Let a -2 -1-3 k- b -4 -3-2 -6 k- and c -3 -6 -3-1 k- be three vectors such that b -2 a and a isperpendicular to c- Then the possible value of -1- -2- -3 is-A- 1-3-1B- 1-2- 4-2C- -1-2- 4-0D- 1-5-1

B=2a 3 λ_2=2λ_1 2λ_1+λ_2=3 ⋯(1) a⊥c So, nbsp;6+6λ_1+3λ_3 3=0 ⇒ 6λ_1+3λ_3+3=0 ⇒ 2λ_1+λ_3+1=0 ⋯(2) So, nbsp;(λ_1,3 2λ_1, 1 2λ_1)≡(λ_1,λ_2,λ_3) Clearly n ...

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Byju's Answer

Standard XII

Mathematics

Dot Product of Two Vectors

Let a =2 î+λ1...

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 the possible value of $(λ_{1}, λ_{2}, λ_{3})$ is :

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

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(1, 5, 1)

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(1, 3, 1)

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

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Solution

The correct option is A $(- \frac{1}{2}, 4, 0)$
$\to b = 2 \to a$
$3 - λ_{2} = 2 λ_{1}$
$2 λ_{1} + λ_{2} = 3 \dots (1)$
$\to a ⊥ \to c$
So, $6 + 6 λ_{1} + 3 λ_{3} - 3 = 0$
$\Rightarrow 6 λ_{1} + 3 λ_{3} + 3 = 0$
$\Rightarrow 2 λ_{1} + λ_{3} + 1 = 0 \dots (2)$
So, $(λ_{1}, 3 - 2 λ_{1}, - 1 - 2 λ_{1}) \equiv (λ_{1}, λ_{2}, λ_{3})$
Clearly $(- \frac{1}{2}, 4, 0)$ is the correct option

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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 the possible value of (λ1,λ2,λ3) is :

The correct option is A (−12,4,0)→b=2→a 3−λ2=2λ1 2λ1+λ2=3 ⋯(1) →a⊥→c So, 6+6λ1+3λ3−3=0 ⇒6λ1+3λ3+3=0 ⇒2λ1+λ3+1=0 ⋯(2) So, (λ1,3−2λ1,−1−2λ1)≡(λ1,λ2,λ3) Clearly (−12,4,0) is the correct option

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 the possible value of $(λ_{1}, λ_{2}, λ_{3})$ is :