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

Category: QuestionsFind the angle between the vectors $$\vec{a}=-\hat{i}+\hat{j}-\hat{k}$$ and $$\vec{b}=\hat{i}-\hat{j}$$
Editor">Editor Staff asked 11 months ago

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
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options
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Editor">Editor Staff answered 11 months ago

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

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

∴$$θ=cos^{-1}⁡-\sqrt{\frac{3}{2}}$$