Let a-2-1-3k- b-4-3-2-6k- and c-3-6-3-1k- be three vectors such that b-2a- and a- is perpendicular to c- Then a possible value of -1- -2- -3 is-

The correct option is B ( 12, 4, 0) b⃗=2a⃗ ⇒4î+(3 λ_2)ĵ+6k̂=4î+2λ_1ĵ+6k̂ ⇒ 3 λ_2=2λ_1⇒ 2λ_1+λ_2=3 ............................ (1) Giv ...

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Byju's Answer

Standard XII

Mathematics

Dot Product of Two Vectors

Let a⃗=2î+λ...

Question

Let $\to a = 2^i + λ_{1}^j + 3^k$ , $\to b = 4^i + (3 - λ_{2})^j + 6^k$ and $\to c = 3^i + 6^j + (λ_{3} - 1)^k$ be three vectors such that $\to b = 2 \to a$ and $\to a$ is perpendicular to $\to c$ . Then a possible value of $(λ_{1}, λ_{2}, λ_{3})$ is?

(\frac{1}{2}, 4, - 2)

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(- \frac{1}{2}, 4, 0)

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(1, 3, 1)

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(1, 5, 1)

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Solution

The correct option is B $(- \frac{1}{2}, 4, 0)$
$\to b = 2 \to a$
$\Rightarrow 4^i + (3 - λ_{2})^j + 6^k = 4^i + 2 λ_{1}^j + 6^k$
$\Rightarrow 3 - λ_{2} = 2 λ_{1} \Rightarrow 2 λ_{1} + λ_{2} = 3$ ............................ $(1)$

Given $\to a \cdot \to c = 0$
$\Rightarrow 6 + 6 λ_{1} + 3 (λ_{3} - 1) = 0$
$\Rightarrow 2 λ_{1} + λ_{3} = - 1$ ....................... $(2)$

Now $(λ_{1}, λ_{2}, λ_{3}) = (λ_{1}, 3 - 2 λ_{1}, - 1 - 2 λ_{1})$
putting $λ_{1} = \frac{- 1}{2} \Rightarrow (\frac{- 1}{2}, 4, 0)$
Now check the options, option $(2)$ is correct.

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Similar questions

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Let →a=2^i+λ1^j+3^k, →b=4^i+(3−λ2)^j+6^k and →c=3^i+6^j+(λ3−1)^k be three vectors such that →b=2→a and →a is perpendicular to →c. Then a possible value of (λ1,λ2,λ3) is?

Let $\to a = 2^i + λ_{1}^j + 3^k$ , $\to b = 4^i + (3 - λ_{2})^j + 6^k$ and $\to c = 3^i + 6^j + (λ_{3} - 1)^k$ be three vectors such that $\to b = 2 \to a$ and $\to a$ is perpendicular to $\to c$ . Then a possible value of $(λ_{1}, λ_{2}, λ_{3})$ is?