Question

Magnitudes of \(\mathrm{A}^{\rightarrow}, \mathrm{B}^{\rightarrow}\) and \(\mathrm{C}^{\rightarrow}\) are 41,40 and 9 respectively, \(\mathrm{A}^{\rightarrow}=\mathrm{B}^{\rightarrow}+\mathrm{C}^{\rightarrow}\) Find the angle between \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) (A) \(\sin ^{-1}(9 / 40)\) (B) \(\cos ^{-1}(9 / 41)\) (C) \(\tan ^{-1}(9 / 41)\) (D) \(\tan ^{-1}(41 / 40)\)

Short answer

The angle between \(\mathrm{A}^{\rightarrow}\) and \(\mathrm{B}^{\rightarrow}\) is found to be \(90^{\circ}\), which corresponds to option (A): \(\sin ^{-1}(\frac{9}{40})\).

Step-by-step solution

Rewrite the given equation using dot product

Since A = B + C, we can write this equation in terms of the dot product: A·A = (B + C)·(B + C)

Expand the dot product equation

Expanding the equation from Step 1, we have: A·A = B·B + 2(B·C) + C·C

Substitute magnitudes of vectors and rewrite the relation

Now, we substitute the magnitudes of the vectors and use the property that the dot product of a vector with itself equals the square of its magnitude: \(41^2 = 40^2 + 2(B \cdot C) + 9^2\)

Simplify the equation

Simplify the equation from Step 3: \(1681 = 1600 + 2(B \cdot C) + 81\)

Find B·C

Solve the simplified equation for B·C: \(B \cdot C = \frac{1681 - 1600 - 81}{2} = 0\)

Find the cosine of the angle between A and B

Since A = B + C, the angle between A and B equals the angle between B and C. We can find the cosine of this angle using the formula for the dot product: \(\cos \theta = \frac{B \cdot C}{|B| |C|} = \frac{0}{40 \cdot 9}\)

Calculate the angle between A and B

Since the cosine of the angle between A and B equals 0, this means the angle between the vectors A and B is 90 degrees, or \(\frac{\pi}{2}\) radians. This corresponds to option (A) in the given choices: \(\sin ^{-1}(\frac{9}{40})\), as the sine of a 90-degree angle is equal to 1, and the ratio of the magnitudes of vectors C and B is \(\frac{9}{40}\). This is the final answer.

Key concepts

Dot Product

The concept of dot product is crucial in vector mathematics. It is a method used to find a scalar from two vectors and signifies a form of multiplication that gives insights into the angle between vectors.

To compute the dot product of two vectors, consider vectors \( \vec{A} \) and \( \vec{B} \). The dot product is calculated using the formula:

• \( \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \)

where \( |\vec{A}| \) and \( |\vec{B}| \) are the magnitudes of vectors \( \vec{A} \) and \( \vec{B} \), and \( \theta \) is the angle between them.

This formula not only provides a relationship between vectors but also helps to understand their alignment. A dot product of zero indicates that two vectors are perpendicular.

Vector Magnitude

Vector magnitude represents the size or length of a vector, which can be understood similar to the concept of absolute value in scalars. Being aware of a vector's magnitude is essential because it provides insights into the vector’s impact or scale.

For a vector \( \vec{A} = (x, y, z) \), its magnitude is determined by:

• \( |\vec{A}| = \sqrt{x^2 + y^2 + z^2} \)

The computation utilizes the Pythagorean theorem, extended into a three-dimensional space. It reflects the true 'distance' from the origin to the point defined by the vector. Recognizing the value of a vector's magnitude is fundamental, especially when simplifying and solving vector equations.

Angle Between Vectors

Understanding the angle between vectors is key when analyzing their spatial relationship. The angle formula connects dots products to vector magnitudes and helps determine orientation.

To find the angle between vectors \( \vec{A} \) and \( \vec{B} \), use the relationship:

• \( \cos \theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}| |\vec{B}|} \)

This equation shows how to derive the angle \( \theta \) using a known dot product and the magnitudes of the vectors involved. In the scenario provided, the angle calculation revealed that vectors \( \vec{A} \) and \( \vec{B} \) are perpendicular, as their dot product equals zero, making their cosine value zero as well. This leads to an angle of \( 90^\circ \) or \( \frac{\pi}{2} \) radians—understanding this aids in solving problems where vector alignment and direction are central.