If the projection Vector O-A- on unit Vector O-B- equals twice the area of triangle - OAB in magnitude- then - AOB in radian is

The correct option is C π4 Area of Δ OAB=12|OA× OB| =12|OA| sinθ where θ is angle between OA amp; OB Projection of OA ...

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Standard XII

Mathematics

Applications of Cross Product

If the projec...

Question

If the projection Vector $\to O A$ on unit Vector $\to O B$ equals twice the area of triangle $Δ O A B$ in magnitude, then $∠ A O B$ in radian is

0

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\frac{π}{2}

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\frac{π}{4}

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π

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Solution

The correct option is C $\frac{π}{4}$
Area of $Δ O A B = \frac{1}{2} | O A \times O B |$
$= \frac{1}{2} | O A | sin θ$
where $θ$ is angle between $O A$ & $O B$
Projection of $O A$ on $O B$ $= | O A | cos θ$
$| O A | cos θ = \frac{12}{2} | O A | sin θ$
$tan θ = 1$
$θ = \frac{π}{4}$

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Similar questions

Q. If

\to O A =^i -^j

and

\to O B =^j -^k

, then a unit vector parpendicular to the plane

A O B

Q. The values of

^{'} a^{'}

and

^{'} b^{'}

so that the function

f (x)

⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} x + a \sqrt{2} sin x, & 0 \leq x < \frac{π}{4} 2 x cot x + b, & \frac{π}{4} \leq x \leq \frac{π}{2} a cos 2 x - b sin x, & \frac{π}{2} < x \leq π \end{matrix}

is continuous at

x = \frac{π}{4}, \frac{π}{2}

are

Q. In

Δ O A B

, if

\to O A = \to a, \to O B = \to b . L

is mid point of

\to O A

and

M

is point on

\to O B

such that

\to O M : \to M B = 2 : 1

. If

P

is mid point of

L M

then

\to A P =

Q. Let

\to O A = \to a

and

\to O B = \to b

. A vector along one of the bisectors of the angle

∠ A O B

Q. Assertion :The resultant of the three vector

\to O A, \to O B

and

\to O C

as shown in the figure is

R (1 + \sqrt{2})

.Here

R

is the radius of the circle.(Here

∠ A O B = ∠ B O C = 45^{\circ}

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\to O A + \to O C

is along

\to O B

and

(\to O A + \to O C) + \to O B

is along

\to O B

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If the projection Vector →OA on unit Vector →OB equals twice the area of triangle ΔOAB in magnitude, then ∠AOB in radian is

The correct option is C π4Area of ΔOAB=12|OA×OB|=12|OA|sinθwhere θ is angle between OA & OBProjection of OA on OB=|OA|cosθ|OA|cosθ=122|OA|sinθtanθ=1θ=π4

If the projection Vector $\to O A$ on unit Vector $\to O B$ equals twice the area of triangle $Δ O A B$ in magnitude, then $∠ A O B$ in radian is

The correct option is C $\frac{π}{4}$
Area of $Δ O A B = \frac{1}{2} | O A \times O B |$
$= \frac{1}{2} | O A | sin θ$
where $θ$ is angle between $O A$ & $O B$
Projection of $O A$ on $O B$ $= | O A | cos θ$
$| O A | cos θ = \frac{12}{2} | O A | sin θ$
$tan θ = 1$
$θ = \frac{π}{4}$