Let a- - b- -k- c-k- - If d is a unit vector such that a-d-0-bcd- then d is equal to-

Name: JEE- Grade 12- Mathematics- Vector Algebra- Session 08- D03
Uploaded: 2022-07-04T10:36:42+05:30
Description: D is perpendicular to nbsp;a and coplanar with nbsp;b and nbsp;c. Hence, nbsp;d is a vector collinear with a× (b×c) ⇒d= ±a× (b×c)|a× (b×c)| Now a× (b×c) ...

D is perpendicular to nbsp;a and coplanar with nbsp;b and nbsp;c. Hence, nbsp;d is a vector collinear with a× (b×c) ⇒d= ±a× (b×c)|a× (b×c)| Now a× (b×c) ...

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Byju's Answer

Standard XII

Mathematics

Linear Dependence and Independence of Vectors

Let a=î -ĵ, b...

Question

Let $\to a =^i -^j, \to b =^j -^k, \to c =^k -^i$ . If $\to d$ is a unit vector such that $\to a \cdot \to d = 0 = [\to b \to c \to d]$ , then $\to d$ is equal to:

\pm (\frac{^i +^j - 2^k}{\sqrt{6}})

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\pm (\frac{^i +^j -^k}{\sqrt{3}})

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\pm (\frac{^i +^j +^k}{\sqrt{3}})

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\pm^k

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Solution

The correct option is A $\pm (\frac{^i +^j - 2^k}{\sqrt{6}})$
$\to d$ is perpendicular to $\to a$ and coplanar with $\to b$ and $\to c$ .
Hence, $\to d$ is a vector collinear with $\to a \times (\to b \times \to c)$
$\Rightarrow \to d = \pm \frac{\to a \times (\to b \times \to c)}{| \to a \times (\to b \times \to c) |}$
Now
$\to a \times (\to b \times \to c) = (\to a \cdot \to c) \to b - (\to a \cdot \to b) \to c$
$= - 1 (^j -^k) + 1 (^k -^i)$
$= -^i -^j + 2^k$
Hence, $\to d = \pm (\frac{^i +^j - 2^k}{\sqrt{6}})$

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Linear Dependence and Independence of Vectors

Standard XII Mathematics

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