If b and c are mutually perpendicular non-collinear unit vectors and a is any vector- then a-bb-a-cc-a-b-c-b-c-b-c-

Name: JEE- Grade 12- Mathematics- Vector Algebra- Session 11- W02
Uploaded: 2022-07-04T10:36:42+05:30
Description: Let î be a unit vector along b and ĵ is a unit vector along c As, b,c are unit vector : ⇒b=î, c=ĵ Now, , b×c=|b||c|sinαk̂⋯(i) where k̂ is a unit vector perpendi ...

Let î be a unit vector along b and ĵ is a unit vector along c As, b,c are unit vector : ⇒b=î, c=ĵ Now, , b×c=|b||c|sinαk̂⋯(i) where k̂ is a unit vector perpendi ...

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Byju's Answer

Standard IX

Mathematics

Triangle Inequality

If b and c ar...

Question

If $\to b$ and $\to c$ are mutually perpendicular non-collinear unit vectors and $\to a$ is any vector, then $(\to a \cdot \to b) \to b + (\to a \cdot \to c) \to c + ⎡ ⎢ ⎣ \frac{\to a \cdot (\to b \times \to c)}{| \to b \times \to c |} ⎤ ⎥ ⎦ (\to b \times \to c) =$

\to a

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\to b

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\to c

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

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Solution

The correct option is A $\to a$
Let $^i$ be a unit vector along $\to b$ and $^j$ is a unit vector along $\to c$
As, $\to b, \to c$ are unit vector :
$\Rightarrow \to b =^i, \to c =^j$
Now, $, \to b \times \to c = | \to b | | \to c | sin α^k \dots (i)$
where $^k$ is a unit vector perpendicular to both $\to b$ and $\to c, α = \frac{π}{2}$ is the angle between $\to b$ and $\to c .$
$\Rightarrow | \to b \times \to c | = sin α = 1 \dots (i i)$
From $(i)$ and $(i i) :^k = \frac{\to b \times \to c}{| \to b \times \to c |} \dots (i i i)$
Any vector can be written as linear combination of three non-coplanar vectors,
$\Rightarrow \to a = a_{1}^i + a_{2}^j + a_{3}^k$
(where $^i,^j,^k$ are three non-coplanar vectors.)
$\Rightarrow \to a \cdot \to b = a_{1}, \to a \cdot \to c = a_{2}$ and
$⎡ ⎢ ⎣ \frac{\to a \cdot (\to b \times \to c)}{| \to b \times \to c |} ⎤ ⎥ ⎦ = \to a \cdot^k = a_{3}$
[from $(i i i)$ ]
Thus, $(\to a \cdot \to b) \to b + (\to a \cdot \to c) \to c + ⎡ ⎢ ⎣ \frac{\to a \cdot (\to b \times \to c)}{| \to b \times \to c |} ⎤ ⎥ ⎦ (\to b \times \to c) = a_{1} \to b + a_{2} \to c + a_{3} (\to b \times \to c) = a_{1}^i + a_{2}^j + a_{3}^k = \to a$

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Triangle Inequality

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