Let \overrightarrow a and \overrightarrow b be two non-zero vectors perpendicular to each other and |\overrightarrow a | = |\overrightarrow b | . If |\overrightarrow a \times \overrightarrow b | = |\overrightarrow a | , then the angle between the vectors ( \overrightarrow a + \overrightarrow b + ( \overrightarrow a \times \overrightarrow b ) ) and \overrightarrow a is equal to :

Vector Algebra PYQ — JEE Main 2021 · 18 Mar · Shift 2 · Q27 with Solution

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Correct answer: D

Given conditions

Let $|\vec a|=|\vec b|=m$ .

Also, $\vec a \perp \vec b$ , so

\vec a\cdot \vec b=0,

and

|\vec a\times \vec b|=|\vec a||\vec b|\sin 90^\circ=m^2.

But it is given that

|\vec a\times \vec b|=|\vec a|=m.

So,

m^2=m \implies m=1

(since vectors are non-zero).

Hence,

|\vec a|=|\vec b|=1,

and because $\vec a\perp \vec b$ ,

|\vec a\times \vec b|=1.

Define the required vector

Let

\vec v=\vec a+\vec b+(\vec a\times \vec b).

We need the angle $\theta$ between $\vec v$ and $\vec a$ .

Using the formula,

\cos\theta=\frac{\vec v\cdot \vec a}{|\vec v|\,|\vec a|}.

Compute $\vec v\cdot \vec a$

\vec v\cdot \vec a=(\vec a+\vec b+\vec a\times \vec b)\cdot \vec a.

Now,

$\vec a\cdot \vec a=|\vec a|^2=1$
$\vec b\cdot \vec a=0$ since $\vec a\perp \vec b$
$(\vec a\times \vec b)\cdot \vec a=0$ since $\vec a\times \vec b$ is perpendicular to $\vec a$

Therefore,

\vec v\cdot \vec a=1.

Compute $|\vec v|$

Since $\vec a$ , $\vec b$ , and $\vec a\times \vec b$ are pairwise perpendicular, and each has magnitude $1$ ,

|\vec v|^2=|\vec a|^2+|\vec b|^2+|\vec a\times \vec b|^2=1+1+1=3.

So,

|\vec v|=\sqrt 3.

Find the angle

Also $|\vec a|=1$ , so

\cos\theta=\frac{1}{\sqrt 3\cdot 1}=\frac{1}{\sqrt 3}.

Hence,

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

Match with options

This is Option D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So, the answer agrees with the stored answer.

More from Vector Algebra

View written solutionFree

Correct answer: D

Given conditions

Let $|\vec a|=|\vec b|=m$ .

Also, $\vec a \perp \vec b$ , so

\vec a\cdot \vec b=0,

and

|\vec a\times \vec b|=|\vec a||\vec b|\sin 90^\circ=m^2.

But it is given that

|\vec a\times \vec b|=|\vec a|=m.

So,

m^2=m \implies m=1

(since vectors are non-zero).

Hence,

|\vec a|=|\vec b|=1,

and because $\vec a\perp \vec b$ ,

|\vec a\times \vec b|=1.

Define the required vector

Let

\vec v=\vec a+\vec b+(\vec a\times \vec b).

We need the angle $\theta$ between $\vec v$ and $\vec a$ .

Using the formula,

\cos\theta=\frac{\vec v\cdot \vec a}{|\vec v|\,|\vec a|}.

Compute $\vec v\cdot \vec a$

\vec v\cdot \vec a=(\vec a+\vec b+\vec a\times \vec b)\cdot \vec a.

Now,

$\vec a\cdot \vec a=|\vec a|^2=1$
$\vec b\cdot \vec a=0$ since $\vec a\perp \vec b$
$(\vec a\times \vec b)\cdot \vec a=0$ since $\vec a\times \vec b$ is perpendicular to $\vec a$

Therefore,

\vec v\cdot \vec a=1.

Compute $|\vec v|$

Since $\vec a$ , $\vec b$ , and $\vec a\times \vec b$ are pairwise perpendicular, and each has magnitude $1$ ,

|\vec v|^2=|\vec a|^2+|\vec b|^2+|\vec a\times \vec b|^2=1+1+1=3.

So,

|\vec v|=\sqrt 3.

Find the angle

Also $|\vec a|=1$ , so

\cos\theta=\frac{1}{\sqrt 3\cdot 1}=\frac{1}{\sqrt 3}.

Hence,

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

Match with options

This is Option D.

Comparison with stored answer

Stored correct answer: D

Our derived answer: D

So, the answer agrees with the stored answer.

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Vector Algebra question

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