A and b are two vectors such that -a-1- -b-4- -c-2-192 and a-b-2- If c-2a-b-3b- then the angle between b and c is

|a|=1, |b|=4, a·b=2 nbsp; c=(2a×b) 3b ⇒c+3b=2a×b ∵a·b=2 |a||b|cosθ=2 ⇒cosθ=2|a||b|=24=12 ⇒θ=π3 |c+3b|^2=|2a×b|^2 ⇒ |c|^2+9|b|^2+2c· 3b=4|a|^2|b|^2 sin^2 θ ⇒ ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XI

Mathematics

Cross Product of Two Vectors

a and b are t...

Question

$\to a$ and $\to b$ are two vectors such that $| \to a | = 1, | \to b | = 4, | \to c |^{2} = 192$ and $\to a . \to b = 2$ . If $\to c = (2 \to a \times \to b) - 3 \to b$ , then the angle between $\to b$ and $\to c$ is

\frac{π}{3}

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

\frac{π}{6}

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

\frac{3 π}{4}

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

\frac{5 π}{6}

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

Open in App

Solution

The correct option is D $\frac{5 π}{6}$
$| \to a | = 1, | \to b | = 4, \to a \cdot \to b = 2$
$\to c = (2 \to a \times \to b) - 3 \to b$
$\Rightarrow \to c + 3 \to b = 2 \to a \times \to b$

$∵ \to a \cdot \to b = 2$
$| \to a | | \to b | cos θ = 2$
$\Rightarrow cos θ = \frac{2}{| \to a | | \to b |} = \frac{2}{4} = \frac{1}{2}$
$\Rightarrow θ = \frac{π}{3}$

$| \to c + 3 \to b |^{2} = | 2 \to a \times \to b |^{2}$
$\Rightarrow | \to c |^{2} + 9 | \to b |^{2} + 2 \to c \cdot 3 \to b = 4 | \to a |^{2} | \to b |^{2} {sin}^{2} θ$
$\Rightarrow | \to c |^{2} + 9 | \to b |^{2} + 6 \to b \cdot \to c = 48$
$\Rightarrow | \to c |^{2} + 96 + 6 (\to b \cdot \to c) = 0 \dots [1]$
$\Rightarrow \to c = 2 \to a \times \to b - 3 \to b$
$\Rightarrow \to b \cdot \to c = 0 - 3 \times 16 = - 48$

Putting value of $\to b \cdot \to c$ in $[1]$ , we have
$| c |^{2} + 96 - 6 \times 48 = 0 \Rightarrow | \to c | = 8 \sqrt{3}$
Putting this value in $[1]$ , we get
$192 + 96 + 6 | \to b | | \to c | cos α = 0$
$\Rightarrow cos α = - \frac{\sqrt{3}}{2}$
or, $α = \frac{5 π}{6}$

Suggest Corrections

Similar questions

\to a

and

\to b

are two vectors such that

| \to a | = 1, | \to b | = 4, | \to c |^{2} = 192

and

\to a . \to b = 2

. If

\to c = (2 \to a \times \to b) - 3 \to b

, then the angle between

\to b

and

\to c

\to a

and

\to b

are two vectors such that

| \to a | = 1, | \to b | = 4, | \to c |^{2} = 192

and

\to a . \to b = 2

. If

\to c = (2 \to a \times \to b) - 3 \to b

, then the angle between

\to b

and

\to c

Q. Let

\to a

and

\to b

be two vectors such that

| \to a | = 1,

| \to b | = 4

and

\to a \cdot \to b = 2.

\to c = (2 \to a \times \to b) - 3 \to b,

then the angle between

\to b

and

\to c

\to a

and

\to b

are two vectors such that

| \to a | = 1, | \to b | = 4

and

\to a \cdot \to b = 2

. If

\to c = (2 \to a \times \to b) - 3 \to b

, then the angle between

\to b

and

\to c

Q. If

\to a

and

\to b

are such that

∣ ∣ \to a ∣ ∣ = 1

∣ ∣ ∣ \to b ∣ ∣ ∣,

\to a . \to b = 2

and

\to c = 2 \to a \times \to b - 3 \to b

, then the angle between

\to b

and

\to c

Join BYJU'S Learning Program

→a and →b are two vectors such that |→a|=1, |→b|=4, |→c|2=192 and →a.→b=2. If →c=(2→a×→b)−3→b, then the angle between →b and →c is

$\to a$ and $\to b$ are two vectors such that $| \to a | = 1, | \to b | = 4, | \to c |^{2} = 192$ and $\to a . \to b = 2$ . If $\to c = (2 \to a \times \to b) - 3 \to b$ , then the angle between $\to b$ and $\to c$ is