Consider r - ai - b - c k- If - - and Y be the projection of r along 2 -5- -2 -5 and k- respectively such that - Y - and- r -1- the number of such ordered triplets a - b - c is less thanA- 5B- 10C- 12D- 15

The correct options are B 10 C 12 D 15r = aî+bĵ+ck̂ α = nbsp;r·(2î+ĵ)√(5) =2a+b√(5) β = nbsp; 2a+b√(5), nbsp; γ = c | α | nbsp; = nbsp;| β | = nbsp; ...

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Byju's Answer

Standard XII

Physics

Acceleration in 2D

Consider r = ...

Question

Consider $\to r = a^i + b^j + c^k$ . If $α$ , $β$ and $γ$ be the projection of $\to r$ along $\frac{2^i +^j}{\sqrt{5}}$ , $\frac{- 2^i +^j}{\sqrt{5}}$ and $^k$ respectively such that $| α | = | β | = | γ |$ and $| \to r | = 1,$ the number of such ordered triplets $(a, b, c)$ is less than

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Solution

The correct options are
B $10$
C $12$
D $15$
$\to r$ = $a^i + b^j + c^k$
$α$ = $\frac{\to r \cdot (2^i +^j)}{\sqrt{5}}$ = $\frac{2 a + b}{\sqrt{5}}$
$β$ = $\frac{- 2 a + b}{\sqrt{5}}, γ = c$
| $α$ | = | $β$ | = | $γ$ |
$| 2 a + b | = | - 2 a + b | = | \sqrt{5} c |$

$\begin{matrix} a = \pm \frac{\sqrt{5}}{3} b = 0 c = \pm \frac{2}{3} \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ \to 4 triplets$

$\begin{matrix} c = \pm \sqrt{\frac{1}{6}} b = \pm \sqrt{\frac{5}{6}} a = 0 \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭ \to 4 triplets$

So, $8$ possible triplets.

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Consider →r=a^i+b^j+c^k. If α, β and γ be the projection of →r along 2^i+^j√5, −2^i+^j√5 and ^k respectively such that |α|=|β|=|γ| and |→r|=1, the number of such ordered triplets (a,b,c) is less than

Consider $\to r = a^i + b^j + c^k$ . If $α$ , $β$ and $γ$ be the projection of $\to r$ along $\frac{2^i +^j}{\sqrt{5}}$ , $\frac{- 2^i +^j}{\sqrt{5}}$ and $^k$ respectively such that $| α | = | β | = | γ |$ and $| \to r | = 1,$ the number of such ordered triplets $(a, b, c)$ is less than