Given - a-k-b-2-k- and - c-2-k- A- unit vector perpendicular to both - a-b- amp- - b-c- is- 2-k-6- - k- -k-3-

The correct option is C k̂ a⃗+b⃗=3ĵ b⃗+c⃗= 2î+4ĵ ( a⃗+b⃗ )× ( b⃗+c⃗ )=6k̂ | ( a⃗+b⃗ )× ( b⃗+c⃗ )|=6 ⇒ ( a⃗+b⃗ )× ( b⃗+c⃗ )| ( a⃗+ ...

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Byju's Answer

Standard XII

Physics

Multiplication with Vectors

Given a⃗...

Question

Given $\to a =^i +^j -^k, \to b = -^i + 2^j +^k$ and $\to c = -^i + 2^i -^k . A$ unit vector perpendicular to both $\to a + \to b$ & $\to b + \to c$ is

\frac{2^i +^j +^k}{\sqrt{6}}

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\frac{^i +^j +^k}{\sqrt{3}}

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Solution

The correct option is C $^k$
$\to a + \to b = 3^j$
$\to b + \to c = - 2^i + 4^j$

$(\to a + \to b) \times (\to b + \to c) = 6^k$

$| (\to a + \to b) \times (\to b + \to c) | = 6$

$\Rightarrow \frac{(\to a + \to b) \times (\to b + \to c)}{| (\to a + \to b) \times (\to b + \to c) |} =^k$

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Similar questions

Q. If

\to a =^i +^j +^k

\to b =^i -^j +^k

\to c =^i +^j -^k

and

\to a =^i -^j -^k

, then

(\to a \times \to b) \times (\to c \times \to d)

is a vector orthogonal to both

Q. Let

\to a =^i +^j +^k, \to b =^i -^j +^k

and

\to c =^i -^j -^k

be three vectors. A vector

\to v

in the plane of

\to a

and

\to b

, whose projection on

\to c

\frac{1}{\sqrt{3}}

, is given by

;

Q. If

\to a =^i +^j +^k, \to b =^i +^j -^k

\to c =^i -^j +^k

and

\to d =^i -^j -^k

, then

(\to a \times \to b) \times (\to c \times \to d) =

Q. Let

\to a =^j -^k

and

\to c =^i -^j -^k

. Then vector

\to b

satisfying

\to a \times \to b + \to c = \to 0

and

\to a \cdot \to b = 3

Q. Let

\to a =^i +^j +^k, \to c =^j -^k

and a vector

\to b

be such that

\to a \times \to b = \to c

and

\to a . \to b = 3

. Then

| \to b |

equals :

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Given →a=^i+^j−^k,→b=−^i+2^j+^k and →c=−^i+2^i−^k.A unit vector perpendicular to both →a+→b & →b+→c is

The correct option is C ^k→a+→b=3^j→b+→c=−2^i+4^j(→a+→b)×(→b+→c)=6^k|(→a+→b)×(→b+→c)|=6⇒(→a+→b)×(→b+→c)|(→a+→b)×(→b+→c)|=^k

Given $\to a =^i +^j -^k, \to b = -^i + 2^j +^k$ and $\to c = -^i + 2^i -^k . A$ unit vector perpendicular to both $\to a + \to b$ & $\to b + \to c$ is

The correct option is C $^k$
$\to a + \to b = 3^j$
$\to b + \to c = - 2^i + 4^j$

$(\to a + \to b) \times (\to b + \to c) = 6^k$

$| (\to a + \to b) \times (\to b + \to c) | = 6$

$\Rightarrow \frac{(\to a + \to b) \times (\to b + \to c)}{| (\to a + \to b) \times (\to b + \to c) |} =^k$