Question

The scalar product of vectors
→
A
and
→
B
is +48.0m2. Vector
→
A
has magnitude 9.00 m and direction 28.0∘ west of south. If vector
→
B
has direction 39.0∘ south of east, what is the magnitude of
→
B
.?

Answer 1

Answer:

B = 27.95 m.

Explanation:

The scalar product between two vectors are

$\vec{A}.\vec{B} = |\vec{A}||\vec{B}|\cos(\theta)$

where Θ is the angle between the vectors.

We need to find the angle between the vectors by using the information given in the question.

If A is directed 28° west of south and if B is directed 39° south of east, then the angle between them is 79°.

Let's plug the variables into the scalar product:

$48 = (9)|\vec{B}|\cos(79)\\|\vec{B}| = 27.95~m$

Answer 2

The magnitude of vector B if the scalar product of vectors is 48m² will be 28.05m

The formula for finding the dot product of two vectors A and B is expressed as:

$\vec A\cdot\vec B=\mid A\mid\mid B\mid cos \theta$

$\theta$ is the angle between vectors A and B

• If  vector  A  has magnitude 9.00 m and direction 28.0 degrees west of south and vector  B has direction 39.0 degrees south of east, the resulting angle between them will be 79 degrees

Given the following parameters

$\vec A \cdot \vec B = +48.0m^2$

$\mid A \mid=9.0m$

Substitute the parameters into the formula to have:

$48.0 =9 \mid \vec B \mid cos 79^0\\\frac{48}{9} = \mid \vec B \mid cos 79^0\\5.33=0.190 \mid \vec B \mid\\\mid \vec B \mid=\frac{5.33}{0.19}\\ \mid \vec B \mid=28.05m$

Hence the magnitude of vector B if the scalar product of vectors is 48m² will be 28.05m

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