Question

Find the angle between the vectors \(\vec{a}=-\hat{i}+\hat{j}-\hat{k}\) and \(\vec{b}=\hat{i}-\hat{j}\)

(a) \(cos^{-1}⁡-\frac{\sqrt{3}}{2}\)

(b) \(cos^{-1}⁡-\frac{2}{\sqrt{3}}\)

(c) \(cos^{-1}⁡-\sqrt{2}\)

(d) \(cos^{-1}⁡-\sqrt{\frac{3}{2}}\)

I have been asked this question in final exam.

My query is from Product of Two Vectors-1 in chapter Vector Algebra of Mathematics – Class 12

Answer 1

The correct option is (d) \(cos^{-1}⁡-\sqrt{\frac{3}{2}}\)

To explain: The angle between two vectors is given by

\(cos⁡θ=\frac{|\vec{a}|.|\vec{b}|}{\vec{a}.\vec{b}}\)

\(|\vec{a}|=\sqrt{(-1)^2+(1)^2+(-1)^2}=\sqrt{3}\)

\(|\vec{b}|=\sqrt{(1)^2+(-1)^2}=\sqrt{2}\)

\(\vec{a}.\vec{b}\)=(-1)(1)+1(-1)+0=-2

\(cos⁡θ=\frac{\sqrt{3}.\sqrt{2}}{-2}=-\sqrt{\frac{3}{2}}\)

∴\(θ=cos^{-1}⁡-\sqrt{\frac{3}{2}}\)