Let a-2-3-6k- b-2-3-6k- and c-2-3-6k- Let a1 be the projection of a on b and a2 be the projection of a1 on c- Then- a1-b is equal to

A_1=[(2î+3ĵ 6k̂) ·(2î 3ĵ+6k̂)7] (2î 3ĵ+6k̂)7 ⇒ a_1= 4149(2î 3ĵ+6k̂) a_2= 4149[(2î 3ĵ+6k̂) ·( 2î+3ĵ+6k̂)7] ( 2î+3ĵ+6k̂)7 ⇒ a_2= 41(49)^2( 4 9+36)( 2î+3ĵ+6k̂) ...

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Byju's Answer

Standard XI

Mathematics

Applications of Dot Product

Let a=2î+3ĵ-6...

Question

Let $\to a = 2^i + 3^j - 6^k, \to b = 2^i - 3^j + 6^k$ and $\to c = - 2^i + 3^j + 6^k$ . Let ${\to a}_{1}$ be the projection of $\to a$ on $\to b$ and ${\to a}_{2}$ be the projection of ${\to a}_{1}$ on $\to c$ . Then, ${\to a}_{1} . \to b$ is equal to

- 41

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- \frac{41}{7}

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41

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287

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Solution

The correct option is A $- 41$
$a_{1} = [(2^i + 3^j - 6^k) \cdot \frac{(2^i - 3^j + 6^k)}{7}] \frac{(2^i - 3^j + 6^k)}{7}$
$\Rightarrow a_{1} = - \frac{41}{49} (2^i - 3^j + 6^k)$

$a_{2} = - \frac{41}{49} [(2^i - 3^j + 6^k) \cdot \frac{(- 2^i + 3^j + 6^k)}{7}] \frac{(- 2^i + 3^j + 6^k)}{7}$

$\Rightarrow a_{2} = - \frac{41}{(49)^{2}} (- 4 - 9 + 36) (- 2^i + 3^j + 6^k)$

$\Rightarrow a_{2} = \frac{943}{(49)^{2}} (2^i - 3^j - 6^k)$

$a_{1} \cdot b = - \frac{41}{49} (2^i - 3^j + 6^k) \cdot (2^i - 3^j + 6^k) = - 41$

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Let →a=2^i+3^j−6^k, →b=2^i−3^j+6^k and →c=−2^i+3^j+6^k. Let →a1 be the projection of →a on →b and →a2 be the projection of →a1 on →c. Then, →a1.→b is equal to

Let $\to a = 2^i + 3^j - 6^k, \to b = 2^i - 3^j + 6^k$ and $\to c = - 2^i + 3^j + 6^k$ . Let ${\to a}_{1}$ be the projection of $\to a$ on $\to b$ and ${\to a}_{2}$ be the projection of ${\to a}_{1}$ on $\to c$ . Then, ${\to a}_{1} . \to b$ is equal to