure.

$\mathrm { OS } ^ { 2 } = \mathrm { ON } ^ { 2 } + \mathrm { SN } ^ { 2 } = ( \mathrm { OP } + \mathrm { PN } ) ^ { 2 } + \mathrm { SN } ^ { 2 }$ $= ( A + B \cos ( \alpha + \beta ) ) ^ { 2 } + ( B \sin ( \alpha + \beta ) ) ^ { 2 }$ $\mathrm { R } ^ { 2 } = \mathrm { A } ^ { 2 } + \mathrm { B } ^ { 2 } + 2 \mathrm { AB } \cos ( \alpha + \beta )$

In

And in $\Delta \mathrm { PSN } , \mathrm { SN } = \mathrm { PS } \sin ( \alpha + \beta ) = \mathrm { B } \sin ( \alpha + \beta )$

$\therefore$

Or $\frac { \mathrm { R } } { \sin ( \alpha + \beta ) } = \frac { \mathrm { B } } { \sin \alpha }$ ….. (i)

Similarly

PM =

$\frac { A } { \sin \beta } = \frac { B } { \sin \alpha }$ …..(ii)

Combining (i) and (ii) we get

$\frac { \mathrm { R } } { \sin ( \alpha + \beta ) } = \frac { \mathrm { A } } { \sin \beta } = \frac { \mathrm { B } } { \sin \alpha }$ …(iii)

From (iii)

$R \sin \beta = A \sin ( \alpha + \beta )$