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(a) Find the scalar product of the vectors \(a \hat{\imath}+b \hat{\jmath}\) and \(b \hat{\imath}-a \hat{\jmath}\) where \(a\) and \(b\) are arbitrary constants. (b) What's the angle between the two vectors?

Short Answer

Expert verified
The scalar product of the vectors is 0 and the angle between them is 90° or \(\frac{\pi}{2}\) radians.

Step by step solution

01

Compute the dot product

Start with the formula for the dot product: \(A \cdot B = a_1b_1 + a_2b_2\). Here, \(a_1 = a, a_2 = b, b_1 = b, b_2 = -a\). So, replacing these values in the scalar product equation gives us: \(A \cdot B = a*b + b*(-a)\). Simplifying this expression we get \(A \cdot B = ab - ab = 0\).
02

Calculate the magnitudes of the vectors

The magnitude or length of a vector \(A = a_1 \hat{\imath} + a_2 \hat{\jmath}\) is given by \(\|A\| = \sqrt{a_1^2 + a_2^2}\). Using this, calculate the magnitude of vectors A and B. For Vector A: \(\|A\| = \sqrt{a^2 + b^2}\) and Vector B: \(\|B\| = \sqrt{b^2 + (-a)^2} = \sqrt{a^2 + b^2}\)
03

Finding the angle

We are now ready to find the angle \( \theta \) between the vectors. Use the formula \(\cos(\theta) = \frac{A \cdot B}{\|A\| \|B\|}\). Substituting the obtained values we get: \(\cos(\theta) = \frac{0}{(\sqrt{a^2 + b^2})(\sqrt{a^2 + b^2})} = 0\). The arccosine (inverse cosine) of zero is \(90^{\circ}\) or \(\frac{\pi}{2}\) radians in standard units hence that's the angle between the vectors.

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