The resultant C- of A- and B- is perpendicular to A- - Also- -A- - -C- The angle between A- and B- is

The correct option is B 3π4 rad | C⃗ #160; | #160; ^ 2 =A⃗ #160; ^ 2 + B ^ 2 +2ABcosθ B ^ 2 +2ABcosθ =0 B+2Acosθ =0 Also, #160; ...

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Byju's Answer

Standard XII

Physics

Vector Addition

The resultant...

Question

The resultant $\to C$ of $\to A$ and $\to B$ is perpendicular to $\to A$ . Also, $| \to A | = | \to C |$ . The angle between $\to A$ and $\to B$ is

\frac{π}{4}

rad

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\frac{3 π}{4}

rad

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\frac{5 π}{4}

rad

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\frac{7 π}{4}

rad

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Solution

The correct option is B $\frac{3 π}{4}$ rad

${∣ ∣ \to C ∣ ∣}^{2} = {\to A}^{2} + B^{2} + 2 A B cos θ$
$B^{2} + 2 A B cos θ = 0$
$B + 2 A cos θ = 0$
Also, $B cos (π - θ) = A \Rightarrow - B cos θ = A$
$B - 2 B {cos}^{2} θ \Rightarrow \frac{1}{2} = {cos}^{2} θ$
$cos θ = \pm \frac{1}{2} \Rightarrow θ = \frac{π}{4}, \frac{3 π}{4}$
$θ$ is obtuse $\Rightarrow θ = \frac{3 π}{4}$

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Similar questions

Q. If

\to a

is perpendicular to

\to b

and

\to c, | \to a | = 2, | \to b | = 3, | \to c | = 4

and the angle between

\to b

and

\to c

\frac{2 π}{3}

, then

[\to a

\to b

\to c]

is equal to

Q. If

\to a, \to b, \to c

are unit vectors and the angles between

\to a, \to b

and

\to b, \to c

and

\to c, \to a

are

\frac{π}{6}, \frac{π}{3}

and

\frac{π}{4}

respectively, then

| \to a + \to b + \to c |

Q. If

| \to a | = 3, ∣ ∣ \to b ∣ ∣ = 4, | \to c | = 5, \to a ⊥ (\to b + \to c), \to b ⊥ (\to c + \to a)

and

\to c ⊥ (\to a + \to b)

then

\sqrt{2} ∣ ∣ \to a + \to b + \to c ∣ ∣

is equal to

Q. Let

\to a, \to b, \to c

be three non-zero vectors such that

\to c

is a unit vectors perpendicular is both

\to a

and

\to b

. If the angle between

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and

\to b

\frac{π}{6}

, prove that

[\to a \to b \to c]^{2} = \frac{1}{4} | \to a |^{2} | \to b |^{2}

Q. If

\to a, \to b, \to c

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, then value of

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The resultant →C of →A and →B is perpendicular to →A . Also, |→A|=|→C|. The angle between →A and →B is

The correct option is B 3π4rad∣∣→C∣∣2=→A2+B2+2ABcosθB2+2ABcosθ=0B+2Acosθ=0Also, Bcos(π−θ)=A⇒−Bcosθ=AB−2Bcos2θ⇒12=cos2θcosθ=±12⇒θ=π4,3π4θ is obtuse ⇒θ=3π4

The resultant $\to C$ of $\to A$ and $\to B$ is perpendicular to $\to A$ . Also, $| \to A | = | \to C |$ . The angle between $\to A$ and $\to B$ is