Let a and b be two non-zero vectors perpendicular to each other and ≤ft|a|=≤ft|b|. If ≤ft|a × b|=≤ft|a|, then the angle between the vectors [a+b+(a × b)] and a is equal to [18 Mar 2021 Shift 2]

Correct Answer is : cos⁡−1(13)^{-1}≤ft( {1}{{3}} )cos−1(31)

Correct Answer is : cos⁡−1(13)^{-1}≤ft( {1}{{3}} )cos−1(31)

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Vector Algebra Part 1

Section: Mathematics

Question : 5 of 100

Marks: +1, -0

Let

a

b

\left|a\right|=\left|b\right|

\left|a \times b\right|=\left|a\right|

[a+b+(a \times b)]

a

Solution:

Given,

a \perp b

...(i)

\left|a\right|=\left|b\right|

...(ii)

and

\left|a \times b\right|=\left|a\right|

\Rightarrow \;\; \left|a\right|\left|b\right| \sin 90^{\circ}=\left|a\right|

[from Eq. (i)]

\Rightarrow \;\; \left|b\right|=1=\left|a\right|

...(iii) [from Eq. (ii)]

From Eq. (iii), we can say that

a \times b

are mutually perpendicular unit vectors.

Let

\;\; a=\hat{i}

and

b=\hat{j}

\therefore \;\; a \times b = \hat{k}

Now,

\;\; [a+b+(a \times b)]=(\hat{i}+\hat{j}+\hat{k})

\therefore \;\; \cos \theta = \frac{ (\hat{i}+\hat{j}+\hat{k}) \cdot \hat{i} }{ \sqrt{3} \sqrt{1} } = \frac{1}{\sqrt{3}}

\therefore \;\; \theta = \cos^{-1}\left( \frac{1}{\sqrt{3}} \right)

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