If a-b-6 - c- is a perpendicular to a- and b- -a-3- -b-4- -c-6- then -a-b-c-

The correct option is C 36 Since c⃗ is perpendicular to a⃗ and b⃗ ∴c⃗=λa⃗×b⃗ | c⃗ |=λ | a⃗×b⃗ | 6=λ× 3× 4×12 λ =1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

YDSE Problems II

If a⃗,b⃗=π6...

Question

If $(\to a, \to b) = \frac{π}{6}$ , $\to c$ is a perpendicular to $\to a$ and $\to b, | \to a | = 3, | \to b | = 4, | \to c | = 6$ , then $| [\to a \to b \to c] | =$

12 \sqrt{3}

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

48 \sqrt{3}

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

36

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

72

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

Open in App

Solution

The correct option is C $36$
Since $\to c$ is perpendicular to $\to a$ and $\to b$
$∴ \to c = λ \to a \times \to b$
$| \to c | = λ ∣ ∣ \to a \times \to b ∣ ∣$
$6 = λ \times 3 \times 4 \times \frac{1}{2}$
$λ = 1$
$[¯ a ¯ b ¯ c] = {∣ ∣ \to a \times \to b ∣ ∣}^{2} = (6)^{2} = 36$

Suggest Corrections

Similar questions

Q. If

\to a, \to b

and

\to c

are perpendicular to

\to b + \to c, \to c + \to a

and

\to a + \to b

respectively and if

| \to a + \to b | = 6, | \to b + \to c | = 8

and

| \to c + \to a | = 10

, the

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

is equal to

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. 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. Magnitudes of vectors

\to a, \to b, \to c

are

3, 4, 5

respectively. If

\to a

and

\to b + \to c

\to b

and

\to c + \to a

\to c

and

\to a + \to b

are mutually perpendicular, then the magnitude of

\to a + \to b + \to c

is,

Q. Let

\to a \times (\to b \times \to c) = \frac{\to b}{3} + \frac{\to c}{2}

and

\to b \times (\to c \times \to a) = - \frac{\to c}{2}

\to a, \to b

and

\to c

are non-collinear pair wise unit vectors, then volume of a parallelopiped, whose coterminous edges are

\to a, \to b

and

\to c

, is

Join BYJU'S Learning Program

If (→a,→b)=π6 , →c is a perpendicular to →a and →b, |→a|=3, |→b|=4, |→c|=6, then |[→a→b→c]|=

The correct option is C 36Since →c is perpendicular to →a and →b∴→c=λ→a×→b|→c|=λ∣∣→a×→b∣∣6=λ×3×4×12λ=1[¯a¯b¯c]=∣∣→a×→b∣∣2=(6)2=36

If $(\to a, \to b) = \frac{π}{6}$ , $\to c$ is a perpendicular to $\to a$ and $\to b, | \to a | = 3, | \to b | = 4, | \to c | = 6$ , then $| [\to a \to b \to c] | =$