Find the angle θ between the vectors {a}={i}-2{j}+3{k} and {b}=3{i}-2{j}+{k} ?

Correct Answer is : cos⁡−1(57)^{-1}≤ft(5/7)cos−1(75)

Correct Answer is : cos⁡−1(57)^{-1}≤ft(5/7)cos−1(75)

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UPSC NDA Math Model Paper 5

Question : 62 of 120

Marks: +1, -0

\vec{a}=\hat{i}-2\hat{j}+3\hat{k}

\vec{b}=3\hat{i}-2\hat{j}+\hat{k} ?

Solution:

Here, we have two vectors

\vec{a}=\hat{i}-2\hat{j}+3\hat{k}

and

\vec{b}=3\hat{i}-2\hat{j}+\hat{k} .

\Rightarrow |\vec{a}|=\sqrt{1^{2}+(-2)^{2}+3^{2}}=\sqrt{14}

and

|\vec{b}|=\sqrt{3^{2}+(-2)^{2}+(1)^{2}}=\sqrt{14}

\Rightarrow \vec{a}\cdot \vec{b}=(\hat{i}-2\hat{j}+3\hat{k})\cdot(3\hat{i}-2\hat{j}+\hat{k})

=3+4+3=10

By, substituting the values of

|\vec{a}|,|\vec{b}|

and

\vec{a}\cdot\vec{b}

\cos \theta = \frac{\overrightarrow{a \cdot \vec{b}}}{\overrightarrow{|a| \times |\vec{b}|}}

, we get

\Rightarrow \cos \theta = \frac{10}{\sqrt{14} \times \sqrt{14}} = \frac{5}{7}

\Rightarrow \theta = \cos^{-1}\left(\frac{5}{7}\right)

Hence, option

B

is the correct answer.

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