Question

The angle $\theta$ between the vectors $\overrightarrow{a}=5\hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{b}=\hat{i}+\hat{j}-\hat{k}$ is:

• (A) ${\cos }^{-1}\left(\frac{1}{3}\right)$
• (B) ${\cos }^{-1}\left(\frac{2}{3}\right)$
• (C) ${\cos }^{-1}\left(\frac{4}{7}\right)$
• (D) None of these

Answer 1

The Correct Option is A

Explanation:
We have two vectors $\overrightarrow{a}=5\hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{b}=\hat{i}+\hat{j}-\hat{k}$We know that the angle $\theta$ between two vectors $\overrightarrow{a}$ and $\overrightarrow{a}$ is:$\theta ={\cos }^{-1}\frac{\overrightarrow{a}\cdot \overrightarrow{b}}{|\overrightarrow{a}|\times |b|}$ ----(1)Therefore,$\overrightarrow{a}\cdot \overrightarrow{b}=(5\hat{i}-\hat{j}+\hat{k})\cdot (\hat{i}+\hat{j}-\hat{k})$$\Rightarrow \overrightarrow{a}\cdot \overrightarrow{b}=5{\hat{i}}^2+5\hat{i}\hat{j}-5\hat{i}\hat{k}-\hat{i}\hat{j}-{\hat{j}}^2+\hat{j}\hat{k}+\hat{i}\hat{k}+\hat{j}\hat{k}-{\hat{k}}^2$We know that, $({\hat{i}}^2={\hat{j}}^2={\hat{k}}^2=1)$ and $(\hat{i}\hat{j}=\hat{j}\hat{k}=\hat{i}\hat{k}=0)$$\therefore \overrightarrow{a}\cdot \overrightarrow{b}=5-1-1=3$We also know that, if $\overrightarrow{a}=(x\hat{i}+y\hat{j}+z\hat{k})$ then,$|\overrightarrow{a}|=\sqrt{x^2+y^2+z^2}$Therefore, according to the question,$\Rightarrow |\overrightarrow{a}|=\sqrt{5^2+(-1{)}^2+1^2}=\sqrt{27}$$\Rightarrow |\overrightarrow{b}|=\sqrt{1^2+1^2+(-1{)}^2}=\sqrt{3}$Putting all the above values in equation (1),$\Rightarrow \theta ={\cos }^{-1}\frac{3}{\sqrt{27}\times \sqrt{3}}$$\Rightarrow \theta ={\cos }^{-1}\frac{1}{3}$Hence, the correct option is (A).

Answer 2

Vectors are the physical quantities that have both magnitude and direction and also they obey triangle law or parallelogram laws of vector addition.

There are two ways of multiplying vectors

• Scalar product
• Vector product

Scalar Product

The scalar product or Dot product of any two vectors A and B denoted as A.B and given by the product of the magnitude of both the vectors multiplied by the cosine of the angle between them.

A.B=AB cos

Vector Product

The vector product or Cross product of any two vectors A and B denoted as AB and given by the product of the magnitude of both the vectors multiplied by the sine of the angle between them.

AB=AB sin