Two vectors A and B have equal magnitudes. If magnitude of A + B is equal to n times the magnitude of A - B, then the angle between A and B is

{cos}^{- 1} (\begin{matrix} \frac{n - 1}{n + 1} \end{matrix})

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

{cos}^{- 1} (\begin{matrix} \frac{n^{2} - 1}{n^{2} + 1} \end{matrix})

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

{sin}^{- 1} (\begin{matrix} \frac{n - 1}{n + 1} \end{matrix})

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

{sin}^{- 1} (\begin{matrix} \frac{n^{2} - 1}{n^{2} + 1} \end{matrix})

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

Open in App

Solution

The correct option is B ${cos}^{- 1} (\begin{matrix} \frac{n^{2} - 1}{n^{2} + 1} \end{matrix})$
$| A + B | = n | A - B |$
$⟹ (A^{2} + B^{2} + 2 A B c o s θ) = n^{2} (A^{2} + B^{2} - 2 A B c o s θ)$
It is given that $A + B$
Hence $c o s θ = \frac{n^{2} - 1}{n^{2} + 1}$
$⟹ θ = c o s^{- 1} (\frac{n^{2} - 1}{n^{2} + 1})$

Suggest Corrections

Similar questions

$= {lim}_{n \to \infty} [\frac{1}{n} + \frac{1}{\sqrt{n^{2} + n}} + \frac{1}{\sqrt{n^{2} + 2 n}} + \dots + \frac{1}{\sqrt{n^{2} + (n - 1) n}}]$ is equal to [RPET 2000]

{lim}_{n \to \infty} (\frac{n}{n^{2}} + \frac{n}{n^{2} + 1} + \frac{n}{n^{2} + 2^{2}} + . . . . . . + \frac{n}{2 n^{2} - 2 n + 1})

is equal to

Q. If the major axis is “n” (n>1) times the minor axis of the ellipse, then eccentricity is

lim n \to \infty (\frac{1}{1 - n^{2}} + \frac{2}{1 - n^{2}} + . . . + \frac{n}{1 - n^{2}})

is equal to

lim n \to \infty {(\frac{n^{2} - n + 1}{n^{2} - n - 1})}^{n (n - 1)}

is equal to

Join BYJU'S Learning Program

Two vectors A and B have equal magnitudes. If magnitude of A + B is equal to n times the magnitude of A - B, then the angle between A and B is

The correct option is B cos−1(n2−1n2+1)|A+B|=n|A−B|⟹(A2+B2+2ABcosθ)=n2(A2+B2−2ABcosθ)It is given that A+BHence cosθ=n2−1n2+1⟹θ=cos−1(n2−1n2+1)