In vector diagram shown in figure where R is the resultant of vectors A and B - If R - B-2 - the value of angle - is -

The correct option is C 45^ 0 From the diagram, we have A⃗ + B⃗ = R⃗ Thus, R^2=A^2+B^2+2ABcosθ or, B^2/2=A^2+B^2+2ABcosθ #160; — (1)Als ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Work Done as Dot Product

In vector dia...

Question

In vector diagram shown in figure where $\to R$ is the resultant of vectors $\to A$ and $\to B$ . If R = $\frac{B}{\sqrt{2}}$ , the value of angle $θ$ is :-

30^{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

45^{0}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

60^{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

75^{0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $45^{0}$
From the diagram, we have $\to A + \to B = \to R$
Thus, $R^{2} = A^{2} + B^{2} + 2 A B c o s θ$
or, $B^{2} / 2 = A^{2} + B^{2} + 2 A B c o s θ$ --- (1)
Also, $\to A$ is perpendicular to $\to R$
Thus, $\to A . (\to A + \to B) = 0$ or, $A^{2} + A B c o s θ = 0$
which gives, $A = - B c o s θ$
Put this in (1) to get
$B^{2} c o s^{2} θ + B^{2} / 2 - 2 B^{2} c o s^{2} θ = 0$
or, $c o s^{2} θ = 1 / 2$ or $c o s θ = \frac{1}{\sqrt{2}}$ , giving $θ = 45^{0}$

Suggest Corrections

Similar questions

Q. In vector diagram shown in fig. R is the resultant of vectors A and B.
If R =

\frac{B}{\sqrt{2}}

, the value of angle

θ

The angle between two vectors $\to A$ and $\to B$ is $θ$ . Vector $\to R$ is the resultant of these two vectors. If $\to R$ makes an angle $\frac{θ}{2}$ with $\to A$ , then

Q. The angle between two vectors A and B is

θ

. Vector R is the resultant of the two vectors. If R makes an angle

\frac{θ}{2}

with A, then

Q. The resultant of two velocity vectors

\to A

and

\to B

is perpendicular to

\to A

. The angle between

\to A

and

\to B

150^{\circ}

. Find

\frac{| \to R |}{| \to B |}

where

\to R

is the resultant vector.

Q. The resultant of two velocity vectors

\to A

and

\to B

is perpendicular to

\to A

. The angle between

\to A

and

\to B

150^{\circ}

. Find

\frac{| \to R |}{| \to B |}

where

\to R

is the resultant vector.

Join BYJU'S Learning Program

In vector diagram shown in figure where →R is the resultant of vectors →A and →B . If R = B√2 , the value of angle θ is :-

In vector diagram shown in figure where $\to R$ is the resultant of vectors $\to A$ and $\to B$ . If R = $\frac{B}{\sqrt{2}}$ , the value of angle $θ$ is :-