If a- and b- are two vectors with magnitude 1 and 3 respectively and 1-3 a-b-2-2 a-b-3a-b-2-98- Then find the angle between a- and b-A- 32 -B- 2 -D- 5 -6

The correct option is B 2π3Solving the given equation 1^2 6a.b+9(a.b)^2 +4|a|^2+4a.b+|b|^2 +2(2a+b)·(3(a×b)) +9|a×b|^2=98 ⇒ 1 2a.b+81cos^2θ+4+9+0 +81sin^2θ=98 ...

107

You visited us 107 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Order of a Differential Equation

If a⃗ and b⃗ ...

Question

If $\to a$ and $\to b$ are two vectors with magnitude $1$ and $3$ respectively and $(1 - 3 \to a . \to b)^{2} + | 2 \to a + \to b + 3 (\to a \times \to b) |^{2} = 98$ .
Then find the angle between $\to a$ and $\to b$

\frac{π}{3}

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

\frac{2 π}{3}

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

\frac{π}{6}

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

\frac{5 π}{6}

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

Open in App

Solution

The correct option is B $\frac{2 π}{3}$
Solving the given equation
$1^{2} - 6 \to a . \to b + 9 (\to a . \to b)^{2} + 4 | \to a |^{2} + 4 \to a . \to b + | \to b |^{2} + 2 (2 \to a + \to b) \cdot (3 (\to a \times \to b)) + 9 | \to a \times \to b |^{2} = 98$
$\Rightarrow 1 - 2 \to a . \to b + 81 {cos}^{2} θ + 4 + 9 + 0 + 81 {sin}^{2} θ = 98$
$(∵ \to a . (\to a \times \to b) = 0 and \to b . (\to a \times \to b) = 0$
$\Rightarrow - 2 \to a . \to b = 3$
$\Rightarrow cos θ = \frac{- 1}{2}$
$\Rightarrow θ = \frac{2 π}{3}$

Suggest Corrections

Similar questions

Q. If

\to a

and

\to b

are two non collinear unit vectors and

| \to a + \to b | = \sqrt{3}

, then

(2 \to a - 5 \to b) . (3 \to a + \to b) =

Q. Vectors

\to a

and

\to b

make an angle

θ = \frac{2 π}{3} .

| \to a | = 1, ∣ ∣ \to b ∣ ∣ = 2

then

{(\to a + 3 \to b) \times (3 \to a - \to b)}^{2} =

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

is a vector such that

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

then the angle between

\to b

and

\to c

Q. Vectors

3 \to a - 5 \to b

and

2 \to a + \to b

are mutually perpendicular. If

\to a + 4 \to b

and

\to b - \to a

are also mutually perpendicular, then cosine of the angle between

\to a

and

\to b

is equal to

Q. Let

\to a

and

\to b

be two vectors such that

| \to a | = 3

| \to b | = 6

and

| \to a + \to b | = 7

. Then the value of

(3 \to a - 2 \to b) . (2 \to a + 5 \to b - 4 \to a \times \to b)

Join BYJU'S Learning Program

If →a and →b are two vectors with magnitude 1 and 3 respectively and (1−3→a.→b)2+|2→a+→b+3(→a×→b)|2=98. Then find the angle between →a and →b

If $\to a$ and $\to b$ are two vectors with magnitude $1$ and $3$ respectively and $(1 - 3 \to a . \to b)^{2} + | 2 \to a + \to b + 3 (\to a \times \to b) |^{2} = 98$ .
Then find the angle between $\to a$ and $\to b$