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- a2 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) ⇒a ...

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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, ${\to a}_{2}$ is equal to

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

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

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\frac{943}{49} (- 2^i + 3^j + 6^k)

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\frac{943}{49^{2}} (- 2^i + 3^j + 6^k)

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Solution

The correct option is B $\frac{943}{49^{2}} (2^i - 3^j - 6^k)$
$_{1} = [(2^i + 3^j - 6^k) \cdot \frac{(2^i - 3^j + 6^k)}{7}] \frac{(2^i - 3^j + 6^k)}{7}$
$\Rightarrow_{1} = - \frac{41}{49} (2^i - 3^j + 6^k)$

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

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

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

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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, →a2 is equal to

The correct option is B 943492(2^i−3^j−6^k) →a1=[(2^i+3^j−6^k)⋅(2^i−3^j+6^k)7](2^i−3^j+6^k)7 ⇒→a1=−4149(2^i−3^j+6^k) →a2=−4149[(2^i−3^j+6^k)⋅(−2^i+3^j+6^k)7](−2^i+3^j+6^k)7 ⇒→a2=−41(49)2(−4−9+36)(−2^i+3^j+6^k) ⇒→a2=943(49)2(2^i−3^j−6^k)

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, ${\to a}_{2}$ is equal to