Question

Let $\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{5}\mathbf{}\mathbf{m}$at ${\mathbf{60}}^{\mathbf{\circ }}$. Let the vector $\overrightarrow{\mathbf{C}}$have the same magnitude as $\overrightarrow{\mathbf{A}}$and a direction angle greater than that of $\overrightarrow{\mathbf{A}}$by ${\mathbf{25}}^{\mathbf{\circ }}$. Let $\overrightarrow{\mathbf{A}}\mathbf{·}\overrightarrow{\mathbf{B}}\mathbf{=}\mathbf{30}\mathbf{}{\mathbf{m}}^{\mathbf{2}}$and $\overrightarrow{\mathbf{B}}\mathbf{·}\overrightarrow{\mathbf{C}}\mathbf{=}\mathbf{35}\mathbf{}{\mathbf{m}}^{\mathbf{2}}$. Find the magnitude and direction of $\overrightarrow{\mathbf{A}}$ .

Short answer

The magnitude of vector $\overrightarrow{\mathrm{A}}$ is $7.05$ and its direction angle is $\mathbf{28}\mathbf{.}{\mathbf{37}}^{\mathbf{\circ }}$.

Step-by-step solution

Given data

The magnitude of vector $\overrightarrow{B}$

$\left|\overrightarrow{B}\right|=5\mathrm{m}$

Direction angle of vector $\overrightarrow{B}$

${\theta }_B={60}^{\circ }$

Difference between directions angles of vectors $\overrightarrow{A}$and $\overrightarrow{C}$

${\theta }_c-{\theta }_A={25}^{\circ }$

Dot product of vectors $\overrightarrow{A}$and $\overrightarrow{B}$

$\overrightarrow{A}·\overrightarrow{B}=30{\mathrm{m}}^2$

Dot product of vectors $\overrightarrow{B}$and $\overrightarrow{C}$

$\overrightarrow{B}·\overrightarrow{C}=35{\mathrm{m}}^2$

Magnitudes of vectors $\overrightarrow{A}$and $\overrightarrow{C}$are equal, that is

$\left|\overrightarrow{A}\right|=\left|\overrightarrow{C}\right|$

Dot product of two vectors

The angle between two vectors $\overrightarrow{A}$and $\overrightarrow{B}$of magnitudes $\left|\overrightarrow{A}\right|,\left|\overrightarrow{B}\right|$and direction angles ${\theta }_A,{\theta }_B$is

$\overrightarrow{\mathbf{A}}\mathbf{·}\overrightarrow{\mathbf{B}}\mathbf{=}\left|\overrightarrow{A}\right|\left|\overrightarrow{B}\right|\mathbf{cos}\left({\theta }_A-{\theta }_B\right).....\left(\mathrm{I}\right)$

Determining the magnitude and direction angle of  A→

From equation (I), the dot product of vectors $\overrightarrow{A}$and $\overrightarrow{B}$is written as

......(II)

From equation (I), the dot product of vectors $\overrightarrow{C}$and $\overrightarrow{B}$is written as

........(III)

Divide equation (II) by equation (III) to get

Divide both sides by localid="1668174912784" $\cos {\theta }_A$to get

Solve the above equation to get

Substitute this in equation (II) to get

Thus the required magnitude is $7.05\mathrm{m}$and the direction angle is $28.{37}^{\circ }$.