Let a- c- be unit vectors and -b-4 - The angle between a- and c- is cos -11-4- Then the positive integral value of - such that b-2 c-a- is

B̅^2=(2c̅+λa̅)^2⇒|b̅|^2=4c̅^2+λ^2a̅^2+4λ(a̅.c̅) ⇒ 16=4+λ^2+λ (a̅.c̅=1/4) λ^2+λ 12=0 (λ + 4)(λ 3)=0 ⇒λ = 4,3 ...

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Byju's Answer

Standard XII

Mathematics

Condition for Coplanarity of Four Points

Let a⃗, c⃗ be...

Question

Let $\to a, \to c$ be unit vectors and $| \to b | = 4$ . The angle between $\to a a n d \to c$ is ${c o s}^{- 1} (\frac{1}{4})$ . Then the positive integral value of $λ$ such that $\to b - 2 \to c = λ \to a$ is

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Solution

${¯ b}^{2} = (2 ¯ c + λ ¯ a)^{2} \Rightarrow {| ¯ b |}^{2} = 4 {¯ c}^{2} + λ^{2} {¯ a}^{2} + 4 λ (¯ a . ¯ c)$

$\Rightarrow 16 = 4 + λ^{2} + λ$ ( $¯ a . ¯ c = \frac{1}{4}$ )

$λ^{2} + λ - 12 = 0$

$(λ + 4) (λ - 3) = 0 \Rightarrow λ = - 4, 3$

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Let →a,→c be unit vectors and |→b|=4. The angle between →a and →c is cos−1(14). Then the positive integral value of λ such that →b−2→c=λ→a is ___

¯b2=(2¯c+λ¯a)2⇒|¯b|2=4¯c2+λ2¯a2+4λ(¯a.¯c) ⇒16=4+λ2+λ (¯a.¯c=14) λ2+λ−12=0 (λ+4)(λ−3)=0⇒λ=−4,3

Let $\to a, \to c$ be unit vectors and $| \to b | = 4$ . The angle between $\to a a n d \to c$ is ${c o s}^{- 1} (\frac{1}{4})$ . Then the positive integral value of $λ$ such that $\to b - 2 \to c = λ \to a$ is

___

${¯ b}^{2} = (2 ¯ c + λ ¯ a)^{2} \Rightarrow {| ¯ b |}^{2} = 4 {¯ c}^{2} + λ^{2} {¯ a}^{2} + 4 λ (¯ a . ¯ c)$

$\Rightarrow 16 = 4 + λ^{2} + λ$ ( $¯ a . ¯ c = \frac{1}{4}$ )

$λ^{2} + λ - 12 = 0$

$(λ + 4) (λ - 3) = 0 \Rightarrow λ = - 4, 3$